These AP 10th Class Maths Chapter Wise Important Questions Chapter 7 Coordinate Geometry will help students prepare well for the exams.

## AP State Syllabus 10th Class Maths 7th Lesson Important Questions and Answers Coordinate Geometry

Question 1.

What do you mean by centroid of a triangle ?

Solution:

“The concurrent point of medians of a triangle is called centroid of the triangle”.

Question 2.

Find the co-ordinates of the point, which divides the line segment joining (2, 0) and (0, 2) in the ratio 1:1.

Solution:

x_{1} = 2 ; x_{2} = 0 ; y_{1} = 0 ; y_{2} = 2

Point divides the line segment joining (2,0) and (0, 2) in the ratio 1 : 1

Question 3.

Find the distance between (a cos θ, 0) and (0, a sin θ).

Solution:

To find the distance between the points (a cos θ, 0) and (0, a sin θ) we use the formula.

\(\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}\)

Where x_{1} = 0, y_{1} = a sin θ and

x_{2} = a cos θ, y_{2} = 0

∴ The distance between above given two points

Question 4.

If A(4,0), B(0, y) and AB = 5, find the possible values of y. d

Solution:

A(4,0), B(0, y) and AB = 5

\(\) = 5

\(\) = 5

16 + y^{2} = 2

y^{2} = 25 – 16 = 9

y = ± \(\sqrt{9}\) = ± 3

Possible values of y are 3 or – 3.

Question 5.

Find the radius of the circle with cen¬tre (3, 2) and passes through (4, – 1).

Solution:

Question 6.

Let the 3 vertices of a triangle ABC are A(3, -2), B(-5, 4) and C(2, – 2). What do you observe for the centroid of this triangle ?

Solution:

centroid of this triangle

= \(\left(\frac{\mathbf{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)\)

= \(\left(\frac{3+(-5)+2}{3}, \frac{-2+4+(-2)}{3}\right)\) = (0,0)

Observed that the centroid is the origin.

Question 7.

Find the centroid of a triangle, whose vertices are (6, 2), (0, 0) and (4, -5).

Solution:

Given vertices are (6, 2) (0, 0) and (4, -5) then centroid

Question 8.

Find the radius of the circle whose cen-tre is (3, 2) and passes through (- 5, 6).

Solution:

Given : A circle with centre A (3, 2) passing through B (- 5, 6).

Radius = AB

[ ∵ Distance of a point from the centre of the circle]

Distance formula

Question 9.

What is the distance between (0, – sin x) and (- cos x, 0) ?

Solution:

Distance between (0, – sin x) and (- cos x, 0)

Question 10.

Where do the points (0, -3) and (-8, 0) lie on co-ordinate axis ?

Solution:

The point (0, – 3) lie on OY

∵ Its x co-ordiante is zero and

y – coordinate is negative.

and the point (-8, 0) lie on OX’

∵ Its y – cordinate is zero and

x-cordinate is negative.

Question 11.

Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Solution:

Let P(x, y); A(7, 1); B(3, 5)

The distance between points (x_{1}, y_{1}) and (x_{2}, y_{2}) is

by the sum, PA = PB

50 – 14x + x^{2} – 2y + y^{2}

= 34 – 6x + x^{2} – 10y + y^{2}

50 – 14x – 2y – 34 + 6x 4- 10y = 0

-8x + 8y + 16 = 0

-x + y + 2 = 0 ,

∴ x = y + 2

Question 12.

Find the value of k, for which the points (7, 2), (5, 1) and (3, k) are collinear.

Solution:

The area of the triangle formed by those points = 0

Area of the triangle

\(\frac { 1 }{ 2 }\){x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1})

+ x_{3}(y_{1} – y_{2})} = 0

\(\frac { 1 }{ 2 }\)|7(1 – k) + 5(k – 2) + 3(2 – 1)| = 0

\(\frac { 1 }{ 2 }\) |-2k| = 0

∴ k = 0

Question 13.

Find the centroid of the triangle, whose vertices are (-4, 4), (-2, 2) and (6,-6).

Solution:

Centroid of the triangle

= (0,0)

Question 14.

If the distance between two points (x, 1) and (-1,5) is ‘5’, find the value of V.

