CBSE Class 10 · Maths
CBSE Class 10 Maths Coordinate Geometry Previous Year Questions
Help your child master CBSE Class 10 Maths Coordinate Geometry Previous Year Questions with this curated collection sourced from real CBSE board papers spanning 2015–2025. Every question comes with a detailed step-by-step solution, building your child’s confidence with the distance formula, section formula, and mid-point theorem — high-weightage topics that consistently appear in the board exam.
CBSE Class 10 Maths Coordinate Geometry — Questions with Solutions
The distance of the point $(-1, 7)$ from the X-axis is:
Solution
Ans. Option (B) is correct.
Explanation:
The distance of a point from the X-axis is the absolute value of its $y$-coordinate. For the point $(-1, 7)$, the $y$-coordinate is $7$. Therefore, the distance from the X-axis is 7 units.
Explanation:
The distance of a point from the X-axis is the absolute value of its $y$-coordinate. For the point $(-1, 7)$, the $y$-coordinate is $7$. Therefore, the distance from the X-axis is 7 units.
The distance of the point $(-6, 8)$ from origin is:
Solution
Ans. Option (D) is correct.
Explanation: Distance between $(-6, 8)$ and $(0, 0)$ is: $$d = \sqrt{(-6 – 0)^2 + (8 – 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
Explanation: Distance between $(-6, 8)$ and $(0, 0)$ is: $$d = \sqrt{(-6 – 0)^2 + (8 – 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
The points $(-4, 0)$, $(4, 0)$ and $(0, 3)$ are the vertices of:
Solution
Ans. Option (B) is correct.
Explanation: Let $A(-4, 0)$, $B(4, 0)$ and $C(0, 3)$ be the vertices.
$$AB = \sqrt{(4-(-4))^2 + (0-0)^2} = \sqrt{64} = 8$$ $$BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{16+9} = \sqrt{25} = 5$$ $$CA = \sqrt{(-4-0)^2 + (0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$$ Since $BC = CA = 5$ but $AB = 8$, the triangle is isosceles.
Explanation: Let $A(-4, 0)$, $B(4, 0)$ and $C(0, 3)$ be the vertices.
$$AB = \sqrt{(4-(-4))^2 + (0-0)^2} = \sqrt{64} = 8$$ $$BC = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{16+9} = \sqrt{25} = 5$$ $$CA = \sqrt{(-4-0)^2 + (0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$$ Since $BC = CA = 5$ but $AB = 8$, the triangle is isosceles.
The point of intersection of the line represented by $3x – y = 3$ and $y$-axis is given by:
Solution
Ans. Option (A) is correct.
Explanation: At the $y$-axis, $x = 0$. Substituting in $3x – y = 3$: $$3(0) – y = 3 \Rightarrow y = -3$$ Hence, the line cuts the $y$-axis at $(0, -3)$.
Explanation: At the $y$-axis, $x = 0$. Substituting in $3x – y = 3$: $$3(0) – y = 3 \Rightarrow y = -3$$ Hence, the line 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 the $x$-axis?
Solution
Ans. Option (A) is correct.
Explanation:
When the line is divided by the $x$-axis, the intersection point is $(x, 0)$. Let the ratio be $k:1$.
Using the section formula for the $y$-coordinate: $$0 = \frac{k \times 6 + 1 \times (-3)}{k + 1} \Rightarrow 6k – 3 = 0 \Rightarrow k = \frac{1}{2}$$ Therefore, the ratio is $\dfrac{1}{2} : 1 = \mathbf{1 : 2}$.
Explanation:
When the line is divided by the $x$-axis, the intersection point is $(x, 0)$. Let the ratio be $k:1$.
Using the section formula for the $y$-coordinate: $$0 = \frac{k \times 6 + 1 \times (-3)}{k + 1} \Rightarrow 6k – 3 = 0 \Rightarrow k = \frac{1}{2}$$ Therefore, the ratio is $\dfrac{1}{2} : 1 = \mathbf{1 : 2}$.
A point $(x, y)$ is at a distance of 5 units from the origin. How many such points lie in the third quadrant?
Solution
Ans. Option (D) is correct.
Explanation: All points at distance 5 from the origin satisfy $x^2 + y^2 = 25$. This is a circle of radius 5 centred at the origin. The arc of this circle passing through the third quadrant (where $x < 0$ and $y < 0$) contains infinitely many points.
Explanation: All points at distance 5 from the origin satisfy $x^2 + y^2 = 25$. This is a circle of radius 5 centred at the origin. The arc of this circle passing through the third quadrant (where $x < 0$ and $y < 0$) contains infinitely many points.
The coordinates of the centre of the circle, $O$, and a point on the circle, $N$, are shown in the figure below. What is the radius of the circle?
Solution
Ans. Option (B) is correct.
Explanation: Given $O(-4, 3)$ and $N(-2.4, 1.8)$. The radius is $ON$: $$ON = \sqrt{(-4-(-2.4))^2 + (3-1.8)^2} = \sqrt{(-1.6)^2 + (1.2)^2} = \sqrt{2.56 + 1.44} = \sqrt{4} = \mathbf{2 \text{ units}}$$
Explanation: Given $O(-4, 3)$ and $N(-2.4, 1.8)$. The radius is $ON$: $$ON = \sqrt{(-4-(-2.4))^2 + (3-1.8)^2} = \sqrt{(-1.6)^2 + (1.2)^2} = \sqrt{2.56 + 1.44} = \sqrt{4} = \mathbf{2 \text{ units}}$$
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 the point B lies in the first quadrant, then all the possible $x$-coordinates of point B are:
Solution
Ans. Option (B) is correct.
Explanation: Given $AB = \sqrt{10^2 + 7^2} = \sqrt{149}$. Since $A(-4, 3)$ lies in the second quadrant and $B$ in the first quadrant, comparing the components: $x = 6$ and $y = 10$. The possible $x$-coordinates of $B$ are multiples of 3 (since $6 = 2 \times 3$).
