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RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

April 20, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry Important Questions and Answers.

RBSE Class 10 Maths Chapter 7 Important Questions Coordinate Geometry

Objective Type Questions—

Question 1.
The distances of the point (3, 5) from y- axis will be—
(A) 1
(B) 4
(C) 2
(D) 3
Answer:
(D) 3

Question 2.
The triangle with vertices (- 2, 1), (2, – 2) and (5, 2) is—
(A) right angled
(B) equilateral
(C) insceles
(D) None of these
Answer:
(C) insceles

Question 3.
The point P divides the line segment joining the points (5, 0) and (0, 4) with the ratio 2 : 3. The coordinates of P are—
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 1
Answer:
(3, \(\frac{8}{5}\))

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 4.
If the points (1, 2), (-1, x) and (2, 3) are collinear, then the value of x will be—
(A) 2
(B) 0
(C) -1
(D) 1
Answer:
(B) 0

Question 5.
If the point (x. y) is equidistant from the points (2, 1) and (1, -2), then—
(A) x + 3y = 0
(B) 3x + y = 0
(C) x + 2y = 0
(D) 2y + 3x = 0
Answer:
(A) x + 3y = 0

Question 6.
If A(4, – 3), B(3, – 2) and C(2, 8) are the vertices of any triangle, then the distances of its centroid from y-axis will be—
(A) 1
(B) 4
(C) 3
(D) 2
Answer:
(C) 3

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 7.
The distance between the two points (0, cos θ) and (sin θ, 0) is—
(A) 1
(B) sin θ + cos θ
(C) \(\frac{1}{2}\)(sinθ + cos θ)
(D) 0
Answer:
(A) 1

Very Short Answer Type Questions—

Question 1.
Write the distance of the point (5, – 2) from the y-axis.
Solution:
5, since the perpendicular distance on y-axis is 5.

Question 2.
Of which type of triangle are the points (- 2, 2), (8, – 2) and (- 4, – 3) are vertices?
Solution:
If the given points are respectively A, B, C then
AB = \(\sqrt {100+16}\) = \(\sqrt {161}\)
BC = \(\sqrt {144+1}\) = \(\sqrt {145}\)
CA = \(\sqrt {4+25}\) = \(\sqrt {29}\)
∵ AB2 + CA2 = BC2
⇒ A, B, C are the vertices of a right angled triangle.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 3.
If (4, 3) and (- 2, – 1) are the opposite vertices of any parellelogram and its third vertex is (1, 0) then what will the product of the coordinates of the fourth vertex?
Answer:
The mid-point of the diagonal joining the given opposite vertices is (\(\frac{4-2}{2}\), \(\frac{3-1}{2}\)) i.e., (1, 1) which is also the mid-point of the other diagonal. Hence if the fourth vertex be (x, y) then
\(\frac{x+1}{2}\) = 1 and \(\frac{y+0}{2}\) = 1
x = 1, y = 2
Therefore product = 1 × 2 = 2

Question 4.
If the distance between the
points (K, 2) and (3, 4) be \(\sqrt {8}\), then write the value of K.
Solution:
According to the question
(\(\sqrt {8}\))2 = (K – 3)2 + (2 – 4)2
8 = (K – 3)2 + 4
4 = (K – 3)2
(K – 3) = ± 2
K = ± 2 + 3 ∴ K = 5 and K = 1

Question 5.
In what ratio does y-axis divide the line segment joining the points (-3, – 4) and (1, – 2)? Write.
Solution:
On y-axis x = 0
Let the line segment be divided in the ratio λ : 1.
By internal division formula
x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
∴ 0 = \(\frac{\lambda \times 1+1 \times(-3)}{\lambda+1}\)
o = λ – 3 ∴ λ = 3
Hence the required ratio is 3 : 1

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 6.
If (a, 0), (0, b) and (1, 1) be collinear, then what will be its condition?
Solution:
For these points to be collinear, the area of the triangle formed by these must be zero.
Area of triangle
= \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
or x1(y2 – y3) + x2(y3 – y2) + x3(y1 – y2) = 0
or a(b – 1) + 0(1 – 0) + 1(0 – b) = 0
or ab – a – b = 0
a + b = ab

Question 7.
Find the coordinates of the mid-point of the line segment joining the points (6, 8) and (2, 4).
Solution:
The coordinates of the mid-point are :
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 2
Therefore the coordinates of the mid-point will be (4, 6).

