# Free Coordinate Geometry 03 Practice Test - 10th Grade

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

A.

(4, – 4)

B.

(– 3, 1)

C.

(– 4, 4)

D.

(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 = -4

and y = my2+ny1m+n = 2(7)+1(2)2+1 = 4

the required point is (-4,4).

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

A.

x+y

B.

x+y

C. x2+y2
D. x2y2

#### SOLUTION

Solution : C

The distance of a point (x,y) from the origin (0,0) is (x0)2+(y0)2=x2+y2

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

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

A.

3:1

B.

2:3

C.

7:2

D.

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+1,4k7k+1)

But C(x,0)=(6k+1k+1,4k7k+1) 4k7k+1=0

4k7 = 0 k = 74

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

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.

A.

5

B. 137
C.

32

D.

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(2k)+(3)(k1)+0(12)|

=12|84k3k+3|=12|117k|

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

|117k|=24

±(117k)=24            117k=24 or (117k)=24

If117k=24, then 7k=2411=13       k=137

(117k)=2411+7k=24

7k=24+11=35    k=357=5

Hence the values of k are : 5,   137

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

A.

6

B.

1

C.

4

D.

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

⇒  12x1( 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.

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

A.

2a2+b2

B.

2b2+c2

C.

22b

D.

2a2c2

#### SOLUTION

Solution : C

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

Distance between the points (x1,y1) and (x2,y2) is
(x2x1)2 + (y2y1)2

AB=(abab)2+(cbbc)2      =(2b)2+(2b)2=8b2AB=22 b

Distance between the points = 22 b

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

(xa1)2+(13)2) = (xa+5)2+(1+5)2
(xa1)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).

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.

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)

A.

a22

B.

a22

C.

a222

D.

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+x22x = 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.