Explanation: Given $AB = \sqrt{10^2 + 7^2} = \sqrt{149}$. Since $A(-4, 3)$ lies in the second quadrant and $B$ in the first quadrant, comparing the components: $x = 6$ and $y = 10$. The possible $x$-coordinates of $B$ are multiples of 3 (since $6 = 2 \times 3$).
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
Ans. Option (C) is correct.
Explanation: For an equilateral triangle, $AB = BC = CA$.
$AB = \sqrt{(3-0)^2 + (\sqrt{3}-0)^2} = \sqrt{9+3} = \sqrt{12}$
$BC = \sqrt{(3-0)^2+(k-0)^2} = \sqrt{9+k^2}$
Setting $AB = BC$: $\sqrt{12} = \sqrt{9+k^2} \Rightarrow k^2 = 3 \Rightarrow k = \pm\sqrt{3}$
Explanation: For an equilateral triangle, $AB = BC = CA$.
$AB = \sqrt{(3-0)^2 + (\sqrt{3}-0)^2} = \sqrt{9+3} = \sqrt{12}$
$BC = \sqrt{(3-0)^2+(k-0)^2} = \sqrt{9+k^2}$
Setting $AB = BC$: $\sqrt{12} = \sqrt{9+k^2} \Rightarrow k^2 = 3 \Rightarrow k = \pm\sqrt{3}$
The distance between the points $(m, -n)$ and $(-m, n)$ is:
Solution
Ans. Option (C) is correct.
Explanation: Let $A(m, -n)$ and $B(-m, n)$. Using the distance formula: $$AB = \sqrt{(m-(-m))^2 + (-n-n)^2} = \sqrt{(2m)^2 + (-2n)^2} = \sqrt{4m^2 + 4n^2} = 2\sqrt{m^2 + n^2}$$
Explanation: Let $A(m, -n)$ and $B(-m, n)$. Using the distance formula: $$AB = \sqrt{(m-(-m))^2 + (-n-n)^2} = \sqrt{(2m)^2 + (-2n)^2} = \sqrt{4m^2 + 4n^2} = 2\sqrt{m^2 + n^2}$$
The centre of a circle whose end points of a diameter are $(-6, 3)$ and $(6, 4)$ is:
Solution
Ans. Option (C) is correct.
Explanation: The centre is the midpoint of the diameter. With $A(-6, 3)$ and $B(6, 4)$: $$O = \left(\frac{-6+6}{2},\ \frac{3+4}{2}\right) = \left(0,\ \frac{7}{2}\right)$$
Explanation: The centre is the midpoint of the diameter. With $A(-6, 3)$ and $B(6, 4)$: $$O = \left(\frac{-6+6}{2},\ \frac{3+4}{2}\right) = \left(0,\ \frac{7}{2}\right)$$
$\triangle ABC$ 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
Ans. Option (B) is correct.
Explanation: Since $A$ lies on the $y$-axis, its coordinates are $(0, y)$. The section formula requires knowing the ratio in which $A$ divides $BC$, which is not directly given (the ratio $AB:BC = 1:2$ tells us side lengths, not the division of segment $BC$). However, since the $x$-coordinate of $A$ is $0$, we can use the distance formula with the known ratio $AB:BC = 1:2$ to find the $y$-coordinate of $A$.
Explanation: Since $A$ lies on the $y$-axis, its coordinates are $(0, y)$. The section formula requires knowing the ratio in which $A$ divides $BC$, which is not directly given (the ratio $AB:BC = 1:2$ tells us side lengths, not the division of segment $BC$). However, since the $x$-coordinate of $A$ is $0$, we can use the distance formula with the known ratio $AB:BC = 1:2$ to find the $y$-coordinate of $A$.
For a point $(3, 5)$, the value of (abscissa + ordinate) is:
Solution
Ans. Option (A) is correct.
Explanation: For the point $(3, 5)$, the abscissa ($x$-coordinate) is $3$ and the ordinate ($y$-coordinate) is $5$. $$\text{Abscissa} + \text{Ordinate} = 3 + 5 = \mathbf{8}$$
Explanation: For the point $(3, 5)$, the abscissa ($x$-coordinate) is $3$ and the ordinate ($y$-coordinate) is $5$. $$\text{Abscissa} + \text{Ordinate} = 3 + 5 = \mathbf{8}$$
The mid-point of a line segment divides the line segment in the ratio:
Solution
Ans. Option (C) is correct.
Explanation: The mid-point of a line segment divides it into two equal halves. Therefore, the ratio is $\mathbf{1 : 1}$.
Explanation: The mid-point of a line segment divides it into two equal halves. Therefore, the ratio is $\mathbf{1 : 1}$.
The mid-point of the line segment joining the points $P(4, 5)$ and $Q(4, 6)$ lies on:
Solution
Ans. Option (D) is correct.
Explanation: Mid-point of $P(4,5)$ and $Q(4,6)$: $$M = \left(\frac{4+4}{2},\ \frac{5+6}{2}\right) = \left(4,\ 5.5\right)$$ Since both coordinates are non-zero, the point $(4, 5.5)$ lies on neither the $x$-axis nor the $y$-axis.
Explanation: Mid-point of $P(4,5)$ and $Q(4,6)$: $$M = \left(\frac{4+4}{2},\ \frac{5+6}{2}\right) = \left(4,\ 5.5\right)$$ Since both coordinates are non-zero, the point $(4, 5.5)$ lies on neither the $x$-axis nor the $y$-axis.
In which quadrant lies the point which divides the line segment joining the points $(8, -9)$ and $(2, 3)$ in ratio $1 : 2$ internally?
Answer
Concept Applied: Section formula to obtain the point and then finding its quadrant.
Sol.
$m = 1,\ n = 2$; Given $(x_1, y_1) = (8, -9)$ and $(x_2, y_2) = (2, 3)$.
