NCERT Solutions For Class 10 Maths: Coordinate Geometry

EXERCISE 7.1

Question 1: Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (– 5, 7), (– 1, 3)
(iii) (a, b), (– a, – b)

Solution-
We use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

1.Let $P(2, 3)$ and $Q(4, 1)$.
Distance $PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2}$.
$PQ = \sqrt{(2)^2 + (-2)^2}$
$PQ = \sqrt{4 + 4} = \sqrt{8}$
$PQ = 2\sqrt{2}$ units.

2.Let $P(– 5, 7)$ and $Q(– 1, 3)$.
Distance $PQ = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$.
$PQ = \sqrt{(-1 + 5)^2 + (-4)^2}$
$PQ = \sqrt{(4)^2 + 16} = \sqrt{16 + 16} = \sqrt{32}$
$PQ = 4\sqrt{2}$ units.

3.Let $P(a, b)$ and $Q(– a, – b)$.
Distance $PQ = \sqrt{(-a - a)^2 + (-b - b)^2}$.
$PQ = \sqrt{(-2a)^2 + (-2b)^2}$
$PQ = \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2 + b^2)}$
$PQ = 2\sqrt{a^2 + b^2}$ units.

Question 2: Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Solution-
1.We find the distance between $O(0, 0)$ and $P(36, 15)$.
We can use the specific distance formula for a point from the origin: $OP = \sqrt{x^2 + y^2}$.
$OP = \sqrt{(36)^2 + (15)^2}$
$OP = \sqrt{1296 + 225} = \sqrt{1521}$
$OP = 39$ units.

2.The situation discussed in Section 7.2 places town A at the origin and town B at the coordinates (36, 15), representing 36 km east and 15 km north.
Since the distance between the points (0, 0) and (36, 15) is 39 units, the distance from town A to town B is 39 km. We used the Pythagoras Theorem concept represented by the distance formula to calculate this distance.

Question 3: Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Solution-
Let the points be $A(1, 5)$, $B(2, 3)$, and $C(– 2, – 11)$. For the points to be collinear, the sum of any two distances must be equal to the third distance.

1.Find the distance AB:
$AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.

2.Find the distance BC:
$BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212}$.

3.Find the distance AC:
$AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$.

4.Check for collinearity:
$AB + BC = \sqrt{5} + \sqrt{212}$
Since $\sqrt{5} + \sqrt{212} \neq \sqrt{265}$, the sum of the distances is not equal to the third distance.
Therefore, the points A, B, and C do not form a line segment; they are not collinear.

Question 4: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Solution-
Let the vertices be $A(5, – 2)$, $B(6, 4)$, and $C(7, – 2)$. A triangle is an isosceles triangle if at least two of its sides are equal in length.

1.Find the length of side AB:
$AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2} = \sqrt{1^2 + (4 + 2)^2} = \sqrt{1 + 36} = \sqrt{37}$ units.

2.Find the length of side BC:
$BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$ units.

3.Find the length of side AC:
$AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$ units.

4.Conclusion:
Since $AB = BC = \sqrt{37}$ units, two sides of the triangle are equal.
Therefore, the points $(5, – 2)$, $(6, 4)$, and $(7, – 2)$ are the vertices of an isosceles triangle.

Question 5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution-
We assume the coordinates based on the conventional graphical representation of Fig. 7.8: A(3, 4), B(6, 7), C(9, 4), and D(6, 1). To check if ABCD is a square, we must determine if all four sides are equal and if the two diagonals are also equal.

1.Calculate the lengths of the sides:
$AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$ units.
$BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}$ units.
$CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}$ units.
$DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$ units.
Since $AB = BC = CD = DA$, all four sides are equal.

2.Calculate the lengths of the diagonals:
Diagonal AC: $AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6$ units.
Diagonal BD: $BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6$ units.
Since $AC = BD$, the diagonals are equal.

