CBSE Class 10 · Maths
CBSE Class 10 Maths Coordinate Geometry Competency Based Questions
Help your child master Coordinate Geometry in CBSE Class 10 Maths with these expert-verified CBSE competency based questions, covering distance formula, section formula, midpoints and area of triangles. Each question includes a detailed step-by-step solution — exactly the way CBSE board exams test this chapter.
CBSE Class 10 Maths Coordinate Geometry — Questions with Solutions
The distance of the point ( $-1,7$ ) from $X$-axis is:
Solution
Option (B) is correct.
The distance of $(-1,7)$ from the $X$-axis equals the absolute value of the $y$-coordinate = 7 units.
The distance of $(-1,7)$ from the $X$-axis equals the absolute value of the $y$-coordinate = 7 units.
The distance of the point $(-6,8)$ from origin is:
Solution
Option (D) is correct.
Distance between $(-6,8)$ and $(0,0)$:
$$\sqrt{(-6-0)^{2}+(8-0)^{2}} = \sqrt{36+64} = \sqrt{100} = 10$$
Distance between $(-6,8)$ and $(0,0)$:
$$\sqrt{(-6-0)^{2}+(8-0)^{2}} = \sqrt{36+64} = \sqrt{100} = 10$$
The points $(-4,0),(4,0)$ and $(0,3)$ are the vertices of a:
Solution
Option (B) is correct.
Let $\mathrm{A}(-4,0), \mathrm{B}(4,0), \mathrm{C}(0,3)$.
$$AB=8,\quad BC=\sqrt{16+9}=5,\quad CA=\sqrt{16+9}=5$$
Since $BC=CA$, the triangle is isosceles.
Let $\mathrm{A}(-4,0), \mathrm{B}(4,0), \mathrm{C}(0,3)$.
$$AB=8,\quad BC=\sqrt{16+9}=5,\quad CA=\sqrt{16+9}=5$$
Since $BC=CA$, the triangle is isosceles.
The point of intersection of the line represented by $3x-y=3$ and $y$-axis is given by
Solution
Option (A) is correct.
At the $y$-axis, $x=0$.
$$3(0)-y=3 \Rightarrow y=-3$$
The line $3x-y=3$ cuts the $y$-axis at $(0,-3)$.
At the $y$-axis, $x=0$.
$$3(0)-y=3 \Rightarrow y=-3$$
The line $3x-y=3$ cuts the $y$-axis at $(0,-3)$.
What is the ratio in which the line segment joining $(2,-3)$ and $(5,6)$ is divided by $x$-axis?
Solution
Option (A) is correct.
Let the ratio be $k:1$ and the point on $x$-axis be $(x,0)$.
$$0 = \frac{6k – 3}{k+1} \Rightarrow 6k=3 \Rightarrow k=\frac{1}{2}$$
So the ratio is $1:2$.
Let the ratio be $k:1$ and the point on $x$-axis be $(x,0)$.
$$0 = \frac{6k – 3}{k+1} \Rightarrow 6k=3 \Rightarrow k=\frac{1}{2}$$
So the ratio is $1:2$.
A point $(x,y)$ is at a distance of 5 units from the origin. How many such points lie in the third quadrant?
Solution
Option (D) is correct.
The equation $x^2+y^2=25$ represents a circle of radius 5. In the third quadrant ($x<0$, $y<0$), there are infinitely many points on this circle.
The equation $x^2+y^2=25$ represents a circle of radius 5. In the third quadrant ($x<0$, $y<0$), there are infinitely many points on this circle.
The distance between two points $A$ and $B$ on a graph is given as $\sqrt{10^{2}+7^{2}}$. The coordinates of $A$ are $(-4,3)$. Given that point $B$ lies in the first quadrant, then all the possible $x$-coordinates of point $B$ are
Solution
Option (B) is correct.
$AB=\sqrt{10^{2}+7^{2}}=\sqrt{149}=\sqrt{(-4-x)^{2}+(3-y)^{2}}$
Since $A(-4,3)$ is in the second quadrant and $B$ is in the first quadrant, comparing gives $x=6$ and $y=0$. So the possible $x$-coordinates of $B$ are multiples of 3.
$AB=\sqrt{10^{2}+7^{2}}=\sqrt{149}=\sqrt{(-4-x)^{2}+(3-y)^{2}}$
Since $A(-4,3)$ is in the second quadrant and $B$ is in the first quadrant, comparing gives $x=6$ and $y=0$. So the possible $x$-coordinates of $B$ are multiples of 3.
The coordinates of the centre $O$ and a point $N$ on the circle are shown in the figure. What is the radius of the circle?
Solution
Option (B) is correct.
Given $O(-4,3)$ and $N(-2.4,1.8)$, radius $= ON$:
$$\sqrt{(-4+2.4)^{2}+(3-1.8)^{2}} = \sqrt{2.56+1.44} = \sqrt{4} = 2 \text{ units}$$
Given $O(-4,3)$ and $N(-2.4,1.8)$, radius $= ON$:
$$\sqrt{(-4+2.4)^{2}+(3-1.8)^{2}} = \sqrt{2.56+1.44} = \sqrt{4} = 2 \text{ units}$$
If $A(3,\sqrt{3})$, $B(0,0)$ and $C(3,k)$ are the three vertices of an equilateral triangle ABC, then the value of $k$ is
Solution
Option (C) is correct.
For an equilateral triangle, $AB=BC$. Using the distance formula and solving gives $k=\pm\sqrt{3}$.
For an equilateral triangle, $AB=BC$. Using the distance formula and solving gives $k=\pm\sqrt{3}$.
Points $A(-1,y)$ and $B(5,7)$ lie on a circle with centre $O(2,-3y)$. The values of $y$ are
Solution
Option (B) is correct.
Since $A$ and $B$ lie on the circle, $AO=BO$.
$$9+16y^{2}=9+9y^{2}+49+42y$$
$$7y^{2}-42y-49=0 \Rightarrow (y-7)(y+1)=0 \Rightarrow y=7,-1$$
Since $A$ and $B$ lie on the circle, $AO=BO$.
$$9+16y^{2}=9+9y^{2}+49+42y$$
$$7y^{2}-42y-49=0 \Rightarrow (y-7)(y+1)=0 \Rightarrow y=7,-1$$
The diagram shows the plans for a sun room built onto the wall of a house. The roof uses four trapezium-shaped clear glass panels and one half-regular-octagon tinted glass panel.
(i) Refer to top view, find the mid-point of the segment joining the points $J(6,17)$ and $I(9,16)$.
(A) $\left(\dfrac{33}{2}, \dfrac{15}{2}\right)$
(B) $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
(C) $\left(\dfrac{15}{2}, \dfrac{33}{2}\right)$
(D) $\left(\dfrac{1}{2}, \dfrac{3}{2}\right)$
(ii) Refer to front view, the distance of point $P$ from the $Y$-axis is
(A) 4 (B) 15 (C) 19 (D) 25
(iii) Refer to front view, the distance between points $A$ and $S$ is
(A) 4 (B) 8 (C) 14 (D) 20
(iv) Refer to front view, find the coordinates of the point dividing the line segment joining $A$ and $B$ in the ratio $1:3$ internally.
(A) $(8.5,2.0)$ (B) $(2.0,9.5)$ (C) $(3.0,7.5)$ (D) $(1.75,8.5)$
(v) Refer to front view, if a point $(x,y)$ is equidistant from $Q(9,8)$ and $S(17,8)$, then
(A) $x+y=13$ (B) $x-13=0$ (C) $y-13=0$ (D) $x-y=13$
(i) Refer to top view, find the mid-point of the segment joining the points $J(6,17)$ and $I(9,16)$.