Question 8.
The opposite vertices of any square are (- 5, – 4) and (3, 2). Write the length of its diagonal.
Solution:
Length of the diagonal of the square
= \(\sqrt{(3+5)^{2}+(2+4)^{2}}\)
= \(\sqrt{(8)^{2}+(6)^{2}}\) = \(\sqrt{64+36}\)
= \(\sqrt{100}\) = 10

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 9.
Find the distance of the point (3, 4) from the origin.
Solution:
Distance from origin
= \(\sqrt{(3-0)^{2}+(-4-0)^{2}}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\) = 5

Question 10.
Find a point on y-axis which is equidistant from points A(6, 5) and B(-4, 3).
Solution:
Point lying on y-axis is (0, y). According to the question, distance of the point (0, y) from A (6, 5).
= Distance of the point (0, y) from B (-4, 3)
∴ (6 – 0)2 + (5 – y)2 = (- 4 – 0)2 + (3 – y)2
or 36 + 25 + y2 – 10y= 16 + 9 + y2 – 6y
or 4y = 36
or y = \(\frac{36}{4}\) = 9
Hence the point on y-axis is (0, 9)

Question 11.
Which figure will be obtained by joining the points (- 2, 0), (2, 0), (2, 2), (0, 4), (- 2, 2) in order.
Solution:
A pentagon is obtained by joining the given points.
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 3

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 12.
In what ratio does the point (3, 4) divide the line segment joining the points (1, 2) and (6, 7)?
Solution:
Let the point (3, 4) divide the line segment joining the points (1, 2) and (6, 7) in the ratio m1 : m2. Then,
3 = \(\frac{m_{1} \times 6+m_{2} \times 1}{m_{1}+m_{2}}\)
3(m1 + m2) = 6m1 + m2
3m1 + 3m2 = 6m1 + m2
3m2 – m2= 6m1 – 3m1
2m2 = 3 m1
\(\frac{m_{1}}{m_{2}}\) = \(\frac{2}{3}\)
Hence m1 : m2 = 2 : 3

Question 13.
If the distances of the point (0, 2) from the points (3, k) and (k, 5) are equal, then find the value of k.
Solution:
According to the question, distance between the point (3, k) and (0, 2) = distance between the points (k, 5) and (0, 2).
∴ (0 – 3)2 + (2 – k)2 = (0 – k)2 + (2 – 5)2
⇒ 9 + 4 – 4k + k2 = k2 + 9
⇒ 4 – 4k = 0
⇒ – 4k = – 4 ∴ k = 1

Question 14.
If the three points (2, 1), (fc, 1) and (2k + 1, 2) are collinear, then find the value of k.
Solution:
Here. x1 = 2, x2 = k, x3 = 2k + 1
and y1 = 1, y2 = 1, y3 = 2
∵ The three points are collinear, therefore
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
⇒ 2(1 – 2) + k(2 – 1) + (2k + 1) (1 – 1) = 0
⇒ – 2 + k + 0 = 0

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 15.
Write the distance of the point (- 2, 9) from the x-axis.
Solution:
Distance from x-axis will be 9 since the perpendicular distance from x-axis is 9.

Question 16.
Write the coordinates of the point dividing the line segment joining the points (4, – 3) and (8, 5) internally in the ratio 3 : 1.
Solution:
Formula of internal division is
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 4
y = 3 is obtained.
Hence the required point of division is P(7, 3)

Question 17.
Write the distance of the point (7, – 3) from the y-axis.
Solution:
7, since the perpendicular distance on y-axis is 7.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 18.
Write the distance of the point (3, -2) from y-axis.
Solution:
3, since the perpendicular distance on y-axis is 3.

