Haryana State Board HBSE 10th Class Maths Important Questions Chapter 7 Coordinate Geometry Important Questions and Answers.

## Haryana Board 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

Short/Long Answer Type Questions

Question 1.

The distance between the point A(5, -3) and B(13, m) is 10 units. Calculate the value of m.

Solution :

Here, x_{1} = 5, y_{1} = – 3, x_{2} = 13, y_{2} = m

Distance (AB) = 10

⇒ \(\sqrt{(13-5)^2+(m+3)^2}\) = 10

⇒ \(\sqrt{8^2+m^2+9+6 m}\) = 10

⇒ \(\sqrt{64+m^2+9+6 m}\) = 10

⇒ \(\sqrt{73+m^2+6 m}\) = 10

⇒ [Squaring both sides]

⇒ m^{2} + 6m + 73 – 100 = 0

⇒ m^{2} + 6m – 27 = 0

⇒ m^{2} + (9m – 3m) – 27 = 0

⇒ m^{2} + 9m – 3m – 27 = 0

⇒ m(m + 9) – 3(m + 9) = 0

⇒ (m + 9) (m – 3) = 0

⇒ m + 9 = 0 or m – 3 = 0

⇒ m = – 9 or m = 3

Question 2.

If the distances of P(x, y) from A(5, 1) and B(-1, 5) are equal, then prove that 3x

Solution :

Since, of from A(5, 1) and B(-1, 5) are equal

So, PA = PB

⇒ PA^{2} = PB^{2}

⇒ (5 – x)^{2} + (1 – y)^{2} = (-1 – x)^{2} + (5 – y)^{2}

⇒ 25 + x^{2} – 10x + 1 + y^{2} – 2y = 1 + x^{2} + 2x + 25 + y^{2} – 10y

⇒ 26 – 10x – 2y = 26 + 2x – 10y

⇒ – 10x – 2x = – 10y + 2y

⇒ – 12x = – 8y

⇒ 3x = 2y Hence proved.

Question 3.

It 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 coordinate of point P be (x, y)

Since k(5, 4) is mid point of P(x, y) and Q(2, 3)

∴ 5 = \(\frac{x+2}{2}\) and 4 = \(\frac{y+3}{2}\)

⇒ 10 = x + 2 and 8 = y + 3

⇒ x = 10 – 2 and y = 8 – 3

⇒ x = 8 and y = 5

∴ coordinates of point P are (8, 5)

Question 4.

If the midpoint of two points A(2, 5) and B(-5, y) is (-\(\frac {7}{2}\), 3). then find the distance between points A and B.

Solution :

Let point C(-\(\frac {7}{2}\), 3) is the midpoint of line AB.

∴ 3 = \(\frac{5+y}{2}\)

⇒ 6 = 5 + y

⇒ y = 6 – 5 = 1

Coordinates of point is (-5, 1)

Length of AB = \(\sqrt{(-5+2)^2+(1-5)^2}\)

= \(\sqrt{(-3)^2+(-4)^2}\)

= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5

Hence, length of AB = 5 units.

Question 5.

In a parallelogram ABCD A(3, 1), B(5, 1), C(a, b) and D(4, 3) are the vertices. Find vertex C(a, b).

Solution :

We know that diagonals of a parallelogram bisects each other so, O is the mid point of AC as well as that of BD

∴ Coordinates of mid point of AC = coordinates of mid point of BD

⇒ (\(\left(\frac{3+a}{2}, \frac{1+b}{2}\right)\)) = (\(\left(\frac{5+4}{2}, \frac{1+3}{2}\right)\))

⇒ (\(\left(\frac{3+a}{2}, \frac{1+b}{2}\right)\)) = (\(\frac {9}{2}\), \(\frac {4}{2}\))

⇒ \(\frac{3+a}{2}\) = \(\frac {9}{2}\) and \(\frac{1+b}{2}\) = \(\frac {4}{2}\)

⇒ 3 + a = 9 and l + b = 4

⇒ a = 9 – 3 and b = 4 – 1

⇒ a = 6 and b = 3

Hence, coordinates of vertex C is (6, 3)

Question 6.