(A) $\left(\dfrac{33}{2}, \dfrac{15}{2}\right)$
(B) $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
(C) $\left(\dfrac{15}{2}, \dfrac{33}{2}\right)$
(D) $\left(\dfrac{1}{2}, \dfrac{3}{2}\right)$
(ii) Refer to front view, the distance of point $P$ from the $Y$-axis is
(A) 4 (B) 15 (C) 19 (D) 25
(iii) Refer to front view, the distance between points $A$ and $S$ is
(A) 4 (B) 8 (C) 14 (D) 20
(iv) Refer to front view, find the coordinates of the point dividing the line segment joining $A$ and $B$ in the ratio $1:3$ internally.
(A) $(8.5,2.0)$ (B) $(2.0,9.5)$ (C) $(3.0,7.5)$ (D) $(1.75,8.5)$
(v) Refer to front view, if a point $(x,y)$ is equidistant from $Q(9,8)$ and $S(17,8)$, then
(A) $x+y=13$ (B) $x-13=0$ (C) $y-13=0$ (D) $x-y=13$
Answer
(i) Option (C) is correct.
Mid-point of $J(6,17)$ and $I(9,16)$:
$$x=\frac{6+9}{2}=\frac{15}{2},\quad y=\frac{17+16}{2}=\frac{33}{2}$$
(ii) Option (A) is correct. The distance of point $P$ from the $Y$-axis $= 4$.
(iii) Option (C) is correct.
$A=(1,8),\ S=(15,8)$
$$AS=\sqrt{(15-1)^{2}+(8-8)^{2}}=\sqrt{196}=14$$
(iv) Option (D) is correct.
$A=(1,8),\ B=(4,10),\ m=1,\ n=3$
$$(x,y)=\left(\frac{4+3}{4},\frac{10+24}{4}\right)=\left(\frac{7}{4},\frac{34}{4}\right)=(1.75,8.5)$$
(v) Option (B) is correct.
$(x-9)^{2}+(y-8)^{2}=(x-17)^{2}+(y-8)^{2} \Rightarrow x-13=0$
Mid-point of $J(6,17)$ and $I(9,16)$:
$$x=\frac{6+9}{2}=\frac{15}{2},\quad y=\frac{17+16}{2}=\frac{33}{2}$$
(ii) Option (A) is correct. The distance of point $P$ from the $Y$-axis $= 4$.
(iii) Option (C) is correct.
$A=(1,8),\ S=(15,8)$
$$AS=\sqrt{(15-1)^{2}+(8-8)^{2}}=\sqrt{196}=14$$
(iv) Option (D) is correct.
$A=(1,8),\ B=(4,10),\ m=1,\ n=3$
$$(x,y)=\left(\frac{4+3}{4},\frac{10+24}{4}\right)=\left(\frac{7}{4},\frac{34}{4}\right)=(1.75,8.5)$$
(v) Option (B) is correct.
$(x-9)^{2}+(y-8)^{2}=(x-17)^{2}+(y-8)^{2} \Rightarrow x-13=0$
Jagdhish has a field in the shape of a right-angled triangle AQC. He wants to leave a square PQRS inside the field for non-wheat space. There is a pole marked as $O$.
(i) Taking $O$ as origin, coordinates of $P$ are $(-200,0)$ and of $Q$ are $(200,0)$. PQRS being a square, what are the coordinates of $R$ and $S$?
(ii) What is the area of square PQRS? OR What is the length of diagonal $PR$?
(iii) If $S$ divides $CA$ in the ratio $K:1$, find $K$ where point $A=(200,800)$.
(i) Taking $O$ as origin, coordinates of $P$ are $(-200,0)$ and of $Q$ are $(200,0)$. PQRS being a square, what are the coordinates of $R$ and $S$?
(ii) What is the area of square PQRS? OR What is the length of diagonal $PR$?
(iii) If $S$ divides $CA$ in the ratio $K:1$, find $K$ where point $A=(200,800)$.
Answer
(i) Coordinates of $R=(200,400)$ and $S=(-200,400)$
(ii) Side of square $PQRS = 400$ units.
$$\text{Area} = (400)^{2} = 160000 \text{ unit}^{2}$$
OR Diagonal $PR = \sqrt{2}\times 400 = 400\sqrt{2}$ units.
(iii) Using section formula:
$$-200=\frac{200K+(-600)}{K+1} \Rightarrow -200K-200=200K-600 \Rightarrow K=1$$
(ii) Side of square $PQRS = 400$ units.
$$\text{Area} = (400)^{2} = 160000 \text{ unit}^{2}$$
OR Diagonal $PR = \sqrt{2}\times 400 = 400\sqrt{2}$ units.
(iii) Using section formula:
$$-200=\frac{200K+(-600)}{K+1} \Rightarrow -200K-200=200K-600 \Rightarrow K=1$$
A craftsman made a tessellation pattern on the Cartesian plane using regular octagons, squares and triangles.
(i) What is the length of the line segment joining points $B$ and $F$?
(ii) The centre $Z$ of the figure is the point of intersection of the diagonals of quadrilateral WXOP. Find the coordinates of $Z$ where $W(-6,2)$, $X(-4,0)$, $O(5,9)$, $P(3,11)$.
(iii) What are the coordinates of the point on $Y$-axis equidistant from $A(-2,2)$ and $G(-4,7)$?
OR What is the area of Trapezium AFGH where $A(-2,2)$, $F(-2,9)$, $G(-4,7)$, $H(-4,4)$?
(i) What is the length of the line segment joining points $B$ and $F$?
(ii) The centre $Z$ of the figure is the point of intersection of the diagonals of quadrilateral WXOP. Find the coordinates of $Z$ where $W(-6,2)$, $X(-4,0)$, $O(5,9)$, $P(3,11)$.
(iii) What are the coordinates of the point on $Y$-axis equidistant from $A(-2,2)$ and $G(-4,7)$?
OR What is the area of Trapezium AFGH where $A(-2,2)$, $F(-2,9)$, $G(-4,7)$, $H(-4,4)$?
Answer
(i) $B(1,2)$ and $F(-2,9)$:
$$BF=\sqrt{(-2-1)^{2}+(9-2)^{2}}=\sqrt{9+49}=\sqrt{58} \text{ units}$$
(ii) WXOP is a rectangle; $Z$ is the midpoint of diagonal $WO$:
$$Z=\left(\frac{-6+5}{2},\frac{2+9}{2}\right)=\left(-\frac{1}{2},\frac{11}{2}\right)$$
(iii) Let point on $y$-axis be $Z(0,y)$. Setting $AZ^2=GZ^2$:
$$(2)^{2}+(y-2)^{2}=(4)^{2}+(y-7)^{2} \Rightarrow 10y=57 \Rightarrow y=5.7$$
The required point is $(0,5.7)$.
OR $GH=3$ units, $AF=7$ units, height $=2$ units.
$$\text{Area}=\frac{1}{2}(7+3)\times2=10 \text{ sq. units}$$
$$BF=\sqrt{(-2-1)^{2}+(9-2)^{2}}=\sqrt{9+49}=\sqrt{58} \text{ units}$$
(ii) WXOP is a rectangle; $Z$ is the midpoint of diagonal $WO$:
$$Z=\left(\frac{-6+5}{2},\frac{2+9}{2}\right)=\left(-\frac{1}{2},\frac{11}{2}\right)$$
(iii) Let point on $y$-axis be $Z(0,y)$. Setting $AZ^2=GZ^2$:
$$(2)^{2}+(y-2)^{2}=(4)^{2}+(y-7)^{2} \Rightarrow 10y=57 \Rightarrow y=5.7$$
The required point is $(0,5.7)$.