Question 19.
If M (4, 5) is the mid-point of line segment AB and the coordinates of A are (3, 4), then find the coordinates of the point B.
Solution:
Let the coordinates of the point B be (x, y).
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 5
∵ M (4, 5) is the mid-point
∴ 4 = \(\frac{3+x}{2}\)
or 3 + x = 8
∴ x = 8 – 3 = 5
and 5 = \(\frac{4+y}{2}\)
or 4 + y = 10
∴ y = 10 – 4 = 6
Hence the coordinates of the point B will be (5, 6)

Question 20.
Write the distance of the point (-5, 4) from x-axis.
Solution:
4, since the perpendicular distance on x-axis is 4.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 21.
If K (5, 4) is the mid-point of line segment PQ and coordinates of Q are (2, 3), then find the coordinates of point P.
Solution:
Let the coordinates of the point P be (x, y).
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 6
∵ Mid-point, 5 = \(\frac{x+2}{2}\)
or x + 2 = 10
∴ x = 10 – 2 = 8
and 4 = \(\frac{y+3}{2}\)
or 8 = y + 3
∴ y = 8 – 3 = 5
Hence the coordinates of the point P will be (8, 5)

Short Answer Type Questions—

Question 1.
If the vertices of a quadrilateral are (1, 4), (- 5, 4), (- 5, – 3) and (1, – 3) then name the type of the quadrilateral.
Solution:
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 7
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 8
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 9
Therefore, AB = CD and BC = AD and diagonal AC = diagonal BD. Hence the given points are the vertices of a rectangle.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 2.
Find the ratio in which the point (- 3, p) divides internally the line segment joining the points (- 5, – 4) and (- 2, 3). Also find the value of p.
Solution:
Let C (- 3 ,p) divide AB internally in the ratio k : 1.
∴ By Section Formula
x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\) and y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)
∴ -3 = \(\frac{-2 k-5}{k+1}\)
or -3k – 3 = -2k – 5
or k = -3 + 5 = 2
i.e., the ratio will be k : 1 or 2 : 1.
and p = \(\frac{2 \times 3+1 \times(-4)}{2+1}\) = \(\frac{2}{3}\)
∴ p = \(\frac{2}{3}\)

Question 3.
In what ratio does the y- axis divide the line segment joining the points (5, – 6) and (- 1, – 4)? Also find the coordinates of this point of intersection.
Solution:
Let the required ratio be k : 1. Then by section formula
x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\), y = \(\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\)
Hence the, coordinates of the point dividing in the ratio k : 1 are :
(\(\frac{-k+5}{k+1}\), \(\frac{-4 k-6}{k+1}\))
Since it is given that this point lies in y- axis and we know that on y-axis the value of x is zero.
Hence, \(\frac{-k+5}{k+1}\) = 0
So, k = 5
i.e., the desired ratio is 5 : 1. On putting the value of k as 5 we obtain the point of intersection as (0, \(\frac{-13}{3}\))

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 4.
Find the value of k of the points A(2, 3), B(4, k) and C(6, – 3) are collinear.
Solution:
Since it is given that the three points are collinear, therefore the area of the triangle formed by these points will be zero.
i.e., \(\frac{1}{2}\)[2(k + 3) + 4(-3 – 3) + 6(3 – k)] = 0
i.e., 2k + 6- 4 × 6 + 18 – 6k = 0
or 2k + 6 – 24 + 18 – 6k = 0
4k = 0 or k = 0
Hence, the required value of k is 0.

Question 5.
Find a point on y-axis which is equidistant from the points A(6, 5) and B(- 4, 3).
Solution:
We know that on y-axis any point is of the form (0, y). Therefore let point P(0, y) be equidistant from points A and B. Then
(6 – 0)2 + (5 – y)2 = (-4 – 0)2 + (3 – y)2
or 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y
or 4y = 36
or y = \(\frac{36}{4}\) = 9
Hence the desired point is (0, 9).

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 6.
One end of a line is (4, 0) and the mid-point is (4, 1), then what will be the coordinates of the other end of the line?
Solution:
Let the coordinates of the other end be (x2, y2).
Mid-point is (4, 1). Therefore,
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 10
⇒ 4 × 2 = 4 + x2
⇒ x2 = 8 – 4 = 4
and 1 = \(\frac{0+y_{2}}{2}\)
⇒ 2 = 0 + y2
⇒ y = 2
∴ (x2, y2) = (4, 2)
Hence, the coordinates of the other end are (4, 2).

