SEBA Class 10 Maths Exercise 7.2 Coordinate Geometry | New Book 2026

In this article, you will get simple and reliable solutions to SEBA Class 10 Maths Exercise 7.2 Coordinate Geometry. These solutions are from the New Book 2026 of SCERT, which is mandatory for SEBA Class 10 students.

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SEBA Class 10 Maths Exercise 7.1 Coordinate Geometry

1. Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.

Solution:

Let $P(x, y)$ be the required point that divides the line segment joining $A(-1, 7)$ and $B(4, -3)$ internally in the ratio $= 2 : 3$.

Here, $x_{1} = -1$, $y_{1} = 7$, $x_{2} = 4$, $y_{2} = -3$, $m_{1} = 2$, and $m_{2} = 3$.

Using the section formula, we have

$(x, y) = \left(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, \frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\right)$

$= \left(\frac{2 \times 4 + 3 \times (-1)}{2 + 3}, \frac{2 \times (-3) + 3 \times 7}{2 + 3}\right)$

$= \left(\frac{8 – 3}{5}, \frac{-6 + 21}{5}\right)$

$= \left(\frac{5}{5}, \frac{15}{5}\right)$

$= (1, 3)$

Therefore, the coordinates of the required point are $(1, 3)$.

2. Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$.

Solution:

Let $P$ and $Q$ be the points of trisection of the line segment joining $A(4, -1)$ and $B(-2, -3)$, so that $AP = PQ = QB$.

$P$ divides $AB$ internally in the ratio $1 : 2$.

Applying the section formula for $P$, we get

$(x, y) = \left[\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, \frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\right]$

$= \left[\frac{1 \times (-2) + 2 \times 4}{1 + 2}, \frac{1 \times (-3) + 2(-1)}{1 + 2}\right]$

$= \left[\frac{-2 + 8}{3}, \frac{-3 – 2}{3}\right]$

$= \left[\frac{6}{3}, \frac{-5}{3}\right]$

$= \left(2, -\frac{5}{3}\right)$

So, the coordinates of $P$ are $(2, -\frac{5}{3})$.

Now, $Q$ is the mid-point of $PB$.

Using the mid-point formula for $Q$, we have

$Q(x, y) = \left[\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right]$

$= \left[\frac{2 + (-2)}{2}, \frac{-\frac{5}{3} + (-3)}{2}\right]$

$= \left[\frac{2 – 2}{2}, \frac{\frac{-5 – 9}{3}}{2}\right]$

$= \left[\frac{0}{2}, \frac{-\frac{14}{3}}{2}\right]$

$= \left[0, -\frac{14}{6}\right]$

$= \left(0, -\frac{7}{3}\right)$

So, the coordinates of $Q$ are $(0, -\frac{7}{3})$.

Therefore, the coordinates of the points of trisection are $(2, -\frac{5}{3})$ and $(0, -\frac{7}{3})$.

3. To conduct Sports Day activities, in your rectangular shaped school ground $ABCD$, lines have been drawn with chalk powder at a distance of $1\text{ m}$ each. $100$ flower pots have been placed at a distance of $1\text{ m}$ from each other along $AD$, as shown in Fig. 7.14. Niharika runs $\frac{1}{4}\text{th}$ the distance $AD$ on the 2nd line and posts a green flag. Padma runs $\frac{1}{5}\text{th}$ the distance $AD$ on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

Total distance $AD = 100 \times 1\text{ m} = 100\text{ m}$.

Taking $A$ as origin $(0, 0)$, $AB$ along the x-axis and $AD$ along the y-axis:

The position of green flag posted by Niharika on the 2nd line is given by $G(2, \frac{1}{4} \times 100 )$

$ = G (2, 25)$

And the position of red flag posted by Padma on the 8th line is $R(8, \frac{1}{5} \times 100)$

$ =R (8, 20)$.

Distance between the two flags using distance formula:

$GR = \sqrt{(8 – 2)^{2} + (20 – 25)^{2}} $

$= \sqrt{6^{2} + (-5)^{2}} $

$= \sqrt{36 + 25} $

$= \sqrt{61}\text{ m}$

Let Rashmi post the blue flag at the mid-point of $GR$ at $B(x, y)$.

So using the mid-point formula, we get

$(x, y) = \left[\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right]$

$= \left[\frac{2 + 8}{2}, \frac{25 + 20}{2}\right]$

$= \left[\frac{10}{2}, \frac{45}{2}\right]$

$= (5, 22.5)$

Therefore, the distance between both flags is $\sqrt{61}\text{ m}$,

and Rashmi should post her blue flag on the 5th line at a distance of $22.5\text{ m}$ along $AD$.