3.Conclusion:
Since all four sides of the quadrilateral ABCD are equal and its diagonals are also equal, ABCD is a square.
Therefore, Champa is correct.

Question 6: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution-
(i) A(– 1, – 2), B(1, 0), C(– 1, 2), D(– 3, 0)

1.Calculate the lengths of the sides:
$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$ units.
$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8}$ units.
$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$ units.
$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$ units.
All sides are equal: $AB = BC = CD = DA$.

2.Calculate the lengths of the diagonals:
Diagonal AC: $AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$ units.
Diagonal BD: $BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$ units.
Diagonals are equal: $AC = BD$.

3.Reason: Since all sides are equal and both diagonals are equal, the quadrilateral formed is a square.

(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)

1.Calculate the lengths of the sides:
Let A(–3, 5), B(3, 1), C(0, 3), D(–1, – 4).
$AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$ units.
$BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$ units.
$CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$ units.
$DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$ units.

2.Reason: Since no sides are equal, and no specific geometric property (like opposite sides equal or diagonals bisecting) is immediately evident, the points form a general quadrilateral.

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

1.Calculate the lengths of the sides:
Let A(4, 5), B(7, 6), C(4, 3), D(1, 2).
$AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
$BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}$ units.
$CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
$DA = \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$ units.
Opposite sides are equal: $AB = CD$ and $BC = DA$. This suggests it is either a parallelogram or a rhombus.

2.Calculate the lengths of the diagonals:
Diagonal AC: $AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$ units.
Diagonal BD: $BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$ units.

3.Reason: Since the opposite sides are equal ($AB=CD$ and $BC=DA$) but the diagonals are unequal ($AC \neq BD$), the quadrilateral formed is a parallelogram.

Question 7: Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Solution-
1.A point on the x-axis has coordinates of the form $(x, 0)$.
Let $P(x, 0)$ be the required point, and $A(2, – 5)$, $B(– 2, 9)$ be the given points.
Since P is equidistant from A and B, we have $AP = BP$, which implies $AP^2 = BP^2$.

2.Apply the distance formula and equate the squares of the distances:
$(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2$.
$(x^2 - 4x + 4) + 25 = (x + 2)^2 + (-9)^2$
$x^2 - 4x + 29 = (x^2 + 4x + 4) + 81$

3.Solve for x:
$x^2 - 4x + 29 = x^2 + 4x + 85$
$-4x + 29 = 4x + 85$
$29 - 85 = 4x + 4x$
$-56 = 8x$
$x = -7$.

4.The required point on the x-axis is (–7, 0).

Question 8: Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Solution-
1.We are given the points $P(2, – 3)$ and $Q(10, y)$, and the distance $PQ = 10$ units.
We use the distance formula such that $PQ^2 = 10^2 = 100$.

2.Apply the distance formula:
$(10 - 2)^2 + (y - (-3))^2 = 100$
$(8)^2 + (y + 3)^2 = 100$
$64 + y^2 + 6y + 9 = 100$
$y^2 + 6y + 73 = 100$
$y^2 + 6y - 27 = 0$

3.Solve the quadratic equation by factorization:
$y^2 + 9y - 3y - 27 = 0$
$y(y + 9) - 3(y + 9) = 0$
$(y - 3)(y + 9) = 0$

4.The possible values for $y$ are $y = 3$ or $y = -9$.

Question 9: If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution-
1.Since $Q(0, 1)$ is equidistant from $P(5, – 3)$ and $R(x, 6)$, we have $QP = QR$, which means $QP^2 = QR^2$.

2.Calculate $QP^2$:
$QP^2 = (5 - 0)^2 + (-3 - 1)^2 = 5^2 + (-4)^2 = 25 + 16 = 41$.

3.Calculate $QR^2$:
$QR^2 = (x - 0)^2 + (6 - 1)^2 = x^2 + 5^2 = x^2 + 25$.

