# Free Coordinate Geometry 01 Practice Test - 10th Grade

If the distance between the points P (x, 2) and Q (3, –­6) is 10 units, then x = _____.

A.

3

B.

-3

C.

9

D.

6

#### SOLUTION

Solution : B and C

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

PQ=(3x)2+(62)2=10(3x)2+(8)2=100(3x)2+64=100(3x)2=363x=±6x=3 or 9

The co-ordinate of the point which divides the line joining the points (–1, 7) and (4, –3) in the ratio 2:3 is ________.

A.

(1, 3)

B.

(-1, 3)

C.

(1, -3)

D.

(-1, -3)

#### SOLUTION

Solution : A

If a point P(x,y)  divides a line segment joining (x1,y1) and (x2,y2) in the ratio m:n, then the coordinates of P are given by:

x=mx2+nx1m+n, y=my2+ny1m+n

Let P(x,y) divide the line segment AB joining A(1,7) and B(4,3) in the ratio 2:3. Then by using section formula, the co-ordinates of P are given by:

(2×4+3×(1)2+3,2×(3)+3×72+3)
=(835,6+215)
=(55,155)

Thus, the coordinates of P is (1,3)

Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is:

A.

(– 7, 6)

B.

(2, – 2)

C.

(7, – 6)

D.

(– 2, 2)

#### SOLUTION

Solution : B

Midpoint of the line segment joining (x1,y1) and (x2,y2) is given by (x1+x22,y1+y22)
Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) = (5+92,482) = (2,-2)

If the points (a, 0), (0, b) and (1, 1) are collinear, then which of the following is true?

A.

1a1b=2

B.

1a1b=1

C.

1a1b=0

D.

1a1b=4

#### SOLUTION

Solution : B

For 3 points to be collinear the area of the triangle formed by these three points should be zero.

Area of a triangle formed by (x1,y1), (x2,y2), (x3,y3) is given by
12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

12[a(b1)+0(10)+1(0b)]=012[a(b1)+1(0b)]=0ab=a+b1a+1b = 1

The point on Y-axis equidistant from (–3, 4) and (7, 6) is  ___

A.

(0, 15)

B.

(0, 14)

C.

(0, 13)

D.

(15 , 0)

#### SOLUTION

Solution : A

Any point on the y-axis has coordinatesof the form (0,y).Since the point is equidistant from bothpoints (3,4) and (7,6) by distance formula, 32+(y4)2=72+(y6)2168y=40+3612y4y=60y=15 the point is (0,15).

The area of a triangle having vertices (a, b+ c), (b, c + a) and (c, a + b) is ___.

A.

0

B.

1

C.

a + b + c

D.

a2 + b2 + c2

#### SOLUTION

Solution : A

Area of a triangle with vertices (x1,y1),(x2,y2) and (x3,y3) is given by
Area=12×|(x1)(y2y3)+(x2)(y3y1)+(x3)(y1y2)|

Thus, Area of the given triangle
=12|a(c+aab)+b(a+bbc)+c(b+cca)|
=12|acab+abbc+bcac|=0

In what ratio the line segment joining the points  (-1, 3) and (4, -7) divided by the point (2, -3)?

A.

3:5

B.

5:3

C.

3:2

D.

2:7

#### SOLUTION

Solution : C

From section formula, if a point (x,y) divides the line segment joining the points (x1,y1) and (x2,y2) internally in the ratio m:n,  then x=mx2+nx1m+n and y=my2+ny1m+n.

So, 2=4m+n(1)m+n

4mn=2m+2n

i.e., 2m=3n

Thus,  mn = 32  or m:n=3:2

Find the area of the triangle formed by joining the mid points of the sides of the triangle formed with coordinates A (-4, 3), B (2, 3) and C (4, 5).

A.

2 sq.units

B.

0.5 sq. units

C.

1 sq. units

D.

1.5 sq. units

#### SOLUTION

Solution : D

The mid-point of the coordinates (x1,y1), (x2,y2) is given by (x1+x22, y1+y22)
The mid-points of AB, BC and CA are (4+223+32), (4+223+52) and (4+425+32) respectively. Let us denote these by D, E and F.
i.e.,  D(-1, 3), E(3, 4) and F(0, 4).

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

Now, let x1 = -1,  y1 = 3, x2 = 3,  y2 = 4 x3 = 0 and  y3 = 4

Required area =12| -1(4 - 4) + 3(4 - 3) + 0(3 - 4)|  = 1.5 square units

Area of the triangle = 1.5 square units

The area of a quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3) is _______.

A.

26 sq. units

B.

30 sq. units

C.

28 sq. units

D.

27 sq. units

#### SOLUTION

Solution : C

Consider the points
A(4,2),B(3,5),C(3,2) and D(2,3).A()=12×|x1(y2y3)+x2(y3y1)+x3(y1y2)|

Area of  ABC

=12|(4)(5+2)3(2+2)+3(2+5)|  =12|2086+15|=212A(ABC)=10.5 sq. units

Similarly, area of   ACD

=12|(4)(23)+3(3+2)+2(2+2)|  =12|20+15|=352A(ACD)=17.5 sq. units

Area of quadrilateral ABCD=A(ABC)+A(ACD)               =(10.5+17.5)=28 sq. units

Find the point on the y-axis which is equidistant from A (3, - 4) and B (- 5, 9).

A.

(0,8128)

B.

(0,3)

C.

(0,8126)

D.

(0,10526)

#### SOLUTION

Solution : C

We have to find a point on the y-axis. Therefore, its x-coordinate will be 0.
Let the point on y-axis be P (0,y)

Distance PA = Distance between (0,y) and (3,4)=(03)2+(y(4))2=(3)2+(y+4)2

Distance PB = Distance between (0,y) and (5,9)=(0(5))2+(y9)2=52+(y9)2

By the given condition, PA = PB, thus
(3)2+(y+4)2=52+(y9)2

Squaring on both sides of the equation

(y+4)2+9=(y9)2+25

y2+16+8y+9=y218y+81+25

26y=10625

26y=81

y=8126

The required point is (0,8126).