CBSE Class 10 Chapter 7 Coordinate Geometry Exercise 7.2

Q1.

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

Ans.

The coordinates $(x, y)$ of the point that divides the line segment joining $(-1, 7)$ and $(4, -3)$ in the ratio $2:3$ can be found using the section formula. The section formula is given by: $$x = \frac{m_1 \cdot x_2 + m_2 \cdot x_1}{m_1 + m_2}$$ $$y = \frac{m_1 \cdot y_2 + m_2 \cdot y_1}{m_1 + m_2}$$ In this case, $m_1 : m_2 = 2:3$. Therefore, $m_1 = 2$ and $m_2 = 3$. Now, substitute the values into the formulas: $$x = \frac{2 \cdot 4 + 3 \cdot (-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5}= 1$$ $$y = \frac{2 \cdot (-3) + 3 \cdot 7}{2 + 3}= \frac{-6 + 21}{5} = \frac{15}{5}= 3$$ So, the coordinates of the point dividing the line segment in the ratio 2:3 are $(1, 3)$.

Q2.

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

Ans.

Assume that $P (x_1, y_1)$ and $Q (x_2, y_2)$ is the points of trisection of the line segment joining the given points $(4, -1)$ and $(-2, -3)$, so we can write $AP = PQ = QB$. Therefore, point $P$ divides $AB$ in the ratio $1:2$. By using section formula: $$x_1 = \frac{1 \cdot (-2) + 2 \cdot 4}{3} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$$ $$y_1 = \frac{1 \cdot (-3) + 2 \cdot (-1)}{3} = \frac{-3 - 2}{3} = \frac{-5}{3}$$ Therefore: $P (x_1, y_1) = P(2, -\frac{5}{3})$ Point $Q$ divides $AB$ in the ratio $2:1$. By using section formula: $$x_2 = \frac{2 \cdot (-2) + 1 \cdot 4}{3} = \frac{-4 + 4}{3} = 0$$ $$y_2 = \frac{2 \cdot (-3) + 1 \cdot (-1)}{3} = \frac{-6 - 1}{3} = -\frac{7}{3}$$ Therefore, the coordinates of the point $Q$ are $(0, -\frac{7}{3})$.

Q3.

To conduct Sports Day activities, in your rectangular shaped school ground $ABCD$, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along $AD$, as shown in Fig. 7.12. Niharika runs $\frac{1}{4}$th the distance $AD$ on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$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?

Ans.

As per question, Niharika posted the green flag at $\frac{1}{4}$th of the distance $AD$, So, $\left(\frac{1}{4} \times 100\right)$ m = 25 m from the starting point of the 2nd line. Therefore, We gets coordinates of this point are (2, 25). Similarly, Preet posted a red flag at $\frac{1}{5}$ th of the distance $AD$, So, $\left(\frac{1}{5} \times 100\right)$ m = 20 m from the starting point of the 8th line. Therefore, We gets coordinates of this point are (8, 20). The distance between these flags: $$\sqrt{{(8 - 2)^2 + (20 - 25)^2}}$$ $$\sqrt{{(6)^2 + ( - 5)^2}} =\sqrt{{36 + 25}}= \sqrt{{61}}$$ So the distance between both flag is $\sqrt{{61}}$ The point at which Rashmi should post her blue flag is the midpoint of the line joining these points. Assume that this point is $P(x, y)$. $$x = \frac{2 + 8}{2} = \frac{10}{2} = 5$$ $$y = \frac{20 + 25}{2} = \frac{45}{2}$$ Hence, $P(x, y) = (5, \frac{45}{2})$. Therefore, Rashmi should post her blue flag at $\frac{45}{2} = 22.5$ m on the 5th line.

Q4.

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

Ans.

Let, the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6) is k:1, then by section formula: $$\left(\frac{kx_{2}+x_{1}}{k+1},\frac{ky_{2}+y_{1}}{k+1}\right)$$ Putting value $$(-1,6)=\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$$ $$-1=\left(\frac{6k-3}{k+1}\right)$$ $$-k-1=6k-3$$ $$-k-6k=-3+1$$ $$7k=2$$ $$k:1 = 2:7$$ $$6=\left(\frac{-8k+10}{k+1}\right)$$ $$6k+6=-8k+10$$ $$6k+8k=10-6$$ $$14k=4$$ both side divided by 2: $$7k=2$$ $$k:1 = 2:7$$

Q5.

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.

Ans.