OR $GH=3$ units, $AF=7$ units, height $=2$ units.
$$\text{Area}=\frac{1}{2}(7+3)\times2=10 \text{ sq. units}$$
Medical science shows that keeping an aquarium reduces stress and promotes creativity. A sketch of an aquarium is drawn on a coordinate plane.
(i) The coordinates of $H$ are:
(A) $(4,2)$ (B) $(4,3)$ (C) $(2,4)$ (D) $(4,8)$
(ii) Distance of point $G$ from the $y$-axis is:
(A) $3$ units (B) $9$ units (C) $5$ units (D) $4$ units
(iii) Length of side $HG$:
(A) $6$ units (B) $7$ units (C) $8$ units (D) $9$ units
(iv) The length of diagonal $FD$ and the value of $x$, respectively, are:
(A) $8$ units, $4$ (B) $8$ units, $5$ (C) $\sqrt{15}$ units, $9$ (D) $\sqrt{61}$ units, $9$
(v) If $Q$ is considered as origin, the coordinates of midpoint of $BC$ are:
(A) $(1,4)$ (B) $(1,6)$ (C) $(6,1)$ (D) $(6,-1)$
(i) The coordinates of $H$ are:
(A) $(4,2)$ (B) $(4,3)$ (C) $(2,4)$ (D) $(4,8)$
(ii) Distance of point $G$ from the $y$-axis is:
(A) $3$ units (B) $9$ units (C) $5$ units (D) $4$ units
(iii) Length of side $HG$:
(A) $6$ units (B) $7$ units (C) $8$ units (D) $9$ units
(iv) The length of diagonal $FD$ and the value of $x$, respectively, are:
(A) $8$ units, $4$ (B) $8$ units, $5$ (C) $\sqrt{15}$ units, $9$ (D) $\sqrt{61}$ units, $9$
(v) If $Q$ is considered as origin, the coordinates of midpoint of $BC$ are:
(A) $(1,4)$ (B) $(1,6)$ (C) $(6,1)$ (D) $(6,-1)$
Answer
(i) Option (C) is correct. $P$ is the origin, so coordinates of $H = (2,4)$.
(ii) Option (B) is correct. Coordinates of $G=(9,4)$. Distance from $y$-axis $= |9| = 9$ units.
(iii) Option (B) is correct. $H=(2,4)$ and $G=(9,4)$:
$$HG=\sqrt{(9-2)^{2}+(4-4)^{2}}=\sqrt{49}=7 \text{ units}$$
(iv) Option (D) is correct. $D=(4,2)$ and $F=(9,8)$:
$$FD=\sqrt{(9-4)^{2}+(8-2)^{2}}=\sqrt{25+36}=\sqrt{61} \text{ units}$$
Given $x=9$.
(v) Option (C) is correct. $B=(4,2)$ and $C=(8,0)$:
$$\text{Midpoint}=\left(\frac{4+8}{2},\frac{2+0}{2}\right)=(6,1)$$
(ii) Option (B) is correct. Coordinates of $G=(9,4)$. Distance from $y$-axis $= |9| = 9$ units.
(iii) Option (B) is correct. $H=(2,4)$ and $G=(9,4)$:
$$HG=\sqrt{(9-2)^{2}+(4-4)^{2}}=\sqrt{49}=7 \text{ units}$$
(iv) Option (D) is correct. $D=(4,2)$ and $F=(9,8)$:
$$FD=\sqrt{(9-4)^{2}+(8-2)^{2}}=\sqrt{25+36}=\sqrt{61} \text{ units}$$
Given $x=9$.
(v) Option (C) is correct. $B=(4,2)$ and $C=(8,0)$:
$$\text{Midpoint}=\left(\frac{4+8}{2},\frac{2+0}{2}\right)=(6,1)$$
The Pacific Ring of Fire is a major area in the basin of the Pacific Ocean where many earthquakes and volcanic eruptions occur.
Fault Lines: Large faults within the Earth’s crust result from the action of plate tectonic forces. Energy release associated with rapid movement on active faults is the cause of most earthquakes.
(i) The distance between Country A and Country B is:
(A) $4$ units (B) $5$ units (C) $6$ units (D) $7$ units
(ii) Find a relation between $x$ and $y$ such that point $(x,y)$ is equidistant from Country C and Country D:
(A) $x-y$ (B) $x+y=2$ (C) $2x-y=2$ (D) $2x+y=2$
(iii) The fault line $3x+y-9=0$ divides the line joining Country $P(1,3)$ and Country $Q(2,7)$ internally in the ratio:
(A) $3:4$ (B) $3:2$ (C) $2:3$ (D) $4:3$
(iv) The distance of Country $M$ from the $x$-axis is:
(A) $1$ unit (B) $2$ units (C) $3$ units (D) $5$ units
(v) The coordinates of the Country at the midpoint of Country $A$ and Country $D$ are:
(A) $(1,3)$ (B) $\left(2,\dfrac{9}{2}\right)$ (C) $\left(4,\dfrac{5}{2}\right)$ (D) $\left(\dfrac{9}{2},2\right)$
Fault Lines: Large faults within the Earth’s crust result from the action of plate tectonic forces. Energy release associated with rapid movement on active faults is the cause of most earthquakes.
(i) The distance between Country A and Country B is:
(A) $4$ units (B) $5$ units (C) $6$ units (D) $7$ units
(ii) Find a relation between $x$ and $y$ such that point $(x,y)$ is equidistant from Country C and Country D:
(A) $x-y$ (B) $x+y=2$ (C) $2x-y=2$ (D) $2x+y=2$
(iii) The fault line $3x+y-9=0$ divides the line joining Country $P(1,3)$ and Country $Q(2,7)$ internally in the ratio:
(A) $3:4$ (B) $3:2$ (C) $2:3$ (D) $4:3$
(iv) The distance of Country $M$ from the $x$-axis is:
(A) $1$ unit (B) $2$ units (C) $3$ units (D) $5$ units
(v) The coordinates of the Country at the midpoint of Country $A$ and Country $D$ are:
(A) $(1,3)$ (B) $\left(2,\dfrac{9}{2}\right)$ (C) $\left(4,\dfrac{5}{2}\right)$ (D) $\left(\dfrac{9}{2},2\right)$
Answer
(i) Option (B) is correct. $A=(1,4)$, $B=(4,0)$:
$$AB=\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=5 \text{ units}$$
(ii) Option (B) is correct. $C=(7,1)$, $D=(3,5)$. Setting $CP=DP$:
$$-14x-2y+50=-6x-10y+34 \Rightarrow x+y=2$$
(iii) Option (A) is correct. Let the line divide $PQ$ in ratio $k:1$. Point of division lies on $3x+y-9=0$:
$$\frac{6k+21+3k+7-9k-9}{k+1}=0 \Rightarrow 4k=3 \Rightarrow k=\frac{3}{4}$$
Ratio = $3:4$.
(iv) Option (C) is correct. $M=(2,3)$. Distance from $x$-axis $=|3|=3$ units.
