# AP 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

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:
x1 = 2 ; x2 = 0 ; y1 = 0 ; y2 = 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 x1 = 0, y1 = a sin θ and
x2 = a cos θ, y2 = 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 + y2 = 2
y2 = 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).

[ ∵ 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 (x1, y1) and (x2, y2) is

by the sum, PA = PB

50 – 14x + x2 – 2y + y2
= 34 – 6x + x2 – 10y + y2
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
$$\frac { 1 }{ 2 }$${x1(y2-y3) + x2(y3-y1)
+ x3(y1 – y2)} = 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 = 32
-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 = $$\frac { 1 }{ 2 }$${x1(y2-y3) + x2(y3-y1)
= $$\frac { 1 }{ 2 }$$|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 cm2

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.

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)

∴ AB2 = 25, BC2 = 50, CA2= 25
BC2 = AB2 + CA2 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) = (x1, y1); C(9, 4) – (x2, y2)

ii) ar ΔABC = $$\frac { 1 }{ 2 }$$| x1 (y2 – y3) + x2(y3 – y1)
+ x3 (y1 -y2)|
= $$\frac { 1 }{ 2 }$$|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 (x1, y1) and (x2, y2) is $$\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$$
Here (x1, y1)=(5,7) and (x2, y2) = (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 (x1, y1) = (x2, y1) then x1 must be equal to x2 and y1 must be equal to y2.
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 + 72
= (p – 7)2 + 52
= p2 – 10p + 25 + 49 = p2 – 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 q1
(5, -7) belongs to fourth quadrant q4.
(-5, 7) belongs to second quadrant q2.
(-5, -7) belongs to third quadrant q3.

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