4. Find the ratio in which the line segment joining the points $(-3, 10)$ and $(6, -8)$ is divided by $(-1, 6)$.

Solution:

Let the point $P(-1, 6)$ divide the line segment joining $A(-3, 10)$ and $B(6, -8)$ internally in the ratio $k : 1$.

We have the section formula,

$(x, y) = \left(\frac{k x_2 + x_1}{k + 1}, \frac{k y_2 + y_1}{k + 1}\right)$

Using the section formula, we get

$(-1, 6) = \left(\frac{k \times 6 + 1 \times (-3)}{k + 1}, \frac{k(-8) + 1 \times 10}{k + 1}\right)$

Equating the x-coordinate:

$-1 = \frac{6k – 3}{k + 1}$

$\implies -k – 1 = 6k – 3$

$\implies -k – 6k = -3 + 1$

$\implies -7k = -2$

$\implies k = \frac{-2}{-7}$

$\implies k = \frac{2}{7}$

Now the ratio $k : 1$ becomes,

$= \frac{2}{7} : 1$

$= 2 : 7$

Therefore, the required ratio is $2 : 7$.

5. Find the ratio in which the line segment joining $A(1, -5)$ and $B(-4, 5)$ is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Let the point $P(x, 0)$ on the x-axis divide the line segment joining $A(1, -5)$ and $B(-4, 5)$ internally in the ratio $k : 1$.

We have the section formula,

$(x, y) = \left(\frac{k x_2 + x_1}{k + 1}, \frac{k y_2 + y_1}{k + 1}\right)$

Using the section formula, we get 

$(x, 0) = \left(\frac{k \times (-4) + 1 \times 1}{k + 1}, \frac{k \times 5 + 1 \times (-5)}{k + 1}\right)$

$(x, 0) = \left(\frac{-4k + 1}{k + 1}, \frac{5k – 5}{k + 1}\right)$

Equating the y-coordinate:

$0 = \frac{5k – 5}{k + 1}$

$\implies 5k – 5 = 0$

$\implies 5k = 5$

$\implies k = \frac{5}{5}$

$\implies k = 1$

Now the ratio $k : 1$ becomes,

$= 1 : 1$

Now, equating the x-coordinate to find the point of division:

$x = \frac{-4k + 1}{k + 1}$

$\implies x = \frac{-4(1) + 1}{1 + 1}$

$\implies x = \frac{-4 + 1}{2}$

$\implies x = -\frac{3}{2}$

Therefore, the required ratio is $1 : 1$ and the coordinates of the point of division are $\left(-\frac{3}{2}, 0\right)$.

6. If $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Solution:

Let $A(1, 2)$, $B(4, y)$, $C(x, 6)$, and $D(3, 5)$ be the vertices of the parallelogram $ABCD$.

We know that the diagonals of a parallelogram bisect each other.

So, coordinates of the mid-point of diagonal $AC =$ coordinates of mid-point of $BD$.

Using the mid-point formula, we get

$\left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right)$

$\left(\frac{1 + x}{2}, \frac{8}{2}\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$

$\left(\frac{1 + x}{2}, 4\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$

Equating the x-coordinate:

$\frac{1 + x}{2} = \frac{7}{2}$

$\implies 1 + x = 7$

$\implies x = 7 – 1$

$\implies x = 6$

Equating the y-coordinate:

$4 = \frac{y + 5}{2}$

$\implies y + 5 = 8$

$\implies y = 8 – 5$

$\implies y = 3$

Therefore, $x = 6$ and $y = 3$.

7. Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose centre is $(2, -3)$ and $B$ is $(1, 4)$.

Solution:

Let the coordinates of point $A$ be $(x, y)$.

The centre $C(2, -3)$ is the mid-point of the diameter $AB$, where $B = (1, 4)$.

Using the mid-point formula, we get

$(2, -3) = \left(\frac{x + 1}{2}, \frac{y + 4}{2}\right)$

Equating the x-coordinate:

$2 = \frac{x + 1}{2}$

$\implies x + 1 = 4$

$\implies x = 4 – 1$

$\implies x = 3$

Equating the y-coordinate:

$-3 = \frac{y + 4}{2}$

$\implies y + 4 = -6$

$\implies y = -6 – 4$

$\implies y = -10$

Therefore, the coordinates of point $A$ are $(3, -10)$.

8. If $A$ and $B$ are $(-2, -2)$ and $(2, -4)$, respectively, find the coordinates of $P$ such that $AP = \frac{3}{7}AB$ and $P$ lies on the line segment $AB$.