Using the section formula: $$\begin{aligned} (x, y) &= \left[\frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n}\right] \\ &= \left[\frac{1 \times 2 + 2 \times 8}{3},\ \frac{1 \times 3 + 2 \times (-9)}{3}\right] \\ &= \left[\frac{18}{3},\ \frac{-15}{3}\right] = (6, -5) \end{aligned}$$ Since $x > 0$ and $y < 0$, the point $(6, -5)$ lies in the IV quadrant.
Sol.
$m = 1,\ n = 2$; Given $(x_1, y_1) = (8, -9)$ and $(x_2, y_2) = (2, 3)$.
Using the section formula: $$\begin{aligned} (x, y) &= \left[\frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n}\right] \\ &= \left[\frac{1 \times 2 + 2 \times 8}{3},\ \frac{1 \times 3 + 2 \times (-9)}{3}\right] \\ &= \left[\frac{18}{3},\ \frac{-15}{3}\right] = (6, -5) \end{aligned}$$ Since $x > 0$ and $y < 0$, the point $(6, -5)$ lies in the IV quadrant.
The distance between point $A(5, -3)$ and $B(13, m)$ is 10 units. Calculate the value of $m$.
Answer
Sol. Given $A(5, -3)$, $B(13, m)$ and $AB = 10$ units.
Using the distance formula: $$\sqrt{(13-5)^2 + (m+3)^2} = 10$$ Squaring both sides: $$64 + (m+3)^2 = 100 \Rightarrow (m+3)^2 = 36 \Rightarrow m + 3 = \pm 6$$ Taking the positive value: $m = 6 – 3 = \mathbf{3}$.
Using the distance formula: $$\sqrt{(13-5)^2 + (m+3)^2} = 10$$ Squaring both sides: $$64 + (m+3)^2 = 100 \Rightarrow (m+3)^2 = 36 \Rightarrow m + 3 = \pm 6$$ Taking the positive value: $m = 6 – 3 = \mathbf{3}$.
A line intersects $y$-axis and $x$-axis at points P and Q, respectively. If $R(2, 5)$ is the mid-point of line segment PQ, then find the coordinates of P and Q.
Answer
Sol.
Since $P$ is on the $y$-axis, its coordinates are $P(0, y_1)$.
Since $Q$ is on the $x$-axis, its coordinates are $Q(x_2, 0)$.
Using the mid-point formula with $R(2, 5)$: $$2 = \frac{0 + x_2}{2} \Rightarrow x_2 = 4 \qquad \text{and} \qquad 5 = \frac{y_1 + 0}{2} \Rightarrow y_1 = 10$$ Therefore, the coordinates of $P$ are $\mathbf{(0, 10)}$ and of $Q$ are $\mathbf{(4, 0)}$.
Since $P$ is on the $y$-axis, its coordinates are $P(0, y_1)$.
Since $Q$ is on the $x$-axis, its coordinates are $Q(x_2, 0)$.
Using the mid-point formula with $R(2, 5)$: $$2 = \frac{0 + x_2}{2} \Rightarrow x_2 = 4 \qquad \text{and} \qquad 5 = \frac{y_1 + 0}{2} \Rightarrow y_1 = 10$$ Therefore, the coordinates of $P$ are $\mathbf{(0, 10)}$ and of $Q$ are $\mathbf{(4, 0)}$.
Find the ratio in which the $Y$-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$.
Answer
Sol.
Let the point $P$ on the $y$-axis be $P(0, y)$, dividing the segment in ratio $m : n$.
Using the section formula for the $x$-coordinate: $$\frac{m \cdot (-1) + n \cdot 5}{m + n} = 0 \Rightarrow -m + 5n = 0 \Rightarrow \frac{m}{n} = \frac{5}{1}$$ Therefore, the $Y$-axis divides the line segment in the ratio $\mathbf{5 : 1}$.
Let the point $P$ on the $y$-axis be $P(0, y)$, dividing the segment in ratio $m : n$.
Using the section formula for the $x$-coordinate: $$\frac{m \cdot (-1) + n \cdot 5}{m + n} = 0 \Rightarrow -m + 5n = 0 \Rightarrow \frac{m}{n} = \frac{5}{1}$$ Therefore, the $Y$-axis divides the line segment in the ratio $\mathbf{5 : 1}$.
$P(-2, 5)$ and $Q(3, 2)$ are two points. Find the coordinates of the point R on PQ such that $PR = 2QR$.
Answer
Sol. Given $P(-2, 5)$ and $Q(3, 2)$. Since $PR = 2QR$:
$$\frac{PR}{QR} = \frac{2}{1} \Rightarrow PR : QR = 2 : 1$$
So R divides PQ in the ratio $2:1$. Using the section formula:
$$R(x, y) = \left(\frac{2 \times 3 + 1 \times (-2)}{2+1},\ \frac{2 \times 2 + 1 \times 5}{2+1}\right) = \left(\frac{4}{3},\ \frac{9}{3}\right) = \left(\frac{4}{3},\ 3\right)$$
Therefore, the coordinates of $R$ are $\mathbf{\left(\dfrac{4}{3},\ 3\right)}$.
Find the ratio in which the line segment joining the points $A(6, 3)$ and $B(-2, -5)$ is divided by the $x$-axis.
Answer
Sol.
Let the ratio be $m : n$. Here $x_1 = 6,\ y_1 = 3,\ x_2 = -2,\ y_2 = -5$. Since the point lies on the $x$-axis, $y = 0$.
Using the section formula for $y$: $$\frac{my_2 + ny_1}{m+n} = 0 \Rightarrow \frac{-5m + 3n}{m+n} = 0$$ $$-5m + 3n = 0 \Rightarrow \frac{m}{n} = \frac{3}{5}$$ Therefore, the ratio is $\mathbf{m : n = 3 : 5}$.
Let the ratio be $m : n$. Here $x_1 = 6,\ y_1 = 3,\ x_2 = -2,\ y_2 = -5$. Since the point lies on the $x$-axis, $y = 0$.
Using the section formula for $y$: $$\frac{my_2 + ny_1}{m+n} = 0 \Rightarrow \frac{-5m + 3n}{m+n} = 0$$ $$-5m + 3n = 0 \Rightarrow \frac{m}{n} = \frac{3}{5}$$ Therefore, the ratio is $\mathbf{m : n = 3 : 5}$.