Solution:

Given points (x, 1) and (-1, 5)

Let A (x, 1) and B (-1, 5)

Distance between

Now, squaring on both sides (-1 -x)^{2} + 16 = 25

(-1 – x)^{2} = 25- 16 = 9

(-1 – x)^{2} = 3^{2}

-1 – x = 3

-1 – 3 = x ⇒ x = -4

Question 15.

Verify whether the following points are collinear or not.

(1,-1), (4, 1), (-2,-3)

Solution:

To show that three points are collinear the area formed by the triangle is zero.

Given points are (1, -1), (4, 1), (-2, -3) Formula for area of triangle

A = \(\frac { 1 }{ 2 }\){x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1})

= \(\frac { 1 }{ 2 }\)|1(1 +3) + 4(-3 + 1)

-2 (-1-1)1

= \(\frac { 1 }{ 2 }\)|4-8 + 4|.

= \(\frac { 1 }{ 2 }\) |8 – 8| = \(\frac { 1 }{ 2 }\)|0| = 0

So the given three points are collinear.

Question 16.

Find the area of a triangle, whose sides are 5 cm, 12 cm and 13 cni, by using Heron’s formula.

Solution:

Let a = 5 cm; b = 12 cm; :c =13 cm

s= \(\frac{a+b+c}{2}=\frac{5+12+13}{2}\) = 15

Area of triangle

(Δ) = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{15(15-5)(15-12)(15-13)}\)

= 30 cm^{2}

Question 17.

Find out whether the points (1,5), (2, 5) and (-2,-1) are collinear using the distance formula.

Solution:

A(1, 5), B(2, 5), C(- 2,-1)

Distance formula,

Question 18.

Read the following graph and answer the questions given below.

i) Write the coordinates of the points A and B.

ii) What is the slope of the line \(\overline{\mathbf{A B}}\) ?

Solution:

i) Coordinates of

point ‘A’ = (0, 2)

point B = (-3, 0)

ii) Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{0-2}{-3-0}=\frac{-2}{-3}=\frac{2}{3}\)

Question 19.

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:

Question 20.

Check whether the points (3, 0), (6,4) and (—1, 3) are the vertices of a right – angled isosceles triangle or not. Also find the area of the triangle.

Solution:

Distance between A(3, O), B(6, 4)

∴ AB^{2} = 25, BC^{2} = 50, CA^{2}= 25

BC^{2} = AB^{2} + CA^{2} and AB = CA

∴ ∆ ABC is an Isosceles right angled triangle

∴ Area of ∆ABC = \(\frac { 1 }{ 2 }\) x AB x AC

= \(\frac { 1 }{ 2 }\) x 5 x 5 = 12.5sq.u

Question 21.

Find the area of the triangle formed by the points (2, 3), (- 1, 3)and (2,-1) using Heron’s formula.

Solution:

To find the area of the triangle formed by (2, 3) (- 1, 3) and (2,-1) using Heron’s formula.

Let the co-ordinates of A = (2, 3) ; B = (- 1, 3); C = (2, – 1) then the sides of AABC are represented by as follows

AB = c, BC = a, CA = b then the formula of the triangle using Heron’s formula

= \(\sqrt{s(s-a)(s-b)(s-c)}\)

where s = \(\frac{a+b+c}{2}\)

Now, we find the sides of ∆ABC, using the formula \(\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\)

∴ CB = a = distance between the points (2, – 1) and (-1,3)

⇒ CB = a = 5 ………………..(1) .

and AB = c the distance between (-1,3) and (2, 3)

∴ AB = c = 3 …………………..(2)

and AC = b the distance between A(2, 3) and C(2, – 1)

AC = b = 4 …………..(3)

∴ Area of above triangle = 6 sq.units.

Question 22.

If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a paral-lelogram, taken in order, then

i) find the value of p.

ii) find the area of ABCD.

Solution:

i) A, B, C, D are the vertices of a parallelogram

Mid point of AC =- Mid point of BD

A(6, 1) = (x_{1}, y_{1}); C(9, 4) – (x_{2}, y_{2})

ii) ar ΔABC = \(\frac { 1 }{ 2 }\)| x_{1} (y_{2} – y_{3}) + x_{2}(y_{3} – y_{1})

+ x_{3} (y_{1} -y_{2})|

= \(\frac { 1 }{ 2 }\)|6(2 – 4) + 8(4 – 1) + 9(1 – 2)|

= \(\frac { 1 }{ 2 }\) |-12+24-9|

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

= \(\frac { 3 }{ 2 }\)|sq. units.