If the coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect each other at (2, – 5), find the coordinates of the other two vertices.

Solution :

Diagonals of a parallelogram bisect each other.

Let coordinates of vertices C be (x_{1}, y_{1}) and d be (x_{2}, y_{2}) O is the mid point of AC as well as that of BD.

∴ coordinates of mid point of AC = (2, – 5)

⇒ (\(\frac{3+x_1}{2}, \frac{2+y_1}{2}\)) = (2, – 5)

⇒ \(\frac{3+x_1}{2}\) = 2, \(\frac{2+y_1}{2}\) = – 5

⇒ 3 + x_{1} = 4, 2 + y_{1} = – 10

⇒ x_{1} = 1, y_{1} = – 12

And coordinates of mid point of BD = (2, – 5)

⇒ (\(\frac{1+x_2}{2}, \frac{0+y_2}{2}\)) = (2, – 5)

⇒ \(\frac{1+x_2}{2}\) = 2, \(\frac{0+y_2}{2}\) = – 5

⇒ 1 + x_{2} = 4, y_{2} = – 10

⇒ x_{2} = 3, y_{2} = – 10

Coordinates of vertice Care (1, -12) and vertice D are (3, – 10).

Question 7.

If the midpoint of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y) and x + y – 10 = 0. Find the value of k.

Solution :

Since, mid point’s coordinates are C(x, y)

∴ x = \(\frac{3+k}{2}\) and y = \(\frac{4+6}{2}\) = 5

Since (x, y) lies the equation x + y – 10, so

substituting the value, A(3, 4) of x, y in this equ., we get

\(\frac{3+k}{2}\) + 5 – 10 = 0

⇒ \(\frac{3+k}{2}\) – 5=0

⇒ \(\frac{3+k-10}{2}\) = 0

⇒ k – 7 = 0

⇒ k = 7

Hence, value of k is 7.

Question 8.

The coordinate of the vertices of ΔABC are A (7, 2), B (9, 10) and C(1, 4). If E and F are the mid points of AB and AC respectvely, prove that EF = \(\frac {1}{2}\)BC

Solution :

Since, E and f are the mid points of AB and AC respectvely coordinates of points are

\(\frac{9+7}{2}, \frac{10+2}{2}\) = (8, 6)

Coordinates of points Fare

\(\frac{1+7}{2}, \frac{4+2}{2}\) = (4, 3)

Question 9.

If the point C(-1, 2) divides in ternally the time segments joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.

Solution :

Hence, x_{1} = 2, y_{1} = 5, x_{2} = x, y_{2} = y, x = – 1, y = 2, m_{1} = 3, m_{2} = – 4

⇒ 3x + 8 = – 7 and 3y + 20 = 14

⇒ 3x = – 7 – 8 and 3y = 14 – 20

⇒ 3x = -15 and 3y = -6

⇒ x = –\(\frac {15}{3}\) and y = – \(\frac {6}{3}\)

⇒ x = – 5 and y = – 2

Hence, coordinate of point Bare (-5, -2)

Question 10.

In what ratio does the point P (\(\frac {24}{11}\), y) divide the line segment joining the points P(2, -2) and Q(3, 7) ? Also find the value y.

Solution :

Let the required ratio be k : 1

Here, x_{1} = 2, y = – 2, x_{2} = 3, y_{2} = 7, x = \(\frac {24}{11}\), y = y

m_{1} = k, m_{2} = 1

By section formula, we have

Question 11.

Find the value of P, if the points A(2, 3) B(4, P) and C(6, – 3) are collinear.

Solution :

The given points are collinear So, the area of ΔABC will be zero.

