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Question

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


  1. (3,4)

  2.  (52, 132)

  3. (72, 132)

  4. (32, 32)


Solution

The correct options are
C

(72, 132)


D (32, 32)

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)

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