(v) Option (B) is correct. $A=(1,4)$, $D=(3,5)$:
$$\text{Midpoint}=\left(\frac{1+3}{2},\frac{4+5}{2}\right)=\left(2,\frac{9}{2}\right)$$
$$AB=\sqrt{(4-1)^{2}+(0-4)^{2}}=\sqrt{9+16}=5 \text{ units}$$
(ii) Option (B) is correct. $C=(7,1)$, $D=(3,5)$. Setting $CP=DP$:
$$-14x-2y+50=-6x-10y+34 \Rightarrow x+y=2$$
(iii) Option (A) is correct. Let the line divide $PQ$ in ratio $k:1$. Point of division lies on $3x+y-9=0$:
$$\frac{6k+21+3k+7-9k-9}{k+1}=0 \Rightarrow 4k=3 \Rightarrow k=\frac{3}{4}$$
Ratio = $3:4$.
(iv) Option (C) is correct. $M=(2,3)$. Distance from $x$-axis $=|3|=3$ units.
(v) Option (B) is correct. $A=(1,4)$, $D=(3,5)$:
$$\text{Midpoint}=\left(\frac{1+3}{2},\frac{4+5}{2}\right)=\left(2,\frac{9}{2}\right)$$
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The midpoint $(0,2)$ is the only point equidistant from $P$ and $Q$.
Reason (R): There are many points $(x,y)$ where $(x+2)^{2}+(y-5)^{2}=(x-2)^{2}+(y+1)^{2}$ are equidistant from $P$ and $Q$.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The midpoint $(0,2)$ is the only point equidistant from $P$ and $Q$.
Reason (R): There are many points $(x,y)$ where $(x+2)^{2}+(y-5)^{2}=(x-2)^{2}+(y+1)^{2}$ are equidistant from $P$ and $Q$.
Answer
Option (D) is correct.
There are several other points equidistant from $P$ and $Q$, so the Assertion is false.
For points $(x,y)$: $(x+2)^{2}+(y-5)^{2}=(x-2)^{2}+(y+1)^{2}$ — many points exist that are equidistant from $P$ and $Q$, so the Reason is true.
There are several other points equidistant from $P$ and $Q$, so the Assertion is false.
For points $(x,y)$: $(x+2)^{2}+(y-5)^{2}=(x-2)^{2}+(y+1)^{2}$ — many points exist that are equidistant from $P$ and $Q$, so the Reason is true.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): If the distance between $(4,p)$ and $(1,0)$ is 5, then the value of $p$ is 4.
Reason (R): The point which divides the line segment joining $(7,-6)$ and $(3,4)$ in ratio $1:2$ internally lies in the fourth quadrant.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): If the distance between $(4,p)$ and $(1,0)$ is 5, then the value of $p$ is 4.
Reason (R): The point which divides the line segment joining $(7,-6)$ and $(3,4)$ in ratio $1:2$ internally lies in the fourth quadrant.
Answer
Option (D) is correct.
Assertion: $5^{2}=(1-4)^{2}+(0-p)^{2} \Rightarrow 25=9+p^{2} \Rightarrow p^{2}=16 \Rightarrow p=\pm4$.
Since $p$ can also be $-4$, Assertion is false.
Reason: $x=\frac{3+14}{3}=\frac{17}{3}$, $y=\frac{4-12}{3}=-\frac{8}{3}$.
Point $\left(\frac{17}{3},-\frac{8}{3}\right)$ lies in the IV quadrant. Reason is true.
Assertion: $5^{2}=(1-4)^{2}+(0-p)^{2} \Rightarrow 25=9+p^{2} \Rightarrow p^{2}=16 \Rightarrow p=\pm4$.
Since $p$ can also be $-4$, Assertion is false.
Reason: $x=\frac{3+14}{3}=\frac{17}{3}$, $y=\frac{4-12}{3}=-\frac{8}{3}$.
Point $\left(\frac{17}{3},-\frac{8}{3}\right)$ lies in the IV quadrant. Reason is true.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): $\triangle ABC$ with $A(-2,0)$, $B(2,0)$, $C(0,2)$ is similar to $\triangle DEF$ with $D(-4,0)$, $E(4,0)$, $F(0,4)$.
Reason (R): The perimeter of $\triangle ABC$ is $4(1+\sqrt{2})$ units.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): $\triangle ABC$ with $A(-2,0)$, $B(2,0)$, $C(0,2)$ is similar to $\triangle DEF$ with $D(-4,0)$, $E(4,0)$, $F(0,4)$.
Reason (R): The perimeter of $\triangle ABC$ is $4(1+\sqrt{2})$ units.
Answer
Option (A) is correct.
Assertion: $AB=4$, $BC=CA=2\sqrt{2}$. For $\triangle DEF$: $DE=8$, $EF=FD=4\sqrt{2}$.
$$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\frac{1}{2}$$
By SSS similarity, $\triangle ABC \sim \triangle DEF$. Assertion is true.
Reason: Perimeter $= 4+2\sqrt{2}+2\sqrt{2}=4+4\sqrt{2}=4(1+\sqrt{2})$ units. Reason is true.
Both A and R are true and R is the correct explanation of A.
Assertion: $AB=4$, $BC=CA=2\sqrt{2}$. For $\triangle DEF$: $DE=8$, $EF=FD=4\sqrt{2}$.
$$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\frac{1}{2}$$
By SSS similarity, $\triangle ABC \sim \triangle DEF$. Assertion is true.
Reason: Perimeter $= 4+2\sqrt{2}+2\sqrt{2}=4+4\sqrt{2}=4(1+\sqrt{2})$ units. Reason is true.
Both A and R are true and R is the correct explanation of A.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): AOBC is a rectangle with vertices $A(0,3)$, $O(0,0)$ and $B(5,0)$. The length of its diagonal is $\sqrt{34}$ units.
Reason (R): The distance between $A$ and $O$ is 3 units.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): AOBC is a rectangle with vertices $A(0,3)$, $O(0,0)$ and $B(5,0)$. The length of its diagonal is $\sqrt{34}$ units.
Reason (R): The distance between $A$ and $O$ is 3 units.
Answer
Option (B) is correct.
Assertion: Diagonal $AB=\sqrt{(5-0)^{2}+(0-3)^{2}}=\sqrt{25+9}=\sqrt{34}$ units. Assertion is true.
Reason: $AO=\sqrt{(0-0)^{2}+(0-3)^{2}}=\sqrt{9}=3$ units. Reason is true.
Both are true but R is not the correct explanation of A (R describes a side, not the diagonal).
Assertion: Diagonal $AB=\sqrt{(5-0)^{2}+(0-3)^{2}}=\sqrt{25+9}=\sqrt{34}$ units. Assertion is true.
Reason: $AO=\sqrt{(0-0)^{2}+(0-3)^{2}}=\sqrt{9}=3$ units. Reason is true.
Both are true but R is not the correct explanation of A (R describes a side, not the diagonal).
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): If the coordinates of mid-points of sides $AB$ and $AC$ of $\triangle ABC$ are $D(3,5)$ and $E(-3,-3)$ respectively, then $BC=20$ units.
Reason (R): The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): If the coordinates of mid-points of sides $AB$ and $AC$ of $\triangle ABC$ are $D(3,5)$ and $E(-3,-3)$ respectively, then $BC=20$ units.
Reason (R): The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
Answer
Option (B) is correct.
Assertion: $DE=\sqrt{(-3-3)^{2}+(-3-5)^{2}}=\sqrt{36+64}=10$ units.