If the point $C(-1, 2)$ divides internally the line segment joining $A(2, 5)$ and $B(x, y)$ in the ratio $3 : 4$, find the co-ordinates of $B$.
Answer
Sol. Using the section formula with $m = 3$, $n = 4$, and the point $C(-1, 2)$:
For the $x$-coordinate: $$-1 = \frac{3x + 4 \times 2}{3 + 4} = \frac{3x + 8}{7} \Rightarrow 3x + 8 = -7 \Rightarrow 3x = -15 \Rightarrow x = -5$$ For the $y$-coordinate: $$2 = \frac{3y + 4 \times 5}{7} = \frac{3y + 20}{7} \Rightarrow 3y + 20 = 14 \Rightarrow 3y = -6 \Rightarrow y = -2$$ Therefore, the co-ordinates of $B$ are $\mathbf{(-5,\ -2)}$.
For the $x$-coordinate: $$-1 = \frac{3x + 4 \times 2}{3 + 4} = \frac{3x + 8}{7} \Rightarrow 3x + 8 = -7 \Rightarrow 3x = -15 \Rightarrow x = -5$$ For the $y$-coordinate: $$2 = \frac{3y + 4 \times 5}{7} = \frac{3y + 20}{7} \Rightarrow 3y + 20 = 14 \Rightarrow 3y = -6 \Rightarrow y = -2$$ Therefore, the co-ordinates of $B$ are $\mathbf{(-5,\ -2)}$.
If the mid-point of the line segment joining the points $A(3, 4)$ and $B(k, 6)$ is $P(x, y)$ and $x + y – 10 = 0$, find the value of $k$.
Answer
Sol. The mid-point $P$ of $A(3,4)$ and $B(k, 6)$ is:
$$x = \frac{3 + k}{2}, \qquad y = \frac{4 + 6}{2} = 5$$
Substituting into $x + y – 10 = 0$:
$$\frac{3+k}{2} + 5 – 10 = 0 \Rightarrow \frac{3+k}{2} = 5 \Rightarrow 3 + k = 10 \Rightarrow k = 7$$
Therefore, the value of $k$ is $\mathbf{7}$.
Find the ratio in which the $Y$-axis divides the line segment joining the points $(6, -4)$ and $(-2, -7)$. Also find the point of intersection.
Answer
Sol. Let the $y$-axis meet the line segment joining $A(6, -4)$ and $B(-2, -7)$ at point $P(0, y)$, dividing it in ratio $k : 1$.
Using the section formula for the $x$-coordinate: $$0 = \frac{k \times (-2) + 1 \times 6}{k + 1} \Rightarrow -2k + 6 = 0 \Rightarrow k = 3$$ The ratio is $\mathbf{3 : 1}$.
For the $y$-coordinate of the intersection point: $$y = \frac{3 \times (-7) + 1 \times (-4)}{3 + 1} = \frac{-21 – 4}{4} = \frac{-25}{4}$$ The point of intersection is $\mathbf{\left(0,\ -\dfrac{25}{4}\right)}$.
Using the section formula for the $x$-coordinate: $$0 = \frac{k \times (-2) + 1 \times 6}{k + 1} \Rightarrow -2k + 6 = 0 \Rightarrow k = 3$$ The ratio is $\mathbf{3 : 1}$.
For the $y$-coordinate of the intersection point: $$y = \frac{3 \times (-7) + 1 \times (-4)}{3 + 1} = \frac{-21 – 4}{4} = \frac{-25}{4}$$ The point of intersection is $\mathbf{\left(0,\ -\dfrac{25}{4}\right)}$.
The line segment joining the points $A(2, 1)$ and $B(5, -8)$ is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by $2x – y + k = 0$, find the value of $k$.
Answer
Sol.
Since P and Q trisect AB: $AP = PQ = QB$, so $P$ divides $AB$ in the ratio $1 : 2$.
Coordinates of $P$: $$P = \left(\frac{1 \times 5 + 2 \times 2}{1+2},\ \frac{1 \times (-8) + 2 \times 1}{1+2}\right) = \left(\frac{9}{3},\ \frac{-6}{3}\right) = (3,\ -2)$$ Since $P(3, -2)$ lies on $2x – y + k = 0$: $$2(3) – (-2) + k = 0 \Rightarrow 6 + 2 + k = 0 \Rightarrow k = -8$$ Therefore, the value of $k$ is $\mathbf{-8}$.
Since P and Q trisect AB: $AP = PQ = QB$, so $P$ divides $AB$ in the ratio $1 : 2$.
Coordinates of $P$: $$P = \left(\frac{1 \times 5 + 2 \times 2}{1+2},\ \frac{1 \times (-8) + 2 \times 1}{1+2}\right) = \left(\frac{9}{3},\ \frac{-6}{3}\right) = (3,\ -2)$$ Since $P(3, -2)$ lies on $2x – y + k = 0$: $$2(3) – (-2) + k = 0 \Rightarrow 6 + 2 + k = 0 \Rightarrow k = -8$$ Therefore, the value of $k$ is $\mathbf{-8}$.
In parallelogram ABCD, $A(3, 1)$, $B(5, 1)$, $C(a, b)$ and $D(4, 3)$ are the vertices. Find vertex $C(a, b)$.
Answer
Sol. In a parallelogram, the diagonals bisect each other — so the midpoint of diagonal $AC$ equals the midpoint of diagonal $BD$.
Midpoint of $BD$ with $B(5,1)$ and $D(4,3)$: $$\left(\frac{5+4}{2},\ \frac{1+3}{2}\right) = \left(\frac{9}{2},\ 2\right)$$ Setting this equal to the midpoint of $AC$ with $A(3,1)$ and $C(a,b)$: $$\frac{3+a}{2} = \frac{9}{2} \Rightarrow a = 6 \qquad \text{and} \qquad \frac{1+b}{2} = 2 \Rightarrow b = 3$$ Therefore, the coordinates of vertex $C$ are $\mathbf{(6,\ 3)}$.