Here, x_{1} = 2, y_{1} = 3, x_{2} = 4, y_{2} = P, x_{3} = 6, y_{3} = – 3

Area of ΔABC = 0

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

⇒ \(\frac {1}{2}\)[2 (P + 3) + 4 (-3 – 3) + 6 (3 – P)]= 0

⇒ \(\frac {1}{2}\)[2P + 6 – 24 + 18 – 6P] = 0

⇒ \(\frac {1}{2}\)[-4P + 0] = 0

⇒ – 4P = 0

⇒ P = 0

Question 12.

Find the area of triangle ABC with A(1, – 4) and the mid points of sides through A being (2, – 1) and (0, – 1).

Solution :

Let the coordinates of vertices B be (x_{2}, y_{2}) and C be (x_{3}, y_{3})

Let E (2, – 1) and f(0, – 1) are mid points of AB and AC respectively,

∴ 2 = \(\frac{1+x_2}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)

⇒ 4 = 1 + x_{2} and – 2 = – 4 + y_{2}

⇒ x_{2} = 3 and y_{2} = 2

∴ (x_{2}, y_{2}) = (3, 2)

Again 0 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)

⇒ 0 = 1 + x_{3} and – 2 = – 4 + y_{3}

⇒ x_{3} = – 1 and y_{3} = 2

∴ (x_{3}, y_{3}) = (-1, 2)

Here, x_{1} = 1, y_{1} = – 4, x_{2} = 3, y_{2} = 2, x_{3} = -1, y_{3} = 2

Area of ABC

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

= \(\frac {1}{2}\)[1(2 – 2) + 3 (2 + 4) + (-1)(-4 – 2)]

= \(\frac {1}{2}\)[1 × 0 + 3 × 6 + (-1) × (- 6)]

= \(\frac {1}{2}\)[0 + 18 + 6]

= \(\frac {1}{2}\) × 24 = 12

Hence, area of ABC = 12sq. units.

Question 13.

Prove that the points (2, -2), (2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the area of this trian.

Solution :

Let the points A (2, -2), B (-2, 1) and C (5, 2) are the vertices of right angled triangle.

AB^{2} = (-2 – 2)^{2} + (1 + 2)^{2}

= 16 + 9 = 25

BC^{2} = (5 + 2)^{2} + (2 – 1)^{2} = 49 + 1 = 50

And AC^{2} = (5 – 2)^{2} + (2 + 2)^{2}

= 9 + 16 = 25

AB^{2} + AC^{2} = 25 + 25 = 50 = BC^{2}

Since, BC^{2} = AB^{2} + AC^{2}, So by converse of phythagores theorem ∠A = 90° there fore point A (2, -2), B (-2, 1) and C (5, 2) are the vertices of a right angled triangle

Here, x_{1} = 2, y_{1} = -2, x_{2} = -2, y_{2} = – 1, x_{3} = 5, y_{3} = 2

Area of ΔABC

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

= \(\frac {1}{2}\)[2(1 – 2) + (-2)(2 + 2) + 5(-2 – 1)]

= \(\frac {1}{2}\)[2 × (-1) + (-2) × 4 + 5 × (-3)]

= \(\frac {1}{2}\)[- 2 – 8 – 15]

= \(\frac {1}{2}\)[-25]

= \(\frac {-25}{2}\)

Hence, area of Δ = \(\frac {25}{2}\)sq. units.

Question 14.

