# Free Coordinate Geometry Subjective Test 02 Practice Test - 10th Grade

Find the midpoint of the line segment joining the points A(-2,8) and B(-6,-4).  [1 MARK]

#### SOLUTION

Solution : Formula : 0.5 Mark
Application : 0.5 Mark

The midpoint of the line segment joining the points A(x1,y1) B(x2,y2) is ,

P(x,y)=(x1+x22),(y1+y22)

P(x,y)=(262),(842)

P(x,y)=(4,2)

Two vertices of a triangle are (3, –5) and (–7, 4). If its centroid is (2, –1). Find the third vertex.  [2 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark

Let the coordinates of the third vertex be (x, y).

We know that the coordinates of the centroid of a triangle whose vertices are

(x1,y1),(x2,y2) and (x3,y3) are

x1+x2+x33,y1+y2+y33

Given points are (3, –5) and (–7, 4) and centroid is (2, –1).

x+373=2 and y5+43=1

x4=6 and y1=3

x=10 and y=2

Thus the coordinates of the third vertex are (10, -2).

Find the distance between the points (-3, -3), (4, 5).  [2 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark

Distance between points A(x1,y1) and B(x2,y2) is,

D=(x2x1)2+(y2y1)2

D=(4(3))2+(5(3))2

D=72+82

D=113 units

Find the value of x, if the distance between the points (x, –1) and (3, 2) is 5.  [2 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark

Let p (x , -1) and Q (3 ,2) be the given points. Then,

PQ = 5 (Given)
Distance between the points is given by
(x1x2)2+(y1y2)2
(x3)2+(12)2=5

(x3)2+9=52     [Squaring both sides]

x26x+18=25

x26x7=0

(x7)(x+1)=0

x=7  or  x=1

Show that the points (1, – 1), (5, 2) and (9, 5) are collinear. [2 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark

A (1 , -1) , B (5 , 2) and C (9 , 5) are the given points. Then, We have
Distance between the points is given by
(x1x2)2+(y1y2)2
AB=(51)2+(2+1)2
So,

AB=16+9=5

BC=(59)2+(25)2

BC=16+9=5

and , AC=(19)2+(15)2

AC=64+36=10

Clearly, AC = AB + BC

Hence A, B, C are collinear points

If the point C (–1, 2) divides internally the line segment joining A (2, 5) and B in ratio 3 : 4, find the coordinates of B. [3 MARKS]

#### SOLUTION

Solution :

Formula: 1 Mark
Concept:1 Mark
Answer: 1 Mark

Let the coordinates of B be (α,β).

It is given that AC:BC = 3:4. So, the coordinates of C are:

C(x,y)=(mx2+nx1m+n),(my2+ny1m+n)

C(1,2)=3α+4×23+4,3β+4×53+4

C(1,2)=3α+87,3β+207

3α+87=1  and  3β+207=2

α=5  and  β=2

Thus, the coordinates of B are (-5, -2).

In what ratio does the x-axis divide the line segment joining the points (2, –3) and (5, 6)? Also, find the coordinates of the point of intersection.
[3 MARKS]

#### SOLUTION

Solution :

Concept: 1 Mark
Application: 1 Mark
Answer: 1 Mark

Let the required ratio be λ:1. Then, the coordinates of the point of division are,

R(5λ+2λ+1,6λ3λ+1) But, it is a point on x-axis on which y-coordinates of every point is zero.

6λ3λ+1=0λ=12

Thus, the required ratio is 12:1 or 1:2.

Substituting λ=12 in the coordinates of R.

R(5λ+2λ+1,0)

R(52+212+1,0)

R(9232,0)

R(3,0)

the coordinates of R are (3, 0)

If A (–2, –1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram. Find the values of a and b. [3 MARKS]

#### SOLUTION

Solution :

Formula: 1 Mark
Application: 1 Mark
Answers: 1 Mark

We know that the diagonals of a parallelogram bisect each other.

The coordinates of the mid-point of AC = The coordinates of the mid-point of BD i.e.

If A(x1,y1),B(x2,y2),C(x3,y3) and D(x4,y4) are the vertices of parallelogram ABCD, Then

x1+x32,y1+y32=x2+x42,y2+y42

(2+42,1+b2)=(a+12,0+22)

(1,b12)=(a+12,1)

a+12=1  and  b12=1

a+1=2  and  b1=2

a=1  and  b=3

Prove that (4, – 1), (6, 0), (7, 2) and (5, 1) are the vertices of a rhombus. Is it a square?  [3 MARKS]

#### SOLUTION

Solution :

Formula: 1 Mark
Application: 1 Mark
Answer: 1 Mark

Let the given points be A(4, – 1), B(6, 0), C(7, 2) and D(5, 1) respectively. Then,

Coordinates of the mid-point of AC are

(4+72,1+22)=(112,12)

Coordinates of the mid-point of BD are

(6+52,0+12)=(112,12)

Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

Now
Distance between the points is given by
(x1x2)2+(y1y2)2
So,
AB = (64)2+(0+1)2=5

BC = (76)2+(20)2=5

AB = BC

So, ABCD is a parallelogram whose adjacent sides are equal.

ABCD is a rhombus.

We have,

AC = (74)2+(2+1)2=32   and ,

BD = (65)2+(01)2=2

Clearly, the diagonals AC ≠ BD.