Midpoint of $BD$ with $B(5,1)$ and $D(4,3)$: $$\left(\frac{5+4}{2},\ \frac{1+3}{2}\right) = \left(\frac{9}{2},\ 2\right)$$ Setting this equal to the midpoint of $AC$ with $A(3,1)$ and $C(a,b)$: $$\frac{3+a}{2} = \frac{9}{2} \Rightarrow a = 6 \qquad \text{and} \qquad \frac{1+b}{2} = 2 \Rightarrow b = 3$$ Therefore, the coordinates of vertex $C$ are $\mathbf{(6,\ 3)}$.
Find the ratio in which $P(4, m)$ divides the line segment joining the points $A(2, 3)$ and $B(6, -3)$. Hence find $m$.
Answer
Sol. Let the ratio be $k : 1$. Using the section formula for the $x$-coordinate:
$$4 = \frac{6k + 2}{k + 1} \Rightarrow 4k + 4 = 6k + 2 \Rightarrow 2k = 2 \Rightarrow k = 1$$
The ratio is $\mathbf{1 : 1}$ (P is the midpoint).
For $m$, using the $y$-coordinate: $$m = \frac{-3 \times 1 + 3 \times 1}{1 + 1} = \frac{0}{2} = 0$$ Therefore, the ratio is $1:1$ and $m = \mathbf{0}$, giving $P(4, 0)$.
For $m$, using the $y$-coordinate: $$m = \frac{-3 \times 1 + 3 \times 1}{1 + 1} = \frac{0}{2} = 0$$ Therefore, the ratio is $1:1$ and $m = \mathbf{0}$, giving $P(4, 0)$.
$A(5, 1)$, $B(1, 5)$ and $C(-3, -1)$ are the vertices of $\triangle ABC$. Find the length of median AD.
Answer
Sol.
$D$ is the mid-point of $BC$: $$D = \left(\frac{1+(-3)}{2},\ \frac{5+(-1)}{2}\right) = (-1,\ 2)$$ Length of median $AD$: $$AD = \sqrt{(5-(-1))^2 + (1-2)^2} = \sqrt{36 + 1} = \sqrt{37} \text{ units}$$
$D$ is the mid-point of $BC$: $$D = \left(\frac{1+(-3)}{2},\ \frac{5+(-1)}{2}\right) = (-1,\ 2)$$ Length of median $AD$: $$AD = \sqrt{(5-(-1))^2 + (1-2)^2} = \sqrt{36 + 1} = \sqrt{37} \text{ units}$$
The three vertices of a rhombus PQRS are $P(2, -3)$, $Q(6, 5)$ and $R(-2, 1)$.
(a) Find the coordinates of the point where both the diagonals PR and QS intersect.
(b) Find the coordinates of the fourth vertex S. Show your steps and give valid reasons.
(a) Find the coordinates of the point where both the diagonals PR and QS intersect.
(b) Find the coordinates of the fourth vertex S. Show your steps and give valid reasons.
Answer
Sol.
(a) The diagonals of a rhombus bisect each other, so the intersection $O$ is the midpoint of $PR$: $$O = \left(\frac{2+(-2)}{2},\ \frac{-3+1}{2}\right) = (0,\ -1)$$ The coordinates of the intersection point are $\mathbf{(0,\ -1)}$.
(b) Since $O$ is also the midpoint of $QS$, with $Q(6, 5)$ and $S(a, b)$: $$\frac{a+6}{2} = 0 \Rightarrow a = -6 \qquad \text{and} \qquad \frac{5+b}{2} = -1 \Rightarrow b = -7$$ Therefore, the coordinates of the fourth vertex $S$ are $\mathbf{(-6,\ -7)}$.
(a) The diagonals of a rhombus bisect each other, so the intersection $O$ is the midpoint of $PR$: $$O = \left(\frac{2+(-2)}{2},\ \frac{-3+1}{2}\right) = (0,\ -1)$$ The coordinates of the intersection point are $\mathbf{(0,\ -1)}$.
(b) Since $O$ is also the midpoint of $QS$, with $Q(6, 5)$ and $S(a, b)$: $$\frac{a+6}{2} = 0 \Rightarrow a = -6 \qquad \text{and} \qquad \frac{5+b}{2} = -1 \Rightarrow b = -7$$ Therefore, the coordinates of the fourth vertex $S$ are $\mathbf{(-6,\ -7)}$.
Raaji and Gagan are finding a treasure that is exactly on the straight line joining them. Raaji’s location is at $(-6, -5)$ and Gagan’s location is at $(10, 11)$. The distance from the treasure to Raaji’s location is three times that of the distance to Gagan’s location. Find the coordinates of the location of the treasure.
Answer
Sol. Let the coordinates of Treasure be $(x, y)$. Since the distance from Raaji to the treasure is 3 times the distance from Gagan, the ratio is $m : n = 3 : 1$.
With $(x_1, y_1) = (-6, -5)$ and $(x_2, y_2) = (10, 11)$: $$x = \frac{3 \times 10 + 1 \times (-6)}{3+1} = \frac{30-6}{4} = \frac{24}{4} = 6$$ $$y = \frac{3 \times 11 + 1 \times (-5)}{4} = \frac{33-5}{4} = \frac{28}{4} = 7$$ Therefore, the coordinates of the treasure are $\mathbf{(6,\ 7)}$.
With $(x_1, y_1) = (-6, -5)$ and $(x_2, y_2) = (10, 11)$: $$x = \frac{3 \times 10 + 1 \times (-6)}{3+1} = \frac{30-6}{4} = \frac{24}{4} = 6$$ $$y = \frac{3 \times 11 + 1 \times (-5)}{4} = \frac{33-5}{4} = \frac{28}{4} = 7$$ Therefore, the coordinates of the treasure are $\mathbf{(6,\ 7)}$.
