# Free Coordinate Geometry 02 Practice Test - 10th Grade

The points (a,a),(a,a) and (3a,3a) are the vertices of a/an  _____ triangle.

A.

scalene

B.

right angled

C.

isosceles right angled

D.

equilateral

#### SOLUTION

Solution : D

Let's name the points as A(a,a), B(a,a) and C(3a,3a).

By distance formula, distance between points (x1,y1) and (x2,y2)

=(x2x1)2+(y2y1)2

BC=(3a+a)2+(3a+a)2      =a2+(3a)22×a×3a+3a)2+a2+2×a×3aBC=8a2=22a units Similarly AB=(a+a)2+(a+a)2      =8a2=22a units AC=(3aa)2+(3aa)2      =8a2=22a units

We get, AB=BC=AC=22a units ABC is an equilateral triangle.

The point C (–1,2) divides the line segment AB in the ratio 3:4, where the coordinates of A is (2, 5). The coordinates of B are___________.

A.

(–5, –2)

B.

(–1, – 2)

C.

(–4, – 2)

D.

(5, – 4)

#### 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 C(1,2) divide the line joining A(2,5) and B(x,y) in the ratio 3:4.

Then,
C(3x+87,3y+207) = C(-1, 2)

3x+87=13x+8=7x=5 3y+207=23y+20=14y=2

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

Midpoint of the line joining the points  (x1,y1) and (x2,y2) is (x1+x22,y1+y22).

A.

True

B.

False

#### SOLUTION

Solution : A

True. Midpoint of the line joining the points (x1,y1) and (x2,y2) is (1(x1)+1(x2)1+1,1(y1)+1(y2)1+1) = (x1+x22,y1+y22).

M (-2, y) is a point equidistant from the coordinates A(5,7) and B(3,-4). Find the value of y.

A. 5722
B. 2
C. 3619
D. 4723

#### SOLUTION

Solution : A

Point M (-2, y) is equidistant from the points A (5, 7) and B(3 , -4).

Distance AM = Distance BM
((xaxm)2+(yaym)2)) = ((xbxm)2+(ybym)2))
(5(2))2 + (7y)2 = (3(2))2 + (4y)2
49+4914y+y2=25+16+8y+y2
22y=57
y=5722

Find the area of the triangle with coordinates A(-2, 1), B(3, 3) and C(1, -2).

A.

10 sq. units

B.

10.5 sq. units

C.

11 sq. units

D.

11.5 sq. units

#### SOLUTION

Solution : B

Area of a triangle having vertices (x1,y1),(x2,y2) and (x3,y3)
=12[x1(y2y3)+x2(y3y1)+x3(y1(y2)]

Given, (x1,y1) = (-2, 1)

(x2,y2) = (3, 3)

(x3,y3) = (1, -2)

=12[2(3+2)+3(21)+1(13)]=10.5

Since area is not a negative quantity, take the absolute value.

Thus, area of the given triangle = 10.5 sq. units.

A(a,b), B(c,d) and C(e,f) are the vertices of a triangle.
i) AB + BC > AC
ii) Area of the triangle = (12)[a(c-e)+c(f-b)+e(b-d)]
Which of the following are true?

A.

Only (i) is true

B.

Only (ii) is true

C.

Both (i) and (ii) is true

D.

Neither (i) nor (ii) is true

#### SOLUTION

Solution : A

Since A, B and C are the vertices of a triangle, it follows that sum of two sides is greater than the third side. So (i) is correct.
And since the area of a triangle with vertices (x1, y1), (x2 , y2 ) and (x3 , y3 ) is
12x1(y2  - y3) +   x2y3 -   y1) + x3(y1y2)]
= 12 [a (d - f) + c (f - b) + e (b - d)] (by putting x1  = a, y1 = b, x2 = c, y2 = d, x3 = e, y3 = f)
Since the above area is not the same as that given in (ii), (ii) is incorrect.

The coordinate of the mid-point of the line segment joining the points (2,1) and (1,-3) is ____.

A.

(32,1)

B.

(3,2)

C.

(3,2)

D.

(4,1)

#### SOLUTION

Solution : A

Let, (x,y) be coordinates of the mid-point.

From the section formula if (x,y) are the midpoints of the line segment joining (x1,y1) and (x2,y2),

then x=x1+x22 and y=y1+y22

x=2+12 and y=3+12

So, (x,y)=(32,1)

The distance between the points (-1, 5) and (2, 1) is ___ units.

#### SOLUTION

Solution :

Distance formula for the points (x1,y1)and(x2,y2)(x2x1)2+(y2y1)2
Distance between the points (-1, 5) and (2, 1),
d = (2(1))2+(15)2
= (2+1)2+(4)2
= 9+16 = 25
= 5 units

The point (2, 0) lies in both the 1st and 4th quadrants.

A. True
B. False

#### SOLUTION

Solution : B

False. Any point that lies on the coordinate axes doesn't belong to any of the quadrants. The point (2,0) lies on the coordinate axes.

The area of a triangle is 5 square units. Two of its vertices are (2, 1) and (3, -2) and the third vertex lies on y = x + 3, the third vertex is

A.

(3,4)

B.

(52, 132)

C.

(72, 132)

D. (32, 32)

#### SOLUTION

Solution : C and D

Given, area of triangle = 5 sq. units.
Let the third vertex be (x,y).
Area of a triangle formed by (x1,y1), (x2,y2) & (x3,y3)=12|x1(y2y3)+x2(y3y1)+x3(y1y2)|

5=12|2(2y)+3(y1)+3x|5=12|42y+3y3+3x|

3x+y7=10 or 3x+y7=10

3x+y=17 or 3x+y=3

Also given that the third vertex lies on y=x+3. It means that the point of intersection of both the lines is the vertex.

Let us now solve y=x+3xy=3 (1) and 3x+y=17 (2).

Multiplying (1) by 3, 3x3y=9 (3)
(2)(3)  3x+y=  173x3y=9–––––––––––––––                                  4y=26                                     y=132.

Substituting this value of y in (1).
x=y3=132+3=72
x=72 and y=132

Now, to solve  3x3y=9(3) and 3x+y=3(4)

(4)(3)  3x+y  =33x3y=9–––––––––––––––                                  4y=6                                     y=32.
Substituting this value of y in (1).
x=y3=323=32
x=32 and y=32.

Vertex are (72,132)  &  (32,32)