
1.

In a right angled triangle, ∠C = 90° and A, B, C are (–1, 0), (4, –5) and (5, –3) respectively. Find the area of the triangle.

A.

57 sq. units

B.

7.05 sq. units

C.

7.5 sq.units

D.

8.75 sq. units

Answer: Option
C
Explanation:
Given, points of a triangle are (–1, 0), (4, –5) and (5, –3)
x _{1} = –1, y _{1} = 0, x _{2} = 4, y _{2} = –5, x _{3} = 5, y _{3} = –3
We know that,
Area of a triangle = 
1 
[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] 
2 
= 
1 
[(–1)(– 5 + 3) + 4(– 3 – 0) + 5(0 + 5)] 
2 
= 
1 
[2 – 12 + 25)] = 
15 
= 7.5 square units 
2 
2 


2.

Find the coordinates of that point which divides the line segment joining the points (1, 3) & (2, 7) in ratio of 3 : 4.

Answer: Option
B
Explanation:
Given points (1, 3) and (2, 7)
x _{1} = 1, y _{1} = 3, x _{2} = 2, y _{2} = 7
And given, m _{1} = 3 and m _{2} = 4
We know that, the coordinates of the point dividing the join of P(x _{1}, y _{1}) and Q(x _{2}, y _{2}) internally in the ratio m _{1} : m _{2} are, then

( 
m_{1}x_{2} + m_{2}x_{1} 
, 
m_{1}y_{2} + m_{2}y_{1} 
) 
m_{1} + m_{2} 
m_{1} + m_{2} 
Required ratio = 
( 
3 × 2 + 4 × 1 
, 
3 × 7 + 4 × 3 
) 
= 
( 
10 
, 
33 
) 
3 + 4 
3 + 4 
7 
7 


3.

If (7, 5), (5, 7) and (–3, 3) are the vertices of triangle then the intersection point of its median is :

A.

(–2, 5)

B.

(3, –5)

C.

(–3, 5)

D.

(3, 5)

Answer: Option
D
Explanation:
Given points (7, 5), (5, 7) and (–3, 3)
x _{1} = 7, y _{1} = 5, x _{2} = 5, y _{2} = 7, x _{3} = –3, y _{3} = 3
We know that, coordinates of intersection point of medians= 
( 
x_{1} + x_{2} + x_{3} 
, 
y_{1} + y_{2} + y_{3} 
) 
3 
3 
= 
( 
7 + 5 – 3 
, 
5 + 7 + 3 
) 
= 
( 
9 
, 
15 
) 
= (3, 5) 
3 
3 
3 
3 


4.

The angle between the lines y = 
3 
x + 8 and y = 
1 
x + 7 is : 
3 

Answer: Option
B
Explanation:
Given equations of line y = 

x + 8 and y = 
1 
x + 7 
3 
Note : The equation of a line in slope intercept form is y = mx + c where m is its slope.
∴ m_{1} = 

and m_{2} = 
1 
3 
We know that, if m _{1}, m _{2} be the slopes of two lines and Θ be the angle between them, then
tan Θ = 
m_{1} – m_{2} 
1 + m_{1}m_{2} 
tan Θ = tan 30°
Θ = 30°


5.

The mid point of points (4, 5) and (–2, –1) is :

A.

(1, 4)

B.

(2, 3)

C.

(1, 2)

D.

(–2, –1)

Answer: Option
C
Explanation:
Given points (4, 5) and (–2, –1)
x _{1} = 4, y _{1} = 5, x _{2} = –2, y _{2} = –1
Note : The coordinates of the midpoint of the line joining (x_{1}, y_{1}) and (x_{2}, y_{2}) are 
( 
x_{1} + x_{2} 
, 
y_{1} + y_{2} 
) 
. 
2 
2 
∴ Coordinates of the mid point = 
( 
4 – 2 
, 
5 – 1 
) 
= 
( 
2 
, 
4 
) 
= (1, 2) 
2 
2 
2 
2 