∴ Area of parallelogram ABCD = 2 x ar AABC 3

= 2 × \(\frac { 3 }{ 2 }\) = 3 sq. units.

Question 23.

Find the value of ‘k’ for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k -2) are collinear.

Solution:

Given points are collinear

Let A (3k – 1, k – 2), B (k, k – 7) and

C (k – 1, -k – 2) are on the same line \(\overleftrightarrow{L N}\)

Then slope of \(\overline{\mathrm{AB}}\) and slope of \(\overline{\mathrm{AC}}\) should be same (∵ they are collinear) Formula for slope

Now (1) = (2)

⇒ \(\frac{-5}{1-2 k}\) = 1

⇒ 1 – 2k = -5

⇒ 1 + 5 = 2k ⇒ 2k = 6

∴ k = 3

Question 24.

Find the co-ordinates of the points of trisection of the line segment joining the points A(2, 1) and B(5, – 8).

Solution:

Let P and Q be the points of trisection of AB, i.e., AP = PQ = QB.

Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordi¬nates of P are (by applying the section formula)

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

Therefore, the coordinates of the points of trisection of the line segment are P(3, – 2) and Q(4, – 5).

Question 25.

Find the distance between the points (5, 7) and (7, 5).

Solution:

Formula for the distance between the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is \(\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\)

Here (x_{1}, y_{1})=(5,7) and (x_{2}, y_{2}) = (7, 5)

∴ The distance between (5, 7) and (7, 5) is

Question 26.

Find the distance between the points (5, 7) and (7, 5) by plotting them in co-ordinate plane with the help of a right angled triangle.

Solution:

A (5, 7), B (7, 5) then O (5, 5)

In the ΔAOB, ∠O = 90

∴ AB is hypotenuse.

AO =(7 – 5)= 2

OB = 7 – 5 = 2

Then from Pythagorus theorem

Hence the distance between (5, 7) and (7, 5) = 2\(\sqrt{2}\) units

Question 27.

Are the points (5, 7) and (7, 5) equal ?

Solution:

No, the given points (5, 7) and (7, 5) are not equal.

Why because, the above two points represent two different points in the co-ordinate plane.

So they are not equal.

For (x_{1}, y_{1}) = (x_{2}, y_{1}) then x_{1} must be equal to x_{2} and y_{1} must be equal to y_{2}.

So to become (5, 7) = (7, 5)

5 should be equal to 7, which is impossible.

So (5, 7) will not bp equal to (7, 5).

Question 28.

Find the point on X-axis which is equi distant from the points (5, 7) and (7,5).

Solution:

Let the point (p, q) i$ on X-axis, which is equi distant from the points (5, 7) and (7, 5).

As this point (p, q) is on X-axis, its y-coordinate q = 0. Then the point is

(p, 0).

The distance between (5, 7) and (p, 0) is \(\sqrt{(p-5)^{2}+(0-7)^{2}}\) …………… (1)

And the distance between (7, 5) and (p,0) is \(\sqrt{(p-7)^{2}+(0-5)^{2}}\)……………. (2)

The above (1) and (2) are equal.

= (p – 5)^{2} + 7^{2}

= (p – 7)^{2} + 5^{2}

= p^{2} – 10p + 25 + 49 = p^{2} – 14p + 49 + 25

∴ -10p + 14p = 0

⇒ 4p = 0

∴ P = o

So (p, 0) = (0, 0) is the point on X-axis, which is equidistant from (5, 7) and (7, 5).

Question 29.

To which quadrants do the following points belong ?

(i) (5, 7)

(ii) (5, – 7)

(iii) (-5, 7)

(iv) (-5, -7)

Solution:

(5, 7) belongs to first quadrant q_{1}

(5, -7) belongs to fourth quadrant q_{4}.

(-5, 7) belongs to second quadrant q_{2}.

(-5, -7) belongs to third quadrant q_{3}.

Question 30.

What will be the lengths of line seg-ments that are parallel to X-axis, as shown in figure ?

Solution:

In the above figure, length of the line segment above the X-axis is 4 – 2 = 2

And length of line segment below X-axis is 4 – 1 = 3