$(1, -1)$, $(0, 4)$ and $(-5, 3)$ are vertices of a triangle. Check whether it is a scalene triangle, isosceles triangle or an equilateral triangle. Also, find the length of its median joining the vertex $(1, -1)$ to the mid-point of the opposite side.
Answer
Sol.
Let $A(1,-1)$, $B(0,4)$ and $C(-5,3)$.
$$AB = \sqrt{(1-0)^2 + (-1-4)^2} = \sqrt{1+25} = \sqrt{26} \text{ units}$$ $$BC = \sqrt{(0-(-5))^2 + (4-3)^2} = \sqrt{25+1} = \sqrt{26} \text{ units}$$ $$AC = \sqrt{(1-(-5))^2 + (-1-3)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13} \text{ units}$$ Since $AB = BC \neq AC$, the triangle is isosceles.
Finding the median from A to mid-point D of BC: $$D = \left(\frac{0+(-5)}{2},\ \frac{4+3}{2}\right) = \left(-\frac{5}{2},\ \frac{7}{2}\right)$$ $$AD = \sqrt{\left(-\frac{5}{2}-1\right)^2 + \left(\frac{7}{2}+1\right)^2} = \sqrt{\left(-\frac{7}{2}\right)^2 + \left(\frac{9}{2}\right)^2} = \sqrt{\frac{49+81}{4}} = \frac{\sqrt{130}}{2} \text{ units}$$
Let $A(1,-1)$, $B(0,4)$ and $C(-5,3)$.
$$AB = \sqrt{(1-0)^2 + (-1-4)^2} = \sqrt{1+25} = \sqrt{26} \text{ units}$$ $$BC = \sqrt{(0-(-5))^2 + (4-3)^2} = \sqrt{25+1} = \sqrt{26} \text{ units}$$ $$AC = \sqrt{(1-(-5))^2 + (-1-3)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13} \text{ units}$$ Since $AB = BC \neq AC$, the triangle is isosceles.
Finding the median from A to mid-point D of BC: $$D = \left(\frac{0+(-5)}{2},\ \frac{4+3}{2}\right) = \left(-\frac{5}{2},\ \frac{7}{2}\right)$$ $$AD = \sqrt{\left(-\frac{5}{2}-1\right)^2 + \left(\frac{7}{2}+1\right)^2} = \sqrt{\left(-\frac{7}{2}\right)^2 + \left(\frac{9}{2}\right)^2} = \sqrt{\frac{49+81}{4}} = \frac{\sqrt{130}}{2} \text{ units}$$
In a society, there is a circular park having two gates placed at points $A(10, 20)$ and $B(50, 50)$. Two fountains are installed at points P and Q on AB such that $AP = PQ = QB$.
(i) Find the coordinates of the center $C$.
(ii) Find the radius of the circular park.
(i) Find the coordinates of the center $C$.
(ii) Find the radius of the circular park.
Answer
(i) Coordinates of the Centre C:
Since $AP = PQ = QB$, points P and Q trisect AB. The center $C$ is the midpoint of the diameter $AB$: $$C = \left(\frac{10+50}{2},\ \frac{20+50}{2}\right) = \left(30,\ 35\right)$$ The coordinates of the center $C$ are $\mathbf{(30,\ 35)}$.
(ii) Radius of the Circular Park:
The radius is the distance from $C(30,35)$ to gate $A(10,20)$: $$r = \sqrt{(30-10)^2 + (35-20)^2} = \sqrt{400+225} = \sqrt{625} = \mathbf{25 \text{ units}}$$
Since $AP = PQ = QB$, points P and Q trisect AB. The center $C$ is the midpoint of the diameter $AB$: $$C = \left(\frac{10+50}{2},\ \frac{20+50}{2}\right) = \left(30,\ 35\right)$$ The coordinates of the center $C$ are $\mathbf{(30,\ 35)}$.
(ii) Radius of the Circular Park:
The radius is the distance from $C(30,35)$ to gate $A(10,20)$: $$r = \sqrt{(30-10)^2 + (35-20)^2} = \sqrt{400+225} = \sqrt{625} = \mathbf{25 \text{ units}}$$
The base BC of an equilateral triangle ABC lies on the $y$-axis. The coordinates of point C are $(0, -3)$. The origin is the mid-point of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that BACD is a rhombus.
Answer
Step 1 — Coordinates of B:
Since the origin is the midpoint of $BC$ and $C = (0,-3)$: $$\frac{y_B + (-3)}{2} = 0 \Rightarrow y_B = 3, \quad x_B = 0$$ Therefore, $B = (0,\ 3)$.
Step 2 — Coordinates of A:
In an equilateral triangle with base $BC$ on the $y$-axis, vertex $A$ lies on the perpendicular bisector of $BC$ (the $x$-axis). The side length $BC = 6$. The height of an equilateral triangle with side 6: $$h = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}$$ Therefore, $A = (3\sqrt{3},\ 0)$ or $A = (-3\sqrt{3},\ 0)$. Taking $A = (3\sqrt{3},\ 0)$.
Step 3 — Coordinates of D for rhombus BACD:
In a rhombus, diagonals bisect each other. The diagonal $AC$ has midpoint: $$\left(\frac{3\sqrt{3}+0}{2},\ \frac{0+(-3)}{2}\right)$$ $D$ is the reflection of $A$ through the midpoint of $AC$, i.e., $D = (-3\sqrt{3},\ 0)$.
Final Answers: $A(3\sqrt{3}, 0)$, $B(0, 3)$, $D(-3\sqrt{3}, 0)$.
Since the origin is the midpoint of $BC$ and $C = (0,-3)$: $$\frac{y_B + (-3)}{2} = 0 \Rightarrow y_B = 3, \quad x_B = 0$$ Therefore, $B = (0,\ 3)$.
Step 2 — Coordinates of A:
In an equilateral triangle with base $BC$ on the $y$-axis, vertex $A$ lies on the perpendicular bisector of $BC$ (the $x$-axis). The side length $BC = 6$. The height of an equilateral triangle with side 6: $$h = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}$$ Therefore, $A = (3\sqrt{3},\ 0)$ or $A = (-3\sqrt{3},\ 0)$. Taking $A = (3\sqrt{3},\ 0)$.
