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—

Answer:

(3, \(\frac{8}{5}\))

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

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 = \(\sqrt {4+25}\) = \(\sqrt {29}\)

∵ AB^{2} + CA^{2} = BC^{2}

⇒ A, B, C are the vertices of a right angled triangle.

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

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

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

or x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{2}) + x_{3}(y_{1} – y_{2}) = 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 :

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

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 + y^{2} – 10y= 16 + 9 + y^{2} – 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.

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 m_{1} : m_{2}. Then,

3 = \(\frac{m_{1} \times 6+m_{2} \times 1}{m_{1}+m_{2}}\)

3(m_{1} + m_{2}) = 6m_{1} + m_{2}

3m_{1} + 3m_{2} = 6m_{1} + m_{2}

3m_{2} – m_{2}= 6m_{1} – 3m_{1}

2m_{2} = 3 m_{1}

\(\frac{m_{1}}{m_{2}}\) = \(\frac{2}{3}\)

Hence m_{1} : m_{2} = 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 + k^{2} = k^{2} + 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. x_{1} = 2, x_{2} = k, x_{3} = 2k + 1

and y_{1} = 1, y_{2} = 1, y_{3} = 2

∵ The three points are collinear, therefore

x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0

⇒ 2(1 – 2) + k(2 – 1) + (2k + 1) (1 – 1) = 0

⇒ – 2 + k + 0 = 0

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

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.

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).

∵ 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.

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).

∵ 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:

Therefore, AB = CD and BC = AD and diagonal AC = diagonal BD. Hence the given points are the vertices of a rectangle.

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, \(\frac{-13}{3}\))

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 + y^{2} – 10y = 16 + 9 + y^{2} – 6y

or 4y = 36

or y = \(\frac{36}{4}\) = 9

Hence the desired point is (0, 9).

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 (x_{2}, y_{2}).

Mid-point is (4, 1). Therefore,

⇒ 4 × 2 = 4 + x_{2}

⇒ x_{2} = 8 – 4 = 4

and 1 = \(\frac{0+y_{2}}{2}\)

⇒ 2 = 0 + y_{2}

⇒ y = 2

∴ (x_{2}, y_{2}) = (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).

Now the distance of the point (4, 6) from the point (1, 2)

Hence, the required distance between the points will be 5.

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.

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}

⇒ x^{2} – 12x + 36 + 25 = x^{2} + 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).

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.

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

\(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

∴ 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.

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

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)

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

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 (x_{1}, y_{1}) and (x_{2}, y_{2})

Solving equations (1), (2) and (3)—

2x_{1} + 2x_{2} + 2x_{3} = 6

or x_{1} + x_{2} + x_{3} = 3 …. (4)

or 2y_{1} + 2y_{2} + 2y_{3} = 0

or y_{1} + y_{2} + y_{3} = \(\frac{0}{2}\) = 0

or y_{1} + y_{2} + y_{3} = 0 ….(5)

From equation (4)

x_{1} + x_{2} + x_{3} = 3

0 + x_{3} = 3

∵ From equation (1) x_{1} + x_{2} = 0

∴ x_{3} = 3

Similarly 2 + x_{2} = 3

x_{2} = 1

Similarly x_{1} + 4 = 3

∴ x_{1} = 3 – 4 = -1

x_{1} = -1

From equation (5) y_{1} + y_{2} + y_{3} = 0

Putting the values of equation (1), (2) and (3) in equation (5) the following values are obtained—

y_{1} + y_{2} + y_{3} = 0

or 4 + y_{2} = 0 ∴ y_{2} = – 4

or – 2 + y_{3} = 0 ∴ y_{3} = 2

y_{1} – 2 = 0 ∴ y_{1} = 2

So we get the following values—

x_{1} = -1, y_{1} = 2

x_{2} = 1, y_{2} = – 4

x_{3} = 3, y_{3} = 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 m_{1} : m_{2}.

Then by using section formula

or -4m_{1} – 4m_{2} = 3m_{1} – 6m_{2}

or -7m_{1} = -2m_{2} or m_{1} : m_{2} = 2 : 7

We shall have to examine whether this ratio satisfies y-coordinate also.

Hence the point (- 4. 6) divides the line segment joining the points A(- 6, 10) and B(3, – 8) in the ratio 2 : 7.

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

This point will be on the line 3x + y = 9 and therefore it will satisfy it. So putting the values

⇒ 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 (x_{1}, y_{1}) and S (x_{2}, y_{2}) be the required points that trisect the line segment joining the points P (-3, 4) and Q (4, 5), i:e., R (x_{1}, y_{1}) divides PQ in the ratio 1 : 2 and S (x_{2}, y_{2}) divides PQ in the ratio 2 : 1.

Hence required points = R(\(-\frac{2}{3}\), \(\frac{13}{3}\)) and S(\(\frac{5}{3}\), \(\frac{14}{3}\))

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