Let the ratio in which the line segment joining points A(1, –5) and B(–4, 5) is divided by the x-axis be k:1. Therefore, the coordinates of the point of division can be denoted as P(x, y) and can be written as: $$\left(\frac{-4k+1}{k+1},\frac{5k-5}{k+1}\right)$$ Since any point on the x-axis has a y-coordinate of 0, we set the y-coordinate expression to 0: $$\frac{5k-5}{k+1}=0$$ $$5k=5$$ $$k=1$$ Thus, the x-axis divides the line segment in the ratio 1:1. Now, we find the coordinates of the point of division, P(x, y): $$P(x,y)=\left(\frac{-4(1)+1}{1+1},\frac{5(1)-5}{1+1}\right)$$ $$P(x,y)=\left(\frac{-4+1}{2},\frac{5-5}{2}\right)$$ $$P(x,y)=\left(\frac{-3}{2},\frac{0}{2}\right)$$ $$P(x,y)=\left(\frac{-3}{2},0\right)$$ Therefore, the x-axis divides the line segment joining A and B in the ratio 1:1, and the coordinates of the point of division, P, are $\left(\frac{-3}{2},0\right)$

Q6.

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

Ans.

We know that the midpoint formula for a line segment with endpoints $\left(x_{1},y_{1}\right)$ and $\left(x_{2},y_{2}\right)$ is: $$Mid Point=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$$ Given points : A(1, 2), B(4, y), C(x, 6) and D(3, 5), So Midpoint of AC: $$\left(\frac{1+x}{2},\frac{2+6}{2}\right)=\left(\frac{1+x}{2},4\right)$$ Midpoint of BD: $$\left(\frac{4+3}{2},\frac{y+5}{2}\right)=\left(\frac{7}{2},\frac{y+5}{2}\right)$$ We also know that in a parallelogram, the midpoint of one diagonal is the same as the midpoint of the other diagonal. So, $$\left(\frac{1+x}{2},4\right)=\left(\frac{7}{2},\frac{y+5}{2}\right)$$ Now, equate corresponding coordinates: For $x$: $$\frac{1+x}{2}=\frac{7}{2}$$ $$2+2x=14$$ $$2x=14-2$$ $$2x=12$$ $$x=6$$ For $y$: $$4=\frac{y+5}{2}$$ $$8=y+5$$ $$8-5=y$$ $$y=3$$ So, the values of $x$ and $y$ that satisfy the conditions are $x=6$ and $y=3$.

Q7.

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).

Ans.

Given: Midpoint of AB is (2,—3) Coordinate of B is (1,4) Let the coordinate of point A is (x,y) We know that midpoint is: $$Mid Point=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$$ Putting value $$(2,-3)=\left(\frac{x+1}{2},\frac{y+4}{2}\right)$$ Now, equate corresponding coordinates: For $x$: $$2=\frac{x+1}{2}$$ $$4=x+1$$ $$4-1=x$$ $$x=3$$ For $y$: $$-3=\frac{y+4}{2}$$ $$-6=y+4$$ $$-6-4=y$$ $$y=-10$$ Hence, The coordinates of $A(3,-10)$.

Q8.

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

Ans.

Given that  $A(-2, -2)$ and $B(2, -4)$. The vector $\vec{AB}$ is given by: $$\vec{AB} = (2 - (-2), (-4) - (-2)) = (4, -2)$$ As per given $AP = \frac{3}{7} AB$, so: $$\vec{AP} = \frac{3}{7} \times \vec{AB}$$ $$\vec{AP} = \frac{3}{7} \times 4,\frac{3}{7} \times (-2) = \left(\frac{12}{7}, -\frac{6}{7}\right)$$ The coordinates of point P, $(x, y)$, are given by: $$(x, y) = (x + \frac{12}{7}, y - \frac{6}{7})$$ $$(x, y) = (-2 + \frac{12}{7}, -2 - \frac{6}{7})$$ $$(x, y) = \left(\frac{-2 \times 7 + 12}{7}, \frac{-2 \times 7 - 6}{7}\right)$$ $$(x, y) = \left(\frac{-14 + 12}{7}, \frac{-20}{7}\right)$$ $$(x, y) = \left(\frac{-2}{7}, -\frac{20}{7}\right)$$ Therefore, the coordinates of point P are $(\frac{-2}{7}, -\frac{20}{7})$. Alternative method: Given : The coordinates of point A and B are (-2,-2) and (2,-4). Since $AP=\frac{3}{7}AB$ Therefore, Ratio between AP and PB = 3:4 So, Point P divides the line segment AB in the ratio 3:4. Then Coordinate of P is: $$\left( x = \frac{m_1 \cdot x_2 + m_2 \cdot x_1}{m_1 + m_2}, y = \frac{m_1 \cdot y_2 + m_2 \cdot y_1}{m_1 + m_2}\right)$$ putting value $$\left( x = \frac{3(2) + 4(-2)}{3 +4}, y = \frac{3(-4) + 4(-2)}{3 + 4}\right)$$ $$\left( x = \frac{6-8}{7}, y = \frac{-12-8}{7}\right)$$ $$\left( x = -\frac{7}{2}, y = -\frac{20}{7}\right)$$ Therefore, the coordinates of point P are $(-\frac{2}{7}, -\frac{20}{7})$.