Question 7.
Find the distance of the point (1, 2) from the mid-point of the line segment joining the points (6, 8) and (2, 4).
Solution:
Coordinates of the mid-point of the line segment joining the given points (6. 8) and (2, 4).
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 11
Now the distance of the point (4, 6) from the point (1, 2)
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 12
Hence, the required distance between the points will be 5.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 8.
Prove that the mid-point of the line segment joining the points (5, 7) and (3, 9) is the same as the mid-point of the line segment joining the points (8, 6) and (0, 10).
Solution:
Mid-point of the line segment joining the points A(5, 7) and B(3, 9).
= (\(\frac{5+3}{2}\), \(\frac{7+9}{2}\))
= (\(\frac{8}{2}\), \(\frac{16}{2}\))
= (4, 8)
and mid-point of the line segment joining the points C(8, 6) and D(0, 10).
= (\(\frac{8+0}{2}\), \(\frac{6+10}{2}\))
= (\(\frac{8}{2}\), \(\frac{16}{2}\))
= (4, 8)
Hence, the mid-points of both the line segments are the same.

Question 9.
If the coordinates of A, B and C are respectively (6, – 1), (1, 3) and (x, 8) then find the value of x when AB = BC.
Solution:
Given that— AB = BC
(AB)2 = (BC)2
=> (6 – 1)2 + (-1 – 3)2 = (x – 1)2 + (8 – 3)2
=> 25 + 16 = (x – 1)2 + 25
=> 16= (x – 1)2
=> ±4 = (x – 1)
Taking positive sign
4 = x – 1 ∴ x = 5
Similarly taking negative sign
– 4= x – 1
∴ – 4 + 1 = x
⇒ x = -3
Hence the value of x will be – 3 or 5.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 10.
In what ratio does the point (2, 1) divide the line segment joining the points (1, – 2) and (4, 7)?
Solution:
Let the point divide the line segment joining the given points in the ratio λ : 1. Therefore by section formula
\(\bar{x}\) = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
∴ 2 = \(\frac{\lambda \times 4+1 \times 1}{\lambda+1}\)
⇒ 2λ + 2 = 4λ + 1
⇒ 2 – 1 = 4λ – 2λ
⇒ 1 = 2λ
∴ λ = \(\frac{1}{2}\)
Hence the required ratio is 1 : 2.
Note : We get the same ratio taking the value of the ordinate.

Question 11.
Find the point on x-axis which is equidistant from the points A(6, 5) and B(- 4, 5).
Solution:
Let the point on x-axis which is equidistant from the points A(6, 5) and B(- 4, 5) be P(x, 0).
According to the question
AP = BP
∴ (AP)2 = (BP)2
Applying the formula of finding the distance between two points.
d = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
∴ (x – 6)2 + (0 – 5)2 = (x + 4)2 + (0 – 5)2
⇒ x2 – 12x + 36 + 25 = x2 + 8x + 16 + 25
⇒ -12x + 36 = 8x + 16
⇒ -12x – 8x= 16 – 36
⇒ – 20x = – 20
x = \(\frac{-20}{-20}\) = 1
Hence that point on x-axis is P(1, 0).

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 12.
If the points A(6, 1), B(8, 2), C(9, 4) and D(x, y) are the vertices of a parallelogram in order, then find the point D(x, y).
Solution:
We know that the diagonals of a parallelogram bisect each other.
Therefore coordinates of the mid-point of diagonal AC = coordinates of the mid-point of diagonal BD.
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 13
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 14
or 5 = 2 + y
∴ y = 5 – 2 = 3
Hence the point D will (7, 3).

Question 13.
Find the area of the triangle formed by the points A(5, 2), B(4, 7) and C(7, -4).
Solution:
The area of △ABC is the numerical value of the expression
\(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
∴ Area of triangle ABC formed by the vertices A(5, 2), B(4, 7) and C(7, – 4) is
= \(\frac{1}{2}\)[5(7 + 4) + 4(- 4 – 2) + 7(2 – 7)]
= \(\frac{1}{2}\)[5 × 11 + 4 × (- 6) + 7 × (-5)]
= \(\frac{1}{2}\)[55 – 24 – 35] = \(\frac{1}{2}\)[55 – 59]
= \(\frac{-4}{2}\) = -2
Since area is a measure, therefore this cannot be negative. Hence we shall take the numerical value of -2, i.e., 2 as the area. Therefore, the area of the triangle is 2 square units.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Long Answer Type Questions—

Question 1.
There are four points P(2, – 1), Q (3, 4), R(- 2, 3) and S(- 3, – 2) in any plane, then prove that PQRS is not a square, it is a rhombus.
Solution:
To prove that PQRS is a rhombus, we shall have to prove :
(i) PQ = QR = RS = SP and PR ≠ QS
(ii) Mid-point of PR = mid-point of QS
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 15
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 16
Thus, mid-point of diagonal PR = mid-point of diagonal QS.
i.e., diagonal of PQRS bisect each other. Therefore these are the diagonals of a rhombus and four sides are also equal and diagonals are not equal to each other. Therefore these are not the vertices of a square.