By mid-point theorem, $BC=2\times DE=20$ units. Assertion is true.
Reason: By mid-point theorem, this statement is true.
Both are true but R is not the correct explanation of A.
Assertion: $DE=\sqrt{(-3-3)^{2}+(-3-5)^{2}}=\sqrt{36+64}=10$ units.
By mid-point theorem, $BC=2\times DE=20$ units. Assertion is true.
Reason: By mid-point theorem, this statement is true.
Both are true but R is not the correct explanation of A.
The line segment joining the points $P(-3,2)$ and $Q(5,7)$ is divided by the $Y$-axis in the ratio
Solution
Option (D) is correct.
Let $M(0,y)$ divide $PQ$ in ratio $k:1$.
$$0=\frac{5k-3}{k+1} \Rightarrow 5k=3 \Rightarrow k=\frac{3}{5}$$
So the ratio is $3:5$.
Let $M(0,y)$ divide $PQ$ in ratio $k:1$.
$$0=\frac{5k-3}{k+1} \Rightarrow 5k=3 \Rightarrow k=\frac{3}{5}$$
So the ratio is $3:5$.
The base $BC$ of an equilateral $\triangle ABC$ lies on the $Y$-axis. The coordinates of $C$ are $(0,-3)$. If the origin is the midpoint of $BC$, what are the coordinates of $A$ and $B$?
Solution
Option (C) is correct.
Since $O$ is the midpoint of $BC$ and $C=(0,-3)$, we get $B=(0,3)$ and $BC=6$ units.
Let $A=(x,0)$. Since $\triangle ABC$ is equilateral, $AB=BC$:
$$\sqrt{x^{2}+9}=6 \Rightarrow x^{2}=27 \Rightarrow x=\pm3\sqrt{3}$$
Coordinates: $A=(\pm3\sqrt{3},0)$, $B=(0,3)$.
Since $O$ is the midpoint of $BC$ and $C=(0,-3)$, we get $B=(0,3)$ and $BC=6$ units.
Let $A=(x,0)$. Since $\triangle ABC$ is equilateral, $AB=BC$:
$$\sqrt{x^{2}+9}=6 \Rightarrow x^{2}=27 \Rightarrow x=\pm3\sqrt{3}$$
Coordinates: $A=(\pm3\sqrt{3},0)$, $B=(0,3)$.
The centre of a circle whose endpoints of a diameter are $(-6,3)$ and $(6,4)$ is
Solution
Option (C) is correct.
The centre is the midpoint of the diameter:
$$O=\left(\frac{-6+6}{2},\frac{3+4}{2}\right)=\left(0,\frac{7}{2}\right)$$
The centre is the midpoint of the diameter:
$$O=\left(\frac{-6+6}{2},\frac{3+4}{2}\right)=\left(0,\frac{7}{2}\right)$$
If the vertices of a parallelogram PQRS taken in order are $P(3,4)$, $Q(-2,3)$ and $R(-3,-2)$, then the coordinates of its fourth vertex $S$ are
Solution
Option (C) is correct.
In a parallelogram, diagonals bisect each other. Midpoint of $PR$ = midpoint of $QS$:
$$\left(\frac{3-3}{2},\frac{4-2}{2}\right)=\left(\frac{x-2}{2},\frac{y+3}{2}\right)$$
$$\Rightarrow x=2,\ y=-1$$
Fourth vertex $S=(2,-1)$.
In a parallelogram, diagonals bisect each other. Midpoint of $PR$ = midpoint of $QS$:
$$\left(\frac{3-3}{2},\frac{4-2}{2}\right)=\left(\frac{x-2}{2},\frac{y+3}{2}\right)$$
$$\Rightarrow x=2,\ y=-1$$
Fourth vertex $S=(2,-1)$.
$\triangle ABC$ is a triangle such that $AB:BC=1:2$. Point $A$ lies on the $Y$-axis and the coordinates of $B$ and $C$ are known. Which of the following formula can definitely be used to find the coordinates of $A$?
(i) Section formula (ii) Distance formula
(i) Section formula (ii) Distance formula
Solution
Option (B) is correct.
Since $A$ lies on the $y$-axis, its coordinates are $(0,y)$. We can use the distance formula (with known $AB:BC=1:2$) to find the $y$-coordinate of $A$. The section formula would require knowing the ratio in which $A$ divides $BC$, which is not directly given.
Since $A$ lies on the $y$-axis, its coordinates are $(0,y)$. We can use the distance formula (with known $AB:BC=1:2$) to find the $y$-coordinate of $A$. The section formula would require knowing the ratio in which $A$ divides $BC$, which is not directly given.
Let $P(4,4)$ and $Q(9,6)$ be points on the parabola $y^{2}=4a(x-b)$. If $R$ is a point on the arc between $P$ and $Q$ such that the area of $\triangle PRQ$ is largest, then $R$ is:
Solution
Option (D) is correct.
For a fixed base $PQ$, the area of $\triangle PRQ$ is maximum when the perpendicular distance of $R$ from line $PQ$ is maximum. For a parabola, this occurs at the midpoint of the arc between the two points. Among the given options, $R=(2,2\sqrt{2})$ satisfies this condition.
For a fixed base $PQ$, the area of $\triangle PRQ$ is maximum when the perpendicular distance of $R$ from line $PQ$ is maximum. For a parabola, this occurs at the midpoint of the arc between the two points. Among the given options, $R=(2,2\sqrt{2})$ satisfies this condition.
The perpendicular bisector of the line segment joining the points $A(1,5)$ and $B(4,6)$ cuts the $Y$-axis at
Solution
Option (A) is correct.
Midpoint $M=\left(\dfrac{5}{2},\dfrac{11}{2}\right)$. Slope of $AB=\dfrac{1}{3}$, so slope of perpendicular bisector $=-3$.
Equation: $y-\dfrac{11}{2}=-3\left(x-\dfrac{5}{2}\right) \Rightarrow y=-3x+13$.
At $x=0$: $y=13$. The point is $(0,13)$.
Midpoint $M=\left(\dfrac{5}{2},\dfrac{11}{2}\right)$. Slope of $AB=\dfrac{1}{3}$, so slope of perpendicular bisector $=-3$.
Equation: $y-\dfrac{11}{2}=-3\left(x-\dfrac{5}{2}\right) \Rightarrow y=-3x+13$.
At $x=0$: $y=13$. The point is $(0,13)$.
The ratio in which the point $(4,5)$ divides the line segment joining the points $(2,3)$ and $(7,8)$ is:
Solution
Option (A) is correct.
Let point $P(4,5)$ divide $A(2,3)$ and $B(7,8)$ in ratio $m:n$.
$$\frac{7m+2n}{m+n}=4 \Rightarrow 7m+2n=4m+4n \Rightarrow 3m=2n \Rightarrow m:n=2:3$$
Let point $P(4,5)$ divide $A(2,3)$ and $B(7,8)$ in ratio $m:n$.
$$\frac{7m+2n}{m+n}=4 \Rightarrow 7m+2n=4m+4n \Rightarrow 3m=2n \Rightarrow m:n=2:3$$
The values of $x$ and $y$, if the distance of the point $(x,y)$ from $(-3,0)$ as well as from $(3,0)$ is $4$, are:
Solution
Option (D) is correct.
$(x+3)^{2}+y^{2}=16$ and $(x-3)^{2}+y^{2}=16$.
Subtracting: $(x+3)^{2}-(x-3)^{2}=0 \Rightarrow 12x=0 \Rightarrow x=0$.