Step 3 — Coordinates of D for rhombus BACD:
In a rhombus, diagonals bisect each other. The diagonal $AC$ has midpoint: $$\left(\frac{3\sqrt{3}+0}{2},\ \frac{0+(-3)}{2}\right)$$ $D$ is the reflection of $A$ through the midpoint of $AC$, i.e., $D = (-3\sqrt{3},\ 0)$.
Final Answers: $A(3\sqrt{3}, 0)$, $B(0, 3)$, $D(-3\sqrt{3}, 0)$.
The base QR of an equilateral triangle PQR lies on the $x$-axis. The coordinates of point Q are $(-4, 0)$, and the origin is the mid-point of the base. Find the coordinates of points P and R.
Answer
Step 1 — Coordinates of R:
Since the origin is the midpoint of $QR$ and $Q = (-4, 0)$: $$\frac{-4 + x_R}{2} = 0 \Rightarrow x_R = 4, \quad y_R = 0$$ Therefore, $R = \mathbf{(4,\ 0)}$.
Step 2 — Coordinates of P:
Side length $QR = |4-(-4)| = 8$. The height of the equilateral triangle: $$h = \frac{\sqrt{3}}{2} \times 8 = 4\sqrt{3}$$ Since the base lies on the $x$-axis, vertex $P$ lies directly above (or below) the origin at height $4\sqrt{3}$.
Therefore, $P = \mathbf{(0,\ 4\sqrt{3})}$ (or $(0, -4\sqrt{3})$).
Since the origin is the midpoint of $QR$ and $Q = (-4, 0)$: $$\frac{-4 + x_R}{2} = 0 \Rightarrow x_R = 4, \quad y_R = 0$$ Therefore, $R = \mathbf{(4,\ 0)}$.
Step 2 — Coordinates of P:
Side length $QR = |4-(-4)| = 8$. The height of the equilateral triangle: $$h = \frac{\sqrt{3}}{2} \times 8 = 4\sqrt{3}$$ Since the base lies on the $x$-axis, vertex $P$ lies directly above (or below) the origin at height $4\sqrt{3}$.
Therefore, $P = \mathbf{(0,\ 4\sqrt{3})}$ (or $(0, -4\sqrt{3})$).
The mid-point $P$ of the line segment joining the points $A(-10, 4)$ and $B(-2, 0)$ lies on the line segment joining the points $C(-9, -4)$ and $D(-4, y)$. Find the ratio in which $P$ divides $CD$. Also, find the value of $y$.
Answer
Step 1 — Find coordinates of P: $$P = \left(\frac{-10+(-2)}{2},\ \frac{4+0}{2}\right) = (-6,\ 2)$$ Step 2 — Find the ratio in which P divides CD:
Let P divide $C(-9,-4)$ and $D(-4,y)$ in ratio $m:n$. Using the $x$-coordinate: $$\frac{m(-4) + n(-9)}{m+n} = -6 \Rightarrow -4m – 9n = -6m – 6n \Rightarrow 2m = 3n \Rightarrow \frac{m}{n} = \frac{3}{2}$$ The ratio is $\mathbf{3 : 2}$.
Step 3 — Find value of y: $$\frac{3y + 2(-4)}{5} = 2 \Rightarrow 3y – 8 = 10 \Rightarrow 3y = 18 \Rightarrow y = \mathbf{6}$$
Let P divide $C(-9,-4)$ and $D(-4,y)$ in ratio $m:n$. Using the $x$-coordinate: $$\frac{m(-4) + n(-9)}{m+n} = -6 \Rightarrow -4m – 9n = -6m – 6n \Rightarrow 2m = 3n \Rightarrow \frac{m}{n} = \frac{3}{2}$$ The ratio is $\mathbf{3 : 2}$.
Step 3 — Find value of y: $$\frac{3y + 2(-4)}{5} = 2 \Rightarrow 3y – 8 = 10 \Rightarrow 3y = 18 \Rightarrow y = \mathbf{6}$$
Case Study — Read the following and answer the questions:
The diagram shows the plans for a sun room built onto the wall of a house. The four walls are square clear glass panels. The roof is made using four clear glass panels (trapezium shaped) and one tinted glass panel (half a regular octagon in shape).
(i) Refer to top view: find the mid-point of the segment joining $J(6, 17)$ 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 segment AB in 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) $x – y = 13$
The diagram shows the plans for a sun room built onto the wall of a house. The four walls are square clear glass panels. The roof is made using four clear glass panels (trapezium shaped) and one tinted glass panel (half a regular octagon in shape).
(i) Refer to top view: find the mid-point of the segment joining $J(6, 17)$ 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 segment AB in 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) Ans. Option (C) is correct.
Mid-point of $J(6,17)$ and $I(9,16)$: $$x = \frac{6+9}{2} = \frac{15}{2}, \qquad y = \frac{17+16}{2} = \frac{33}{2}$$ Mid-point $= \left(\dfrac{15}{2},\ \dfrac{33}{2}\right)$.
(ii) Ans. Option (A) is correct.
The distance of point P from the $Y$-axis $= \mathbf{4}$ (the $x$-coordinate of P).
(iii) Ans. Option (C) is correct.
$A = (1, 8)$ and $S = (15, 8)$. $$AS = \sqrt{(15-1)^2 + (8-8)^2} = \sqrt{196} = 14$$
(iv) Ans. Option (D) is correct.
$A = (1, 8)$, $B = (4, 10)$, ratio $1:3$: $$\left(\frac{1 \times 4 + 3 \times 1}{4},\ \frac{1 \times 10 + 3 \times 8}{4}\right) = \left(\frac{7}{4},\ \frac{34}{4}\right) = (1.75,\ 8.5)$$
(v) Ans. Option (B) is correct.