Solution:

Given,

$AP = \frac{3}{7}AB$

$\implies 7AP = 3AB$

$\implies 7AP = 3(AP + PB)$

$\implies 7AP – 3AP = 3PB$

$\implies 4AP = 3PB$

$\implies \frac{AP}{PB} = \frac{3}{4}$

$\therefore AP : PB = 3 : 4$

Let the coordinates of point $P$ be $P(x, y)$

So, point $P$ divides the line segment joining $A(-2, -2)$ and $B(2, -4)$ internally in the ratio $ 3 : 4$

We know the section formula

$(x, y) = \left(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, \frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\right)$

Using the section formula, we have

$(x, y) = \left(\frac{3 \times 2 + 4 \times (-2)}{3 + 4}, \frac{3 \times (-4) + 4 \times (-2)}{3 + 4}\right)$

$ = \left(\frac{6 – 8}{7}, \frac{-12 – 8}{7}\right)$

$ = \left(-\frac{2}{7}, -\frac{20}{7}\right)$

Therefore, the coordinates of $P$ are $\left(-\frac{2}{7}, -\frac{20}{7}\right)$.

9. Find the coordinates of the points which divide the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts.

Solution:

Let $P$, $Q$, and $R$ be the three points dividing the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts such that $AP = PQ = QR = RB$.

$Q$ is the mid-point of $AB$:

Using the mid-point formula, we get

Coordinates of $Q = \left(\frac{-2 + 2}{2}, \frac{2 + 8}{2}\right)$

$ = \left(\frac{0}{2}, \frac{10}{2}\right)$

$ = (0, 5)$

$P$ is the mid-point of $AQ$:

Here, $A = (-2, 2)$ and $Q = (0, 5)$.

So, coordinates of $P = \left(\frac{-2 + 0}{2}, \frac{2 + 5}{2}\right)$

$ = \left(\frac{-2}{2}, \frac{7}{2}\right)$

$ = \left(-1, \frac{7}{2}\right)$

$R$ is the mid-point of $QB$:

Here, $Q = (0, 5)$ and $B = (2, 8)$.

So, coordinates of $R = \left(\frac{0 + 2}{2}, \frac{5 + 8}{2}\right)$

$ = \left(\frac{2}{2}, \frac{13}{2}\right)$

$ = \left(1, \frac{13}{2}\right)$

Therefore, the coordinates of the points are $(-1, \frac{7}{2})$, $(0, 5)$, and $(1, \frac{13}{2})$.

10. Find the area of a rhombus if its vertices are $(3, 0)$, $(4, 5)$, $(-1, 4)$ and $(-2, -1)$ taken in order.

Solution:

Let $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$, and $D(-2, -1)$ be the vertices of the rhombus taken in order.

Using the distance formula, we have

Diagonal$AC = \sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}$

$ = \sqrt{(-1 – 3)^{2} + (4 – 0)^{2}}$

$= \sqrt{(-4)^{2} + (4)^{2}}$

$= \sqrt{16 + 16}$

$= \sqrt{32}$

$= 4\sqrt{2}\text{ units}$

Diagonal $BD = \sqrt{(-2 – 4)^{2} + (-1 – 5)^{2}}$

$= \sqrt{(-6)^{2} + (-6)^{2}}$

$= \sqrt{36 + 36}$

$= \sqrt{72}$

$= 6\sqrt{2}\text{ units}$

Now, $\text{Area of rhombus} = \frac{1}{2} \times (\text{Product of diagonals})$

$\implies \text{Area} = \frac{1}{2} \times AC \times BD$

$ = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2}$

$ = \frac{1}{2} \times 24 \times 2$

$ = 24\text{ sq. units}$

Therefore, the area of the rhombus is $24\text{ sq. units}$.

11. If the line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at a point $P$ in the ratio $1 : 2$ and it lies on the line $3x – 18y + k = 0$, then the value of $k$ is:

(a) 12

(b) 15

(c) 19

(d) 22

Solution:

Let $P(x, y)$ divide $AB$ internally in the ratio $1 : 2$.

Using the section formula, we have

$(x, y) = \left(\frac{1 \times 5 + 2 \times 3}{1 + 2}, \frac{1 \times 1 + 2 \times 2}{1 + 2}\right)$

$= \left(\frac{5 + 6}{3}, \frac{1 + 4}{3}\right)$

$= \left(\frac{11}{3}, \frac{5}{3}\right)$

Since point $P\left(\frac{11}{3}, \frac{5}{3}\right)$ lies on the line $3x – 18y + k = 0$, it satisfies the equation:

$3\left(\frac{11}{3}\right) – 18\left(\frac{5}{3}\right) + k = 0$

$\implies 11 – 6(5) + k = 0$

$\implies 11 – 30 + k = 0$

$\implies -19 + k = 0$

$\implies k = 19$

Therefore, the correct answer is (c) 19.

12. Point $P$ divides the line segment joining the points $A(2, 1)$ and $B(5, -8)$, such that $\frac{AP}{AB} = \frac{1}{3}$. If $P$ lies on the line $2x – y + k = 0$, find the value of $k$.