Find the area of that triangle whose vertices are (-3, -2), (5, -2) and (5, 4). Also prove that it is a right-angled triangle

Solution :

Let the points of vertices be A(-3, – 2), B (5, -2) and C (5, 4)

Here, x_{1} = – 3, y_{1} = – 2, x_{2} = 5, y_{2} = – 2, x_{3} = 5 and y_{3} = 4

Area of D ABC

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

= \(\frac {1}{2}\)[(-3)(-2 – 4) + 5 (4 + 2) + 5(-2 + 2)]

= \(\frac {1}{2}\)[(-3) × (-6) + 5 × 6 + 5 × 0)]

= \(\frac {1}{2}\)[18 + 30 + 0]

= \(\frac {1}{2}\) × 48 = 24 sq. units

AB^{2} = (5 + 3)^{2} + (-2 + 2)^{2} = 8^{2} + 0^{2} = 64

BC^{2} = (5 – 5)^{2} + (4 + 2)^{2} = 0^{2} + 6^{2} = 36

AC^{2} = (5 + 3)^{2} + (4 +2)^{2} = 8^{2} + 6^{2} = 64 + 36 = 100

AB^{2} + BC^{2} = 64 + 36 = 100 = AC^{2}

By converse of phythagoras theorem ∠B = 90° Hence, ΔABC is a right angled triangle.

Question 15.

In a ΔABC, A is (1, – 4). If E (0, -1) and Δ(2,-1) are the mid points of AB and AC. Calculate the area of ΔABC.

Solution :

Let the coordinates of vertices B is (x_{2}, y_{2}) and C is (x_{3}, y_{3}) since, E is the mid point of AB

∴ 0 = \(\frac{1+x_2}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)

⇒ 0 = \(\frac{1+x_2}{2}\) and – 2 = – 4 + y_{2}

⇒ x_{2} = -1 and y_{2} = – 2 + 4 = 2

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

Since, D is the midpoint of AC

∴ 2 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)

⇒ 0 = 1 + x_{2} and – 2 = – 4 + y_{2}

⇒ x_{2} = -1 and y_{2} = – 2 + 4 = 2

∴ (x_{3}, y_{3}) = (-1, 2)

Since, D is the midpoint of AC.

∴ 2 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_3}{2}\)

4 = 1 + x_{3} and -2 = – 4 + y_{3}

x_{3} = 3 and y_{3} = 4 – 2 = 2

∴ (x_{3}, y_{3}) = (3, 2)

Here, x_{1} = 1, y_{1} = – 4, x_{2} = – 1, y_{2} = 2, x_{3} = 3, y_{3} = 2

Area of ΔABC

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

= \(\frac {1}{2}\)[1(2 – 2) + (-1) (2 + 4) + 3(-4 – 2)]

= \(\frac {1}{2}\)[1 × 0 + (-1) × 6 + 3 × (-6)]

= \(\frac {1}{2}\)[0 – 6 – 18]

= \(\frac {1}{2}\) × (-24) = – 12

Hence, area of ΔABC = 12 sq. units.

Question 16.

Find the area of the triangle formed by joining the mid points of the sides of a triangle, whose coordinates of vertices are (0, -1), (2, 1) and (0, 3).

Solution :

Let vertices of triangle be A (0, -1), B (2, 1) and (0, 3) in which of sides AB, BC and AC respectvely.

∴ Coordinates of point

(D) = \(\frac{1+0}{2}, \frac{1-1}{2}\) = (1, 0)

Coordinates of point

E = \(\frac{2+0}{2}, \frac{1+3}{2}\) = (1, 2)

And Coordinates of point

F = \(\frac{0+0}{2}, \frac{3-1}{2}\) = (0, 1)

Here, x_{1} = – 1, y_{1} = 0, x_{2} = 1, y_{2} = 2, x_{3} = 0, y_{3} = 1

Area of ΔDEF

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

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

= \(\frac {1}{2}\)[1 × 1 + 1 × 1 + 0]

= \(\frac {1}{2}\) × (1 + 1)

= \(\frac {1}{2}\) × 2 = 1 sq. unit.

Hence, area of ΔDEF = 1sq. unit.

Fill in the Blanks

Question 1.

The point of intersection of two axes is called ……….

Solution :

origin

Question 2.

The distance of a point from the y axis is known its ………….

Solution :

abscissa

Question 3.