4.Equate $QP^2$ and $QR^2$:
$41 = x^2 + 25$
$x^2 = 41 - 25 = 16$
$x = \pm 4$.
The values of $x$ are 4 and – 4.

5.Find the distances QR and PR for $x=4$ and $x=-4$:
Case 1: $R(4, 6)$.
Distance $QR = \sqrt{QR^2} = \sqrt{41}$ units.
Distance $PR = \sqrt{(5 - 4)^2 + (-3 - 6)^2} = \sqrt{1^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}$ units.

Case 2: $R(-4, 6)$.
Distance $QR = \sqrt{QR^2} = \sqrt{41}$ units.
Distance $PR = \sqrt{(5 - (-4))^2 + (-3 - 6)^2} = \sqrt{9^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$ units.

Question 10: Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Solution-
1.Let $P(x, y)$ be the point equidistant from $A(3, 6)$ and $B(– 3, 4)$.
We require $AP = BP$, which means $AP^2 = BP^2$.

2.Apply the distance formula and equate the squares of the distances:
$(x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2$.
$(x^2 - 6x + 9) + (y^2 - 12y + 36) = (x + 3)^2 + (y - 4)^2$
$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$

3.Simplify the equation (cancelling $x^2$ and $y^2$ from both sides):
$-6x - 12y + 45 = 6x - 8y + 25$

4.Collect the terms to find the required relation:
$0 = 6x + 6x - 8y + 12y + 25 - 45$
$0 = 12x + 4y - 20$
Dividing by 4:
$3x + y - 5 = 0$.

5.The required relation between $x$ and $y$ is $3x + y = 5$. This equation represents the perpendicular bisector of the line segment joining the two given points.

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EXERCISE 7.2

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

Solution-
1.Let $A(x_1, y_1) = (–1, 7)$ and $B(x_2, y_2) = (4, – 3)$. The ratio is $m_1 : m_2 = 2 : 3$.
We use the Section Formula for internal division:
$P(x, y) = \left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)$.

2.Calculate the x-coordinate:
$x = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$.

3.Calculate the y-coordinate:
$y = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$.

4.The coordinates of the required point are (1, 3).

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

Solution-
1.Let $A(4, – 1)$ and $B(– 2, – 3)$ be the end points. Trisection means the line segment is divided into three equal parts. Let the points of trisection be $P$ and $Q$.

2.Point $P$ divides $AB$ internally in the ratio $1 : 2$.
$x_P = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$.
$y_P = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 - 2}{3} = \frac{-5}{3}$.
The coordinates of P are $\mathbf{(2, -5/3)}$.

3.Point $Q$ divides $AB$ internally in the ratio $2 : 1$.
$x_Q = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = \frac{0}{3} = 0$.
$y_Q = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3}$.
The coordinates of Q are $\mathbf{(0, -7/3)}$.

4.The coordinates of the points of trisection are $\mathbf{(2, -5/3)}$ and $\mathbf{(0, -7/3)}$.

Question 3: 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 $1/4$th the distance AD on the 2nd line and posts a green flag. Preet runs $1/5$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?

Solution-
1.Determine the coordinates of Niharika's flag (N) and Preet's flag (P):
Total distance along AD is 100 m (100 pots at 1m distance each). The horizontal lines represent x-coordinates (line numbers), and the distance covered along AD represents the y-coordinate.

Niharika (Green Flag): On the 2nd line ($x_1 = 2$). Distance covered is $1/4$ of $100$ m.
$y_1 = \frac{1}{4} \times 100 = 25$ m.
Coordinates of N are (2, 25).

Preet (Red Flag): On the 8th line ($x_2 = 8$). Distance covered is $1/5$ of $100$ m.
$y_2 = \frac{1}{5} \times 100 = 20$ m.
Coordinates of P are (8, 20).

2.Find the distance between the two flags (NP):
Using the distance formula:
$NP = \sqrt{(8 - 2)^2 + (20 - 25)^2}$
$NP = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61}$ m.
The distance between the two flags is $\mathbf{\sqrt{61} \text{ m}}$.