Substituting: $9+y^{2}=16 \Rightarrow y^{2}=7 \Rightarrow y=\pm\sqrt{7}$.
$(x+3)^{2}+y^{2}=16$ and $(x-3)^{2}+y^{2}=16$.
Subtracting: $(x+3)^{2}-(x-3)^{2}=0 \Rightarrow 12x=0 \Rightarrow x=0$.
Substituting: $9+y^{2}=16 \Rightarrow y^{2}=7 \Rightarrow y=\pm\sqrt{7}$.
The distance between the points $(3,4)$ and $(8,-6)$ is
Solution
Option (D) is correct.
$$d=\sqrt{(8-3)^{2}+(-6-4)^{2}}=\sqrt{25+100}=\sqrt{125}=5\sqrt{5} \text{ units}$$
$$d=\sqrt{(8-3)^{2}+(-6-4)^{2}}=\sqrt{25+100}=\sqrt{125}=5\sqrt{5} \text{ units}$$
Alia and Shagun are friends living on the same street in Patel Nagar. The school is at origin $O$, Alia’s house is at $A$, Shagun’s house is at $B$ and the library is at $C$.
(i) How far is Alia’s house from Shagun’s house?
(ii) How far is the library from Shagun’s house?
(iii) Show that for Shagun, the school is farther compared to Alia’s house and the library.
OR
Show that Alia’s house, Shagun’s house and the library form an isosceles right triangle.
(i) How far is Alia’s house from Shagun’s house?
(ii) How far is the library from Shagun’s house?
(iii) Show that for Shagun, the school is farther compared to Alia’s house and the library.
OR
Show that Alia’s house, Shagun’s house and the library form an isosceles right triangle.
Answer
(i) $A(2,3)$ and $B(2,1)$:
$$BA=\sqrt{(2-2)^{2}+(3-1)^{2}}=\sqrt{4}=2 \text{ units}$$
(ii) $B(2,1)$ and $C(4,1)$:
$$BC=\sqrt{(4-2)^{2}+(1-1)^{2}}=\sqrt{4}=2 \text{ units}$$
(iii) $BO=\sqrt{(2-0)^{2}+(1-0)^{2}}=\sqrt{5}$ units.
Since $\sqrt{5}>2$, school $O$ is farther from Shagun’s house than both $A$ and $C$. Hence proved.
OR $AB=BC=2$ and $AC=\sqrt{(4-2)^{2}+(1-3)^{2}}=2\sqrt{2}$.
$AB^{2}+BC^{2}=4+4=8=AC^{2}$, and $AB=BC$.
Therefore $\triangle ABC$ is an isosceles right triangle. Hence proved.
$$BA=\sqrt{(2-2)^{2}+(3-1)^{2}}=\sqrt{4}=2 \text{ units}$$
(ii) $B(2,1)$ and $C(4,1)$:
$$BC=\sqrt{(4-2)^{2}+(1-1)^{2}}=\sqrt{4}=2 \text{ units}$$
(iii) $BO=\sqrt{(2-0)^{2}+(1-0)^{2}}=\sqrt{5}$ units.
Since $\sqrt{5}>2$, school $O$ is farther from Shagun’s house than both $A$ and $C$. Hence proved.
OR $AB=BC=2$ and $AC=\sqrt{(4-2)^{2}+(1-3)^{2}}=2\sqrt{2}$.
$AB^{2}+BC^{2}=4+4=8=AC^{2}$, and $AB=BC$.
Therefore $\triangle ABC$ is an isosceles right triangle. Hence proved.
Some college students started a “NO SMOKING” awareness campaign. Students prepared campaign banners in the shape of triangle $PQR$ as shown in the figure.
(i) Find the coordinates of the midpoint of $Q$ and $R$.
(ii) Find the area of triangle $PQR$.
(iii) Find the point on the $x$-axis equidistant from $Q$ and $R$.
OR
Find the centroid of triangle $PQR$.
(i) Find the coordinates of the midpoint of $Q$ and $R$.
(ii) Find the area of triangle $PQR$.
(iii) Find the point on the $x$-axis equidistant from $Q$ and $R$.
OR
Find the centroid of triangle $PQR$.
Answer
(i) $Q(2,3)$ and $R(9,3)$:
$$\text{Midpoint}=\left(\frac{2+9}{2},\frac{3+3}{2}\right)=\left(\frac{11}{2},3\right)$$
(ii) $P(6,9)$, $Q(2,3)$, $R(9,3)$:
$$\text{Area}=\frac{1}{2}|6(3-3)+2(3-9)+9(9-3)|=\frac{1}{2}|0-12+54|=21 \text{ sq. units}$$
(iii) Let the point be $(x,0)$. Setting equal distances:
$(x-2)^{2}+9=(x-9)^{2}+9 \Rightarrow 14x=77 \Rightarrow x=\dfrac{11}{2}$
The point is $\left(\dfrac{11}{2},0\right)$.
OR Centroid $=\left(\dfrac{6+2+9}{3},\dfrac{9+3+3}{3}\right)=\left(\dfrac{17}{3},5\right)$.
$$\text{Midpoint}=\left(\frac{2+9}{2},\frac{3+3}{2}\right)=\left(\frac{11}{2},3\right)$$
(ii) $P(6,9)$, $Q(2,3)$, $R(9,3)$:
$$\text{Area}=\frac{1}{2}|6(3-3)+2(3-9)+9(9-3)|=\frac{1}{2}|0-12+54|=21 \text{ sq. units}$$
(iii) Let the point be $(x,0)$. Setting equal distances:
$(x-2)^{2}+9=(x-9)^{2}+9 \Rightarrow 14x=77 \Rightarrow x=\dfrac{11}{2}$
The point is $\left(\dfrac{11}{2},0\right)$.
OR Centroid $=\left(\dfrac{6+2+9}{3},\dfrac{9+3+3}{3}\right)=\left(\dfrac{17}{3},5\right)$.
On Annual Sports Day, parallel lines are drawn 1 m apart in a rectangular playground. 80 cones are placed 1 m apart along $PY$.
Pushpendra runs $\dfrac{1}{4}$ the distance $PY$ on the 3rd line and posts a yellow flag.
Pankaj runs $\dfrac{1}{5}$ the distance $PY$ on the 7th line and posts a blue flag.
(i) Find the coordinates of the yellow flag.
(ii) What is the distance between both flags?
(iii) If Raman posts a green flag exactly halfway between the two flags, where should he post it?
OR
Raman posts a green flag that divides the segment joining the two flags in ratio $1:2$ internally. Find its coordinates.
Pushpendra runs $\dfrac{1}{4}$ the distance $PY$ on the 3rd line and posts a yellow flag.
Pankaj runs $\dfrac{1}{5}$ the distance $PY$ on the 7th line and posts a blue flag.
(i) Find the coordinates of the yellow flag.
(ii) What is the distance between both flags?
(iii) If Raman posts a green flag exactly halfway between the two flags, where should he post it?
OR
Raman posts a green flag that divides the segment joining the two flags in ratio $1:2$ internally. Find its coordinates.
Answer
(i) $\dfrac{1}{4}\times80=20$ m on the 3rd line.
Yellow flag coordinates: $(3,20)$.
(ii) $\dfrac{1}{5}\times80=16$ m on the 7th line.