Q9.

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

Ans.

Given: the coordinates of the points that divide the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts. Let the points dividing the line segment be $P_1$, $P_2$, and $P_3$. The section formula for dividing a line segment into $n$ equal parts is: $$P_i = \left(\frac{{(n - i)x_1 + ix_2}}{{n}}, \frac{{(n - i)y_1 + iy_2}}{{n}}\right)$$ where $i = 1, 2, \ldots, n-1$. For dividing the segment AB into four equal parts, $n = 4$, and we have three points to find ($P_1$, $P_2$, and $P_3$). 1. Coordinates of $P_1$: $$P_1 = \left(\frac{{3(-2) + 1(2)}}{4}, \frac{{3(2) + 1(8)}}{4}\right) = (-1, 3.5)$$ 2. Coordinates of $P_2$: $$P_2 = \left(\frac{{2(-2) + 2(2)}}{4}, \frac{{2(2) + 2(8)}}{4}\right) = (0, 5)$$ 3. Coordinates of $P_3$: $$P_3 = \left(\frac{{1(-2) + 3(2)}}{4}, \frac{{1(2) + 3(8)}}{4}\right) = (1, 6.5)$$ Therefore, the coordinates of the points dividing the line segment AB into four equal parts are $P_1(-1, 3.5)$, $P_2(0, 5)$, and $P_3(1, 6.5)$. Alternative Method: As per figure, we can be observed that points P, Q, R are dividing the line segment AB in a ratio 1:3, 1:1, 3:1, respectively. Coordinate of P: $$P=\left(\frac{1(2)+3(-2)}{1+3},\frac{1(8)+3(2)}{1+3}\right)$$ $$P=\left(\frac{2-6}{4},\frac{8+6}{4}\right)$$ $$P=\left(\frac{-4}{4},\frac{14}{4}\right)$$ $$P=\left(-1,\frac{7}{2}\right)=\left(-1,3.5\right)$$ Coordinate of Q: $$Q=\left(\frac{2(1)+2(1)}{1+1},\frac{2(1)+8(1)}{1+1}\right)$$ $$Q=\left(\frac{2-2}{2},\frac{2+8}{2}\right)$$ $$Q=\left(\frac{0}{2},\frac{10}{2}\right)$$ $$Q=\left(0,5\right)$$ Coordinate of R: $$R=\left(\frac{3(2)+1(-2)}{3+1},\frac{3(8)+1(2)}{3+1}\right)$$ $$R=\left(\frac{6-2}{4},\frac{24+2}{4}\right)$$ $$R=\left(\frac{4}{4},\frac{26}{4}\right)$$ $$R=\left(0,\frac{13}{2}\right)=\left(0,6.5\right)$$ Therefore, the coordinates of the points dividing the line segment AB into four equal parts are $P_1(-1, 3.5)$, $P_2(0, 5)$, and $P_3(1, 6.5)$.

Q10.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint : Area of a rhombus$=\frac{1}{2}$ (product of its diagonals)].

Ans.

The vertices of the rhombus are given as $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$, and $D(-2, -1)$. The formula for the area of a rhombus using the lengths of its diagonals is: $$\text{Area} = \frac{1}{2} \times \text{product of diagonals}$$ By the distance formula: Length of diagonal $AC$: $$AC=\sqrt{\left(x_{C}-x_{A}\right)+\left(y_{C}-y_{A}\right)}$$ $$AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2}$$ $$AC=\sqrt{(-4)^{2}+4^{2}}=\sqrt{16+16}=\sqrt{32}$$ Length of diagonal $BD$: $$BD=\sqrt{\left(x_{D}-x_{B}\right)+\left(y_{D}-y_{B}\right)}$$ $$BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{72}$$ $$BD=\sqrt{(-6)^{2}+(-6)^{2}}=\sqrt{36+36}=\sqrt{72}$$ Now, calculate the product of the diagonals: $$\text{Area} = \frac{1}{2} \times \sqrt{32} \times \sqrt{72}$$ $$\text{Area} = \frac{1}{2} \times \sqrt{32 \times 72}$$ $$\text{Area} = \frac{1}{2} \times \sqrt{2304}$$ $$\text{Area} = \frac{1}{2} \times 48=24$$ Therefore, the area of the rhombus is $24$ square units.