Equidistant from $Q(9,8)$ and $S(17,8)$: $PQ^2 = PS^2$ $$(x-9)^2 + (y-8)^2 = (x-17)^2 + (y-8)^2 \Rightarrow x^2 – 18x + 81 = x^2 – 34x + 289$$ $$16x = 208 \Rightarrow x = 13 \Rightarrow x – 13 = 0$$
Mid-point of $J(6,17)$ and $I(9,16)$: $$x = \frac{6+9}{2} = \frac{15}{2}, \qquad y = \frac{17+16}{2} = \frac{33}{2}$$ Mid-point $= \left(\dfrac{15}{2},\ \dfrac{33}{2}\right)$.
(ii) Ans. Option (A) is correct.
The distance of point P from the $Y$-axis $= \mathbf{4}$ (the $x$-coordinate of P).
(iii) Ans. Option (C) is correct.
$A = (1, 8)$ and $S = (15, 8)$. $$AS = \sqrt{(15-1)^2 + (8-8)^2} = \sqrt{196} = 14$$
(iv) Ans. Option (D) is correct.
$A = (1, 8)$, $B = (4, 10)$, ratio $1:3$: $$\left(\frac{1 \times 4 + 3 \times 1}{4},\ \frac{1 \times 10 + 3 \times 8}{4}\right) = \left(\frac{7}{4},\ \frac{34}{4}\right) = (1.75,\ 8.5)$$
(v) Ans. Option (B) is correct.
Equidistant from $Q(9,8)$ and $S(17,8)$: $PQ^2 = PS^2$ $$(x-9)^2 + (y-8)^2 = (x-17)^2 + (y-8)^2 \Rightarrow x^2 – 18x + 81 = x^2 – 34x + 289$$ $$16x = 208 \Rightarrow x = 13 \Rightarrow x – 13 = 0$$
Case Study — Read the following and answer the questions:
Jagdhish has a field in the shape of a right-angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat, and the remaining for growing vegetables. There is a pole marked as O in the field.
(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 in square PQRS?
(iii) If S divides CA in the ratio $K:1$, what is the value of $K$, where point $A$ is $(200, 800)$?
Jagdhish has a field in the shape of a right-angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat, and the remaining for growing vegetables. There is a pole marked as O in the field.
(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 in square PQRS?
(iii) If S divides CA in the ratio $K:1$, what is the value of $K$, where point $A$ is $(200, 800)$?
Answer
(i) Side of square $PQRS = PQ = 400$ units. Since the square lies above the $x$-axis:
$$R = (200,\ 400) \qquad \text{and} \qquad S = (-200,\ 400)$$
(ii) Area of square PQRS: $$\text{Area} = (\text{side})^2 = (400)^2 = 160000 \text{ unit}^2$$ OR Length of diagonal PR: $$PR = \sqrt{2} \times \text{side} = 400\sqrt{2} \text{ units}$$
(iii) Finding K:
$S = (-200, 400)$, $C$ is taken as $(-600, 0)$ and $A = (200, 800)$. Using the section formula for the $x$-coordinate with ratio $K:1$: $$-200 = \frac{200K + 1 \times (-600)}{K + 1} \Rightarrow -200K – 200 = 200K – 600 \Rightarrow 400 = 400K \Rightarrow K = 1$$ Therefore, $K = \mathbf{1}$ (S is the mid-point of CA).
$$R = (200,\ 400) \qquad \text{and} \qquad S = (-200,\ 400)$$
(ii) Area of square PQRS: $$\text{Area} = (\text{side})^2 = (400)^2 = 160000 \text{ unit}^2$$ OR Length of diagonal PR: $$PR = \sqrt{2} \times \text{side} = 400\sqrt{2} \text{ units}$$
(iii) Finding K:
$S = (-200, 400)$, $C$ is taken as $(-600, 0)$ and $A = (200, 800)$. Using the section formula for the $x$-coordinate with ratio $K:1$: $$-200 = \frac{200K + 1 \times (-600)}{K + 1} \Rightarrow -200K – 200 = 200K – 600 \Rightarrow 400 = 400K \Rightarrow K = 1$$ Therefore, $K = \mathbf{1}$ (S is the mid-point of CA).
Frequently Asked Questions
What does the Coordinate Geometry chapter cover in CBSE Class 10 Maths? ⌄
The Coordinate Geometry chapter in CBSE Class 10 covers three main concepts: the distance formula (to find the length between two points), the section formula (to find a point dividing a segment in a given ratio), and the mid-point formula. Students also apply these to verify properties of geometric figures such as triangles, parallelograms, and rhombuses — all of which appear regularly in previous year papers.
How many marks does Coordinate Geometry carry in the CBSE Class 10 board exam? ⌄
Coordinate Geometry is part of the Geometry unit in CBSE Class 10, which typically carries around 15 marks in the board exam. Questions from this chapter appear across multiple formats — 1-mark MCQs, 2-mark short answers, 3-mark problems involving section or distance formula, and 5-mark case studies — making it one of the most mark-rich chapters for your child to practise.
What are the most important topics in CBSE Class 10 Coordinate Geometry? ⌄
The highest-priority topics are: applying the distance formula to find lengths and classify triangles, using the section formula internally to divide a segment in a given ratio, finding mid-points, and locating where the axes divide a segment. Questions where the student must find the coordinates of a missing vertex (of a parallelogram, rhombus, or equilateral triangle) using these formulas are especially frequent in CBSE board papers.
What common mistakes do students make in Coordinate Geometry questions? ⌄
A very common error is mixing up the order of coordinates when applying the section formula — substituting $y$-values in the $x$-formula and vice versa. Students also frequently forget to set the correct coordinate to zero when a point lies on the $x$-axis ($y = 0$) or $y$-axis ($x = 0$). Practising a variety of previous year questions with step-by-step solutions helps your child avoid these mistakes under exam pressure.
How does Angle Belearn help students score well in Coordinate Geometry? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks directly from real board papers, each paired with clear, step-by-step solutions that mirror the structured working expected in board exams. Regular practice with these verified questions builds both speed and accuracy, so your child walks into the exam confident in applying the distance formula, section formula, and mid-point theorem correctly — every time.
