# Free Coordinate Geometry 03 Practice Test - 10th Grade

### Question 1

The co-ordinates of the points which divides the line joining (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is _____.

(4, – 4)

(– 3, 1)

(– 4, 4)

(1, – 3)

#### SOLUTION

Solution :C

If a point P(x,y) divides the line joining (x1,y1) and (x2,y2) in the ratio m:n

then x = mx2+nx1m+n and y = my2+ny1m+n

∴x = mx2+nx1m+n = 2(−5)+1(−2)2+1 = -4and y = my2+ny1m+n = 2(7)+1(−2)2+1 = 4

∴ the required point is (-4,4).

### Question 2

The distance of any point (x,y) from the origin is =

x+y

√x+y

#### SOLUTION

Solution :C

The distance of a point (x,y) from the origin (0,0) is √(x−0)2+(y−0)2=√x2+y2

### Question 3

What is the x coordinate of point C?

#### SOLUTION

Solution :Point C lies on the line 'l' which is parallel to Y-axis. So the x-coordinate of any point on this line will be constant.

Therefore its x coordinate is 4

### Question 4

The ratio in which the line segment joining the points (1, – 7) and (6, 4) is divided by x-axis is _____.

3:1

2:3

7:2

7:4

#### SOLUTION

Solution :D

Let C(x,0) divide AB in the ratio k:1.

By section formula, the coordinates of C are given by: (6k+1k+1,4k−7k+1)

But C(x,0)=(6k+1k+1,4k−7k+1) ⇒4k−7k+1=0

⇒4k−7 = 0 ⇒k = 74

i.e., the x-axis divides AB in the ratio 7:4.

### Question 5

The co-ordinates of the vertices of Triangle ABC are A(4,1),B(–3,2) and C(0,k). Given that the area of △ABC is 12 sq. units. Find the value(s) of k.

5

32

4

#### SOLUTION

Solution :A and B

Area of a triangle formed by (x1,y1), (x2,y2) (x3,y3) = 12|x1(y2 - y3) + x2(y3 - y1)+ x3(y1 - y2)|

Thus, the area of ΔABC formed by the given points A(4, 1), B(-3, 2) and C(0, k) is=12|4(2−k)+(−3)(k−1)+0(1−2)|

=12|8−4k−3k+3|=12|11−7k|

But given that the area of ΔABC=12 sq. unit

⇒|11−7k|=24

±(11−7k)=24 ⇒11−7k=24 or −(11−7k)=24

If11−7k=24, then −7k=24−11=13 ⇒k=−137

−(11−7k)=24⟹−11+7k=24

⇒7k=24+11=35 ⇒k=357=5

Hence the values of k are : 5, −137

### Question 6

The value of p for which the points (–1, 3), (2, p), (5, –1) are collinear is

6

1

4

8

#### SOLUTION

Solution :B

The given points A(-1, 3), B(2, p), C(5, -1) are collinear.

⇒ Area ΔABC formed by these points should be zero.

⇒ The area of ΔABC = 0

⇒ 12( x1( y2- y3) + x2( y3- y1) + x3( y1 - y2)) = 0

⇒ -1(p + 1) + 2 (-1 - 3) + 5(3 -p) = 0

⇒ -p - 1 - 8 + 15 - 5p = 0

⇒ -6p +15 - 9 = 0 ⇒ -6p = -6 ⇒ p = 1

Hence the value of p is 1.

### Question 7

The distance between the points (a + b, b + c) and (a – b, c – b) is :

2√a2+b2

2√b2+c2

2√2b

2√a2−c2

#### SOLUTION

Solution :C

Let, A be (a+b, b+c) and B be (a−b, c−b)

∵ Distance between the points (x1,y1) and (x2,y2) is

√(x2−x1)2 + (y2−y1)2

∴AB=√(a−b−a−b)2+(c−b−b−c)2 =√(−2b)2+(−2b)2=√8b2AB=2√2 b

∴ Distance between the points = 2√2 b

### Question 8

Points P and Q are two points in the coordinate plane. The x-coordinate of the point equidistant from the points P and Q and lying on the line y = -1 is

#### SOLUTION

Solution :The required point is equidistant from the points P and Q, and also lies on the line y = -1.

Let the point be A, it lies on y = -1

Therefore consider the point A(x_a, -1)

Distance AP = Distance AQ

√(xa−1)2+(−1−3)2) = √(xa+5)2+(−1+5)2

(xa−1)2 + (−4)2 = (xa+5)2 + (4)2

-2xa + 1 = 10xa + 25

-12xa = 24

xa = -2

The coordinates of the point A is (-2, -1).

### Question 9

In the figure below, is any side of ΔABC parallel to X axis? If so, the side is

#### SOLUTION

Solution :Any line whose y-coordinate is a constant will be parallel to the X axis. Observe that B

and C have the same y coordinate, c, and hence this side BC is parallel to the X-axis.Similarly, A and B have the same x – coordinates, -a, hence this side is parallel to the Y axis.

### Question 10

A, B, C are the three vertices of a triangle such that they are equidistant from the origin. A and B lie on the positive X and Y axes respectively and C lies on the line y + x = a. If AB = √2a, find the area of the triangle. (C lies in the first quadrant)

a2√2

a22

a22√2

0

#### SOLUTION

Solution :D

A and B are equidistant from origin and AB is √2a.

If we assume the points A and B to be (x,0) and (0,x) respectively,

AB= √x2+x2 = √2x = √2a

x=a

So A(0,a) and B(a,0) are the vertices.

Observe that A,B both lie on the line x+y=a and since C also lies on the line y+x = a, they are collinear.

Hence area of the triangle formed by the three points is 0.