So, ABCD is not a square.

Find the lengths of the medians of a triangle ABC whose vertices are A(7, –3), B(5, 3) and C(3, –1).  [4 MARKS]

#### SOLUTION

Solution :

Formula: 1 Mark
Concept: 1 Mark
Answer: 2 Marks

Let D, E, F be the mid-points of the sides BC, CA and AB respectively.

Then, the coordinates of D, E and F are

D(5+32,312)=D(4,1),

E(3+72,132)=E(5,2)

and F(7+52,3+32)=F(6,0) Distance between the points is given by
(x1x2)2+(y1y2)2
AD=(74)2+(31)2

AD=9+16=5 units

BE=(55)2+(23)2

BE=0+25=5 units

and, CF=(63)2+(0+1)2

CF=9+1=10 units

If A (1,2), B (4, y), C (x,6) and D (3,5) are vertices of a parallelogram ABCD, find the values of x and y. [4 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Consider a  parallelogram ABCD .

Let O be the point of intersection of the diagonals AC and BD We know that diagonals of a parallelogram bisect each other.

O is the midpoint of both the digonals AC and BD

Now, coordinates of O as mid-point of BD are

O(x,y)=(x1+x22),(y1+y22)

O(a,b)=4+32,y+52(1)

Also, coordinates of O as mid-point of AC are

O(a,b)=1+x2,2+62(2)

From (1) and (2), we have

4+32=1+x2

72=1+x2x=6

And y+52=2+62

y+52=82y=3

If the coordinates of the mid points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4) Find its centroid. [4 MARKS]

#### SOLUTION

Solution :

Formula for centroid and mid-point: 1 Mark each

Let A(x1,y1),B(x2,y2) and C(x3,y3) be the vertices of triangle ABC.

Let P(1,1), Q(2,-3), R(3,4) be the mid-points of sides AB, BC and CA respectively.

Then, P is the mid-point of BC

x1+x22=1,y1+y22=1

x1+x2=2 and y1+y2=2(1)

Q is the mid-point of AB

x2+x32=2,y2+y32=3

x2+x3=4 and y2+y3=6(2)

R is the mid-point of AC

x1+x32=3,y1+y32=4

x1+x3=6 and y1+y3=8(3)

Adding (1),(2) and (3) we get,

x1+x2+x2+x3+x1+x3=2+4+6

and y1+y2+y2+y3+y1+y3=26+8

x1+x2+x3=6(4)

and y1+y2+y3=2(5)

The coordinates of the centroid of ABC are

(x1+x2+x33,y1+y2+y33)=63,23=2,23 [Using (4) and (5)]

For what value of k are the points A(k, 2 – 2k) ,B(–k + 1, 2k) and C(–4 – k, 6 – 2k) are collinear ?  [4 MARKS]

#### SOLUTION

Solution :

Formula: 1 Mark
Calculations: 2 Marks
Answer: 1 Mark

For points A,B,C to be collinear, area of triangle ABC formed by these points should be zero.

The area of  ΔABC=0

12[x1(y2y3)+x2(y3y1)+x3(y1y2)]=0

12[k(2k6+2k)+(k+1)(62k2+2k)+(4k)(22k2k)]=0

12[k(4k6)+(k+1)(4)+(4k)(24k)]=0

12[4k26k4k+48+16k2k+4k2]=0

8k2+4k4=0

8k2+4k4=0

2k2+k1=0

(k+1)(2k1)=0

k=1 or k=12

The two opposite vertices of a square are (–1, 2) and (3, 2). Find the co-ordinates of the other two vertices. [4 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Let ABCD be a square and two opposite vertices of it are A(-1, 2) and C(3, 2). ABCD is a square. Since ABCD is a square.

AB=BC

AB2=BC2
[Distance between the points is given by
(x1x2)2+(y1y2)2]

(x+1)2+(y2)2=(x3)2+(y2)2

x2+2x+1=x26x+9

2x+6x=91=8

8x=8x=1

ABC is right Δ at B, then

AC2=AB2+BC2 (Pythagoras theorem)

(3+1)2+(22)2=(x+1)2+(y2)2+(x3)2+(y2)2

16=2(y2)2+(1+1)2+(13)2

16=2(y2)2+4+42(y2)2=168=8

(y2)2=4y2=±2y=4 and 0

i.e when x = 1 then y = 4 and 0

Co-ordinates of the opposite vertices are: B(1,0) or D(1,4)

Find the area of the quadrilateral ABCD whose vertices are respectively A (1, 1), B (7, –3) C (12, 2) and D (7,21). [4 MARKS]

#### SOLUTION

Solution :

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Area of quadrilateral ABCD=|Area of ΔABC|+|Area of ΔACD|

We have, Area of ΔABC=12|(1×3+7×2+12×1)(7×1+12×(3)+1×2)|

ΔABC=12|(3+14+12)(736+2)|

ΔABC=12|23+27|=25sq.units

Also, we have Area of ΔACD=12|(1×2+12×21+7×1)(12×1+7×2+1×21)|

ΔACD=12|(2+252+7)(12+14+21)|

ΔACD=12|26147|=107sq.units

Area of quadrilateral ABCD = 25 + 107 = 132 sq. units