Question 2.
Prove that in right angled triangle AOB the mid-point C of the hypotenuse is equidistant from the vertices O, A and B.
Solution:
Let A = (2a, 0) and B = (0, 2b)
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 17
OC = AC = BC
i.e., the point C is equidistant from O, A and B.

Question 3.
Find the lengths of the medians of a triangle whose vertices are (1, – 1), (0, 4) and (- 5, 3).
Solution:
Coordinates of mid point D of BC
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 18
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 19
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 20

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 4.
If the mid-points of the sides of a triangle are (1, 2), (0, – 1) and (2, – 1), then find the coordinates of the vertices of the triangle.
Solution:
Mid-point of line joining the points (x1, y1) and (x2, y2)
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 21
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 22
Solving equations (1), (2) and (3)—
2x1 + 2x2 + 2x3 = 6
or x1 + x2 + x3 = 3 …. (4)
or 2y1 + 2y2 + 2y3 = 0
or y1 + y2 + y3 = \(\frac{0}{2}\) = 0
or y1 + y2 + y3 = 0 ….(5)
From equation (4)
x1 + x2 + x3 = 3
0 + x3 = 3
∵ From equation (1) x1 + x2 = 0
∴ x3 = 3
Similarly 2 + x2 = 3
x2 = 1
Similarly x1 + 4 = 3
∴ x1 = 3 – 4 = -1
x1 = -1
From equation (5) y1 + y2 + y3 = 0
Putting the values of equation (1), (2) and (3) in equation (5) the following values are obtained—
y1 + y2 + y3 = 0
or 4 + y2 = 0 ∴ y2 = – 4
or – 2 + y3 = 0 ∴ y3 = 2
y1 – 2 = 0 ∴ y1 = 2
So we get the following values—
x1 = -1, y1 = 2
x2 = 1, y2 = – 4
x3 = 3, y3 = 2
Therefore the required points are (- 1. 2), (1, – 4) and (3, 2)

Question 5.
In what ratio does the point (- 4, 6) divide the line segment joining the points A(- 6, 10) and B(3, – 8)?
Solution:
Let (-4, 6) divide the line segment AB internally in the ratio of m1 : m2.
Then by using section formula
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 23
or -4m1 – 4m2 = 3m1 – 6m2
or -7m1 = -2m2 or m1 : m2 = 2 : 7
We shall have to examine whether this ratio satisfies y-coordinate also.
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 24
Hence the point (- 4. 6) divides the line segment joining the points A(- 6, 10) and B(3, – 8) in the ratio 2 : 7.

RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry

Question 6.
Find in what ratio does the line 3x + y = 9 divide the line segment joining the points (1, 3) and (2, 7)?
Solution:
Let the point of division divide the line segment joining the points (1, 3) and (2, 7) in the ratio λ : 1. Then the coordinates of the point of division are
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 25
This point will be on the line 3x + y = 9 and therefore it will satisfy it. So putting the values
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 26
⇒ 6λ + 3 – 7λ + 3 = 9(λ + 1)
⇒ 13λ + 6 = 9λ + 9
⇒ 13 λ – 9λ = 9 – 6
⇒ 4λ = 3
∴ λ = \(\frac{3}{4}\)
∴ Required ratio = \(\frac{3}{4}\) : 1 or 3 : 4

Question 7.
Find the coordinates of the points trisecting the line segment joining the points P (-3, 4) and Q (4, 5).
Solution:
Let R (x1, y1) and S (x2, y2) be the required points that trisect the line segment joining the points P (-3, 4) and Q (4, 5), i:e., R (x1, y1) divides PQ in the ratio 1 : 2 and S (x2, y2) divides PQ in the ratio 2 : 1.
RBSE Class 10 Maths Important Questions Chapter 7 Coordinate Geometry 27
Hence required points = R(\(-\frac{2}{3}\), \(\frac{13}{3}\)) and S(\(\frac{5}{3}\), \(\frac{14}{3}\))

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