3.Find the position of Rashmi's blue flag (R), which is exactly halfway (the mid-point) between N and P.
We use the Mid-point formula:
$x_R = \frac{2 + 8}{2} = \frac{10}{2} = 5$.
$y_R = \frac{25 + 20}{2} = \frac{45}{2} = 22.5$.
The coordinates of R are (5, 22.5).

4.Rashmi should post her blue flag on the 5th line at a distance of 22.5 m from the starting line along AD.

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

Solution-
1.Let $A(– 3, 10)$ and $B(6, – 8)$ be the given points, and let $P(– 1, 6)$ divide the line segment $AB$ internally in the ratio $k : 1$.

2.Apply the Section Formula for the x-coordinate:
$x = \frac{k x_2 + 1 x_1}{k + 1}$
$-1 = \frac{k(6) + 1(-3)}{k + 1}$

3.Solve for k:
$-1(k + 1) = 6k - 3$
$-k - 1 = 6k - 3$
$3 - 1 = 6k + k$
$2 = 7k$
$k = \frac{2}{7}$.

4.Check with the y-coordinate (optional verification based on source context):
$y = \frac{k y_2 + 1 y_1}{k + 1}$
$6 = \frac{k(-8) + 1(10)}{k + 1}$
$6k + 6 = -8k + 10$
$14k = 4$
$k = \frac{4}{14} = \frac{2}{7}$.

5.The required ratio $k : 1$ is $\mathbf{2 : 7}$.

Question 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-
1.Let the x-axis divide the line segment joining $A(1, – 5)$ and $B(– 4, 5)$ in the ratio $k : 1$.
A point lying on the x-axis has its ordinate (y-coordinate) equal to 0. Let the point of division be $P(x, 0)$.

2.Use the y-coordinate of the Section Formula and set it to 0:
$y = \frac{k y_2 + 1 y_1}{k + 1} = 0$
$0 = \frac{k(5) + 1(-5)}{k + 1}$
$0 = 5k - 5$
$5k = 5$
$k = 1$.

3.Ratio of division:
The ratio $k : 1$ is $\mathbf{1 : 1}$.

4.Find the coordinates of the point of division P(x, 0):
Since the ratio is $1:1$, P is the mid-point of AB.
$x = \frac{x_1 + x_2}{2} = \frac{1 + (-4)}{2} = \frac{-3}{2}$.
The coordinates of the point of division are $\mathbf{(-3/2, 0)}$.

Question 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-
1.Let the vertices be $A(1, 2)$, $B(4, y)$, $C(x, 6)$, and $D(3, 5)$.
We know that the diagonals of a parallelogram bisect each other.
Therefore, the coordinates of the mid-point of diagonal AC must be equal to the coordinates of the mid-point of diagonal BD.

2.Find the mid-point of AC:
Midpoint of $AC = \left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{1 + x}{2}, 4\right)$.

3.Find the mid-point of BD:
Midpoint of $BD = \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$.

4.Equate the corresponding coordinates:
Equating the x-coordinates:
$\frac{1 + x}{2} = \frac{7}{2}$
$1 + x = 7$
$x = 6$.

Equating the y-coordinates:
$4 = \frac{y + 5}{2}$
$8 = y + 5$
$y = 3$.

5.The values are $\mathbf{x = 6}$ and $\mathbf{y = 3}$.

Question 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-
1.In a circle, the centre is the mid-point of its diameter.
Let the coordinates of point A be $(x, y)$. The centre $C$ is $(2, – 3)$, and $B$ is $(1, 4)$.

2.Use the Mid-point formula where C is the mid-point of AB:
$C(2, -3) = \left(\frac{x + 1}{2}, \frac{y + 4}{2}\right)$.

3.Equate the coordinates:
For the x-coordinate:
$2 = \frac{x + 1}{2}$
$4 = x + 1$
$x = 3$.