Blue flag coordinates: $(7,16)$.
$$\text{Distance}=\sqrt{(7-3)^{2}+(16-20)^{2}}=\sqrt{16+16}=4\sqrt{2} \text{ m}$$
(iii) Midpoint of $(3,20)$ and $(7,16)$:
$$\left(\frac{3+7}{2},\frac{20+16}{2}\right)=(5,18)$$
Raman should post at $18$ m on the 5th line.
OR Using section formula with ratio $1:2$:
$$x=\frac{7+6}{3}=\frac{13}{3},\quad y=\frac{16+40}{3}=\frac{56}{3}$$
Green flag coordinates: $\left(\dfrac{13}{3},\dfrac{56}{3}\right)$.
Yellow flag coordinates: $(3,20)$.
(ii) $\dfrac{1}{5}\times80=16$ m on the 7th line.
Blue flag coordinates: $(7,16)$.
$$\text{Distance}=\sqrt{(7-3)^{2}+(16-20)^{2}}=\sqrt{16+16}=4\sqrt{2} \text{ m}$$
(iii) Midpoint of $(3,20)$ and $(7,16)$:
$$\left(\frac{3+7}{2},\frac{20+16}{2}\right)=(5,18)$$
Raman should post at $18$ m on the 5th line.
OR Using section formula with ratio $1:2$:
$$x=\frac{7+6}{3}=\frac{13}{3},\quad y=\frac{16+40}{3}=\frac{56}{3}$$
Green flag coordinates: $\left(\dfrac{13}{3},\dfrac{56}{3}\right)$.
The Chief Minister of Delhi launched the “Switch Delhi” electric vehicle mass awareness campaign. Charging stations are set up along a straight line at:
$$A\left(-\frac{7}{3},0\right),\ B(0,4),\ C(3,4),\ D(7,7),\ E(x,y)$$
The distance between $C$ and $E$ is 10 units.
(i) The distance $DE$ is:
(A) $5$ units (B) $10$ units (C) $4$ units (D) $6$ units
(ii) The value of $x+y$ is:
(A) $20$ (B) $21$ (C) $22$ (D) $23$
(iii) Which of the following is true?
(A) $C$, $D$, $E$ are vertices of a triangle
(B) $C$, $D$, $E$ are collinear
(C) $C$, $D$, $E$ lie on a circle
(D) None of these
(iv) The ratio in which point $B$ divides $AC$ is:
(A) $9:7$ (B) $4:7$ (C) $7:4$ (D) $7:9$
$$A\left(-\frac{7}{3},0\right),\ B(0,4),\ C(3,4),\ D(7,7),\ E(x,y)$$
The distance between $C$ and $E$ is 10 units.
(i) The distance $DE$ is:
(A) $5$ units (B) $10$ units (C) $4$ units (D) $6$ units
(ii) The value of $x+y$ is:
(A) $20$ (B) $21$ (C) $22$ (D) $23$
(iii) Which of the following is true?
(A) $C$, $D$, $E$ are vertices of a triangle
(B) $C$, $D$, $E$ are collinear
(C) $C$, $D$, $E$ lie on a circle
(D) None of these
(iv) The ratio in which point $B$ divides $AC$ is:
(A) $9:7$ (B) $4:7$ (C) $7:4$ (D) $7:9$
Answer
(i) Option (A) is correct.
$CD=\sqrt{(7-3)^{2}+(7-4)^{2}}=\sqrt{16+9}=5$ units.
$DE=CE-CD=10-5=5$ units.
(ii) Option (B) is correct.
Since $CD=DE$, $D$ is the midpoint of $CE$:
$$\frac{x+3}{2}=7 \Rightarrow x=11,\quad \frac{y+4}{2}=7 \Rightarrow y=10$$
$$x+y=21$$
(iii) Option (B) is correct. Since $D$ is the midpoint of $CE$, points $C$, $D$ and $E$ are collinear.
(iv) Option (D) is correct. Let $B$ divide $AC$ in ratio $k:1$:
$$\frac{7}{4}=\frac{4k+0}{k+1} \Rightarrow 7k+7=16k \Rightarrow k=\frac{7}{9}$$
Ratio = $7:9$.
$CD=\sqrt{(7-3)^{2}+(7-4)^{2}}=\sqrt{16+9}=5$ units.
$DE=CE-CD=10-5=5$ units.
(ii) Option (B) is correct.
Since $CD=DE$, $D$ is the midpoint of $CE$:
$$\frac{x+3}{2}=7 \Rightarrow x=11,\quad \frac{y+4}{2}=7 \Rightarrow y=10$$
$$x+y=21$$
(iii) Option (B) is correct. Since $D$ is the midpoint of $CE$, points $C$, $D$ and $E$ are collinear.
(iv) Option (D) is correct. Let $B$ divide $AC$ in ratio $k:1$:
$$\frac{7}{4}=\frac{4k+0}{k+1} \Rightarrow 7k+7=16k \Rightarrow k=\frac{7}{9}$$
Ratio = $7:9$.
Alia and Shagun are friends living on the same street in Patel Nagar. School is at origin $O$, Alia’s house is at $A(2,3)$, Shagun’s house is at $B(2,1)$ and library is at $C(4,1)$.
(i) How far is Alia’s house from Shagun’s house?
(A) $3$ units (B) $4$ units (C) $5$ units (D) $2$ units
(ii) How far is the library from Shagun’s house?
(A) $3$ units (B) $2$ units (C) $5$ units (D) $4$ units
(iii) How far is the library from Alia’s house?
(A) $2$ units (B) $3$ units (C) $4$ units (D) None of these
(iv) Which of the following is true?
(A) $\triangle ABC$ is scalene (B) $\triangle ABC$ is isosceles (C) $\triangle ABC$ is equilateral (D) None of these
(v) How far is the school from Alia’s house?
(A) $\sqrt{13}$ units (B) $\sqrt{5}$ units (C) $(\sqrt{13}+\sqrt{5})$ units (D) $(\sqrt{13}-\sqrt{5})$ units
(i) How far is Alia’s house from Shagun’s house?
(A) $3$ units (B) $4$ units (C) $5$ units (D) $2$ units
(ii) How far is the library from Shagun’s house?
(A) $3$ units (B) $2$ units (C) $5$ units (D) $4$ units
(iii) How far is the library from Alia’s house?
(A) $2$ units (B) $3$ units (C) $4$ units (D) None of these
(iv) Which of the following is true?
(A) $\triangle ABC$ is scalene (B) $\triangle ABC$ is isosceles (C) $\triangle ABC$ is equilateral (D) None of these
(v) How far is the school from Alia’s house?
(A) $\sqrt{13}$ units (B) $\sqrt{5}$ units (C) $(\sqrt{13}+\sqrt{5})$ units (D) $(\sqrt{13}-\sqrt{5})$ units
Answer
(i) Option (D) is correct. $AB=\sqrt{(2-2)^{2}+(1-3)^{2}}=2$ units.
(ii) Option (B) is correct. $BC=\sqrt{(4-2)^{2}+(1-1)^{2}}=2$ units.
(iii) Option (D) is correct. $AC=\sqrt{(4-2)^{2}+(1-3)^{2}}=2\sqrt{2}$ units.
(iv) Option (B) is correct. $AB=BC=2\neq AC=2\sqrt{2}$, so $\triangle ABC$ is isosceles.
(v) Option (A) is correct. $OA=\sqrt{2^{2}+3^{2}}=\sqrt{13}$ units.
(ii) Option (B) is correct. $BC=\sqrt{(4-2)^{2}+(1-1)^{2}}=2$ units.