The ……… of a line segment divides the line segment in the ratio 1 : 1

Solution :

midpoint

Question 4.

If the area of a triangle is ……… sequare unit, then its vertices will be collinear.

Solution :

zero

Question 5.

The distance of a point from the x axis is called its …….

Solution :

ordinate

Question 6.

The distance of a point P(x, y) from the ………. is \(\sqrt{x^2+y^2}\)

Solution :

origin.

Multiple Choice Questions

Question 1.

The coordinates of the point which is reflection of point (-3, 5) in x axis are :

(a) (3, 5)

(b) (3, -5)

(c) (-3, -5)

(d) (-3, 5)

Solution :

(d) (-3, 5)

In graph paper we observe reflection of (-3, 5) in axis is (-3, -5).

Question 2.

The point P on x axis equidistant from the A(-1, 0) and B(5, 0) is:

(a) (2, 0)

(b) (0, 2)

(c) (3, 0)

(d) (2, 2)

Solution :

(a) (2, 0)

Since point P on x axis. So the coordinates of P(x, 0)

∵ Point P is equidistant from point A and point B.

∴ PA = PB

⇒ PA^{2} = PB^{2}

⇒ (-1 – x)^{2} + (0 – 0)^{2} = (5 – x)^{2} + (0 – 0)^{2}

⇒ 1 + x^{2} + 2x + 0 = 25 + x^{2} – 10x + 0

⇒ x^{2} + 2x – x^{2} + 10x = 25 – 1

⇒ 12x = 24

⇒ x = \(\frac {24}{12}\) = 2

Coordinates of point P(2, 0).

Question 3.

The distance between the points (acosθ + b sinθ, 0) and (0, asinθ – bcosθ), is :

(a) a^{2} + b^{2}

(b) a^{2} – b^{2}

(c) \(\sqrt{a^2+b^2}\)

(d) \(\sqrt{a^2-b^2}\)

Solution :

(c) \(\sqrt{a^2+b^2}\)

Let the given points A (a cosθ + b sinθ, 0) and (o, a sinθ – b cosθ)

The distance AB

Question 4.

If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is:

(a) 4

(b) 3

(c) 2

(d) 1

Solution :

(d) 1

The point P(6, 2) divides the line segment joining A(6, 2) and B(4, y) in the ratio 3 : 1

Here x_{1} = 6, y_{1} = 5, x_{2} = 4, y_{2} = y, m_{1} = 3, m_{2} = 1, x = 6, y = 2

By section formula

Question 5.

If the point P(k, 0) divides the line segment joining the points A(2, – 2) and B(-7, 4) in the ratio 1 : 2 then the value of k is :

(a) 1

(b) 2

(c) – 2

(d) – 1

Solution :

(d) – 1

Here, x_{1} = 2, y_{1} = – 2, x_{2} = – 7, y_{2} = 4, x = k, y = 0, m_{1} = 1, m_{2} = 2.

By section formula,

Question 6.

The value of P, for which the points A(3, 1), B(5, P) and C (7, – 5) are collinear is :

(a) – 2

(b) 2

(c) – 1

(d) 1.

Solution :

(a) – 2

the given points are collinear. So, area of ΔABC is zero.

Here, x_{1} = 3, y_{1} = – 1, x_{2} = 5, y_{2} = P, x_{3} = 7, y_{3} = – 5

Area of ΔABC = 0

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

⇒ \(\frac {1}{2}\)[3(P + 5) + 5(-5 – 1) + 7(1 – P)] = 0

⇒ \(\frac {1}{2}\)[3P + 15 + 5(-6) + 7 – 7P] = 0

⇒ \(\frac {1}{2}\)[3P + 15 – 30 + 7 – 7P] = 0

⇒ \(\frac {1}{2}\)[-4P – 8] = 0

⇒ -4P = 8

⇒ P = \(\frac {8}{- 4}\)

⇒ P = – 2