Solution:

Given,

$\frac{AP}{AB} = \frac{1}{3}$

$\implies \frac{AP}{AP + PB} = \frac{1}{3}$

$\implies 3AP = AP + PB$

$\implies 2AP = PB$

$\implies \frac{AP}{PB} = \frac{1}{2}$

$\therefore AP : PB = 1 : 2$

So, point $P$ divides the line segment joining $A(2, 1)$ and $B(5, -8)$ internally in the ratio $m_{1} : m_{2} = 1 : 2$.

Using the section formula, we have

$(x, y) = \left(\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}, \frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\right)$

$ = \left(\frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-8) + 2 \times 1}{1 + 2}\right)$

$ = \left(\frac{5 + 4}{3}, \frac{-8 + 2}{3}\right)$

$ = \left(\frac{9}{3}, \frac{-6}{3}\right)$

$= (3, -2)$

Since point $P(3, -2)$ lies on the line $2x – y + k = 0$, it satisfies the equation:

$2(3) – (-2) + k = 0$

$\implies 6 + 2 + k = 0$

$\implies 8 + k = 0$

$\implies k = -8$

Therefore, the value of $k$ is $-8$.

13. Haradhan has a field $PQR$, which is in the shape of right angled triangle. He wants to leave a space in the form of a square $AQBC$ inside the field for growing wheat and the remaining for growing vegetables. In the field, there is a pole marked as $O$. (Fig. 7.15)

Based on the above information, answer the following questions:

(i) Taking $O$ as origin, coordinates of $B$ are $(-200, 0)$ and $Q$ are $(200, 0)$. $AQBC$ being a square, what are the co-ordinates of $A$ and $C$?

(ii) What is the area of square $AQBC$?

(iii) What is the length of diagonal $BA$ of square $AQBC$?

(iv) If $C$ divides $RP$ in the ratio $K : 1$, what is the value of $K$, where point $P$ is $(200, 800)$?

Solution:

(i)

Given coordinates are $B(-200, 0)$ and $Q(200, 0)$.

Using the distance formula:

$\text{Distance} = \sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}$

Side length of the square:

$BQ = \sqrt{(200 – (-200))^{2} + (0 – 0)^{2}}$

$\implies BQ = \sqrt{(200 + 200)^{2} + 0^{2}}$

$\implies BQ = \sqrt{(400)^{2}}$

$\implies BQ = 400\text{ units}$

Since $AQBC$ is a square, all its sides are equal to $400$ units

$\text{Side } QA = 400$ and $\text{Side } BC = 400$

Since $AQBC$ is a square of side $400$ units lying above the x-axis:

Point $A$ lies directly above $Q(200, 0)$ at a distance of $400$ units.

$\implies \text{Coordinates of } A = (200, 400)$

Point $C$ lies directly above $B(-200, 0)$ at a distance of $400$ units.

$\implies \text{Coordinates of } C = (-200, 400)$

(ii)

$\text{Area of square } AQBC = \text{Side}^{2}$

$ = (400)^{2}$

$\therefore \text{Area} = 160000\text{ sq. units}$

(iii)

Here, $B = (-200, 0)$ and $A = (200, 400)$.

Using the distance formula:

$BA = \sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}$

$\implies BA = \sqrt{(200 – (-200))^{2} + (400 – 0)^{2}}$

$= \sqrt{(200 + 200)^{2} + (400)^{2}}$

$ = \sqrt{(400)^{2} + (400)^{2}}$

$ = \sqrt{160000 + 160000}$

$ = \sqrt{320000}$

$\therefore BA = 400\sqrt{2}\text{ units}$

Therefore, the length of diagonal $BA$ is $400\sqrt{2}\text{ units}$.

(iv)

Let the coordinates of point $R$ lying on the x-axis be $(x, 0)$. Given that $C(-200, 400)$ divides the line segment joining $R(x, 0)$ and $P(200, 800)$ internally in the ratio $K : 1$.

Using the section formula:

$(x, y) = \left(\frac{k x_2 + x_1}{k + 1}, \frac{k y_2 + y_1}{k + 1}\right)$

$(-200, 400) = \left(\frac{K \times 200 + 1 \times x}{K + 1}, \frac{K \times 800 + 1 \times 0}{K + 1}\right)$

$(-200, 400) = \left(\frac{200K + x}{K + 1}, \frac{800K}{K + 1}\right)$

Equating the y-coordinate:

$400 = \frac{800K}{K + 1}$

$\implies 400(K + 1) = 800K$

$\implies 400K + 400 = 800K$

$\implies 800K – 400K = 400$

$\implies 400K = 400$

$\implies K = \frac{400}{400}$

$\implies K = 1$

Therefore, the value of $K$ is $1$.