(iii) Option (D) is correct. $AC=\sqrt{(4-2)^{2}+(1-3)^{2}}=2\sqrt{2}$ units.
(iv) Option (B) is correct. $AB=BC=2\neq AC=2\sqrt{2}$, so $\triangle ABC$ is isosceles.
(v) Option (A) is correct. $OA=\sqrt{2^{2}+3^{2}}=\sqrt{13}$ units.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The distance between $(\cos\theta,\sin\theta)$ and $(\sin\theta,-\cos\theta)$ is $2$ units.
Reason (R): The distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is $AB=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The distance between $(\cos\theta,\sin\theta)$ and $(\sin\theta,-\cos\theta)$ is $2$ units.
Reason (R): The distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is $AB=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$.
Answer
Option (D) is correct.
$$AB=\sqrt{(\sin\theta-\cos\theta)^{2}+(-\cos\theta-\sin\theta)^{2}}$$
$$=\sqrt{2(\sin^{2}\theta+\cos^{2}\theta)}=\sqrt{2}$$
Since $\sin^{2}\theta+\cos^{2}\theta=1$, the distance is $\sqrt{2}$ (not 2).
Assertion (A) is false; Reason (R) is true.
$$AB=\sqrt{(\sin\theta-\cos\theta)^{2}+(-\cos\theta-\sin\theta)^{2}}$$
$$=\sqrt{2(\sin^{2}\theta+\cos^{2}\theta)}=\sqrt{2}$$
Since $\sin^{2}\theta+\cos^{2}\theta=1$, the distance is $\sqrt{2}$ (not 2).
Assertion (A) is false; Reason (R) is true.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The point $(0,4)$ lies on the $Y$-axis.
Reason (R): The $x$-coordinate of the point on the $Y$-axis is zero.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The point $(0,4)$ lies on the $Y$-axis.
Reason (R): The $x$-coordinate of the point on the $Y$-axis is zero.
Answer
Option (A) is correct.
The $x$-coordinate of every point on the $Y$-axis is $0$. The point $(0,4)$ has $x=0$, so it lies on the $Y$-axis. Assertion (A) is true.
The $x$-coordinate of any point on the $Y$-axis is indeed zero. Reason (R) is true and is the correct explanation of A.
The $x$-coordinate of every point on the $Y$-axis is $0$. The point $(0,4)$ has $x=0$, so it lies on the $Y$-axis. Assertion (A) is true.
The $x$-coordinate of any point on the $Y$-axis is indeed zero. Reason (R) is true and is the correct explanation of A.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The coordinates of the points dividing the segment joining $A(2,-8)$ and $B(-3,-7)$ into three equal parts are $\left(\dfrac{1}{3},-\dfrac{23}{3}\right)$ and $\left(-\dfrac{4}{3},-\dfrac{22}{3}\right)$.
Reason (R): The points dividing $AB$ in ratio $1:3$ and $3:1$ are called trisection points.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The coordinates of the points dividing the segment joining $A(2,-8)$ and $B(-3,-7)$ into three equal parts are $\left(\dfrac{1}{3},-\dfrac{23}{3}\right)$ and $\left(-\dfrac{4}{3},-\dfrac{22}{3}\right)$.
Reason (R): The points dividing $AB$ in ratio $1:3$ and $3:1$ are called trisection points.
Answer
Option (C) is correct.
Assertion: Using section formula with ratio $1:2$ and $2:1$:
$$P=\left(\frac{1}{3},-\frac{23}{3}\right),\quad Q=\left(-\frac{4}{3},-\frac{22}{3}\right)$$
Assertion is true.
Reason: Trisection points divide a line segment in ratios $1:2$ and $2:1$ (not $1:3$ and $3:1$). Reason is false.
Assertion: Using section formula with ratio $1:2$ and $2:1$:
$$P=\left(\frac{1}{3},-\frac{23}{3}\right),\quad Q=\left(-\frac{4}{3},-\frac{22}{3}\right)$$
Assertion is true.
Reason: Trisection points divide a line segment in ratios $1:2$ and $2:1$ (not $1:3$ and $3:1$). Reason is false.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The coordinates of the centroid of a triangle with vertices $(0,6)$, $(8,12)$ and $(8,0)$ are $\left(\dfrac{17}{3},5\right)$.
Reason (R): The centroid of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The coordinates of the centroid of a triangle with vertices $(0,6)$, $(8,12)$ and $(8,0)$ are $\left(\dfrac{17}{3},5\right)$.
Reason (R): The centroid of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$.
Answer
Option (D) is correct.
Assertion: Centroid $=\left(\dfrac{0+8+8}{3},\dfrac{6+12+0}{3}\right)=\left(\dfrac{16}{3},6\right)$, not $\left(\dfrac{17}{3},5\right)$. Assertion is false.
Reason: The centroid formula stated is correct. Reason is true.
Assertion: Centroid $=\left(\dfrac{0+8+8}{3},\dfrac{6+12+0}{3}\right)=\left(\dfrac{16}{3},6\right)$, not $\left(\dfrac{17}{3},5\right)$. Assertion is false.
Reason: The centroid formula stated is correct. Reason is true.
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The point $(-1,6)$ divides the line segment joining $(-3,10)$ and $(6,-8)$ in the ratio $2:7$ internally.
Reason (R): Given three points $A$, $B$, and $C$ form an equilateral triangle, then $AB=BC=AC$.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Assertion (A): The point $(-1,6)$ divides the line segment joining $(-3,10)$ and $(6,-8)$ in the ratio $2:7$ internally.
Reason (R): Given three points $A$, $B$, and $C$ form an equilateral triangle, then $AB=BC=AC$.
Answer
No complete solution was provided for this question in the source data. Students should verify the assertion using the section formula and confirm the reason independently.
Frequently Asked Questions
What is Coordinate Geometry in CBSE Class 10 Maths? ⌄
Coordinate Geometry (Chapter 7) teaches students how to locate points on a plane using an ordered pair $(x, y)$. The chapter covers the distance formula, section formula, midpoint formula, and the area of a triangle using coordinates — all fundamental skills that connect algebra and geometry.
How many marks does Coordinate Geometry carry in the Class 10 CBSE board exam? ⌄
Coordinate Geometry typically carries 6 marks in the Class 10 CBSE board examination. Questions appear in MCQ, short answer, and case study (competency-based) formats, so practising all question types is important for full marks.
What are the most important topics in Class 10 Coordinate Geometry for CBSE boards? ⌄
The four most tested topics are: (1) Distance Formula — finding the length between two points; (2) Section Formula — internal and external division of a line segment; (3) Midpoint Formula — a special case of section formula in ratio 1:1; and (4) Area of a Triangle — using the coordinate formula. Case study questions frequently combine two or more of these concepts.
What common mistakes should students avoid in Coordinate Geometry? ⌄
The most common errors include: mixing up $x$ and $y$ coordinates when applying formulas, forgetting the square root in the distance formula, confusing internal and external division in the section formula, and sign errors with negative coordinates. Regular practice with varied question types — especially case studies — helps students avoid these mistakes under exam pressure.
How does Angle Belearn help students prepare Coordinate Geometry for CBSE boards? ⌄
Angle Belearn provides CBSE-aligned competency based questions with detailed, step-by-step solutions for every question in Coordinate Geometry. The questions cover all question formats — MCQ, case study, and assertion-reason — exactly as they appear in board exams. Parents can use these resources to track their child’s understanding and target weak areas before the examination.