For the y-coordinate:
$-3 = \frac{y + 4}{2}$
$-6 = y + 4$
$y = -10$.

4.The coordinates of point A are (3, – 10).

Question 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-
1.We are given $A(– 2, – 2)$ and $B(2, – 4)$. P lies on the segment AB and $AP = \frac{3}{7} AB$.

2.Determine the ratio in which P divides AB:
If $AP = \frac{3}{7} AB$, then $AP$ corresponds to 3 parts out of 7 total parts.
The remaining length is $PB = AB - AP = AB - \frac{3}{7} AB = \frac{4}{7} AB$.
The ratio of division $AP : PB$ is $\frac{3}{7} AB : \frac{4}{7} AB$, or $3 : 4$.
Here, $m_1 = 3$ and $m_2 = 4$.

3.Apply the Section Formula to find the coordinates of $P(x, y)$:
$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{3(2) + 4(-2)}{3 + 4} = \frac{6 - 8}{7} = \frac{-2}{7}$.

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{3(-4) + 4(-2)}{3 + 4} = \frac{-12 - 8}{7} = \frac{-20}{7}$.

4.The coordinates of P are $\mathbf{(-2/7, -20/7)}$.

Question 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-
1.Let $A(– 2, 2)$ and $B(2, 8)$. To divide AB into four equal parts, we need three points, $P_1, P_2$, and $P_3$.
$P_1$ divides AB in ratio $1:3$.
$P_2$ divides AB in ratio $2:2$ (mid-point).
$P_3$ divides AB in ratio $3:1$.

2.Coordinates of $P_2$ (Mid-point of AB, ratio 1:1):
$x_2 = \frac{-2 + 2}{2} = 0$.
$y_2 = \frac{2 + 8}{2} = \frac{10}{2} = 5$.
$P_2$ is (0, 5).

3.Coordinates of $P_1$ (Ratio 1:3):
$x_1 = \frac{1(2) + 3(-2)}{1 + 3} = \frac{2 - 6}{4} = \frac{-4}{4} = -1$.
$y_1 = \frac{1(8) + 3(2)}{1 + 3} = \frac{8 + 6}{4} = \frac{14}{4} = \frac{7}{2}$.
$P_1$ is $\mathbf{(-1, 7/2)}$.

4.Coordinates of $P_3$ (Ratio 3:1):
$x_3 = \frac{3(2) + 1(-2)}{3 + 1} = \frac{6 - 2}{4} = \frac{4}{4} = 1$.
$y_3 = \frac{3(8) + 1(2)}{3 + 1} = \frac{24 + 2}{4} = \frac{26}{4} = \frac{13}{2}$.
$P_3$ is $\mathbf{(1, 13/2)}$.

5.The coordinates of the points that divide the line segment into four equal parts are $\mathbf{(-1, 7/2), (0, 5), \text{ and } (1, 13/2)}$.

Question 10: 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)]

Solution-
1.Let the vertices of the rhombus be $A(3, 0)$, $B(4, 5)$, $C(– 1, 4)$, and $D(– 2, – 1)$.
The Area of a rhombus is given by $\frac{1}{2}$ (product of its diagonals). We need to find the lengths of the diagonals AC and BD using the distance formula.

2.Find the length of diagonal AC ($d_1$):
$d_1 = \sqrt{(-1 - 3)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}$
$d_1 = 4\sqrt{2}$ units.

3.Find the length of diagonal BD ($d_2$):
$d_2 = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72}$
$d_2 = 6\sqrt{2}$ units.

4.Calculate the area of the rhombus:
Area $= \frac{1}{2} \times d_1 \times d_2$.
Area $= \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2})$
Area $= \frac{1}{2} \times (4 \times 6 \times (\sqrt{2})^2)$
Area $= \frac{1}{2} \times (24 \times 2)$
Area $= 24$ square units.

5.The area of the rhombus is 24 square units.