  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 (3,4)  (52, 132) (72, 132) (−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(y2−y3)+x2(y3−y1)+x3(y1−y2)| 5=12|2(−2−y)+3(y−1)+3x|5=12|−4−2y+3y−3+3x| 3x+y−7=10 or 3x+y−7=−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+3⇒x−y=−3 −−−(1) and 3x+y=17 −−−(2). Multiplying (1) by 3, ⇒3x−3y=−9 −−−(3) (2)−(3)⇒  3x+y=  17−3x∓3y=∓9–––––––––––––––––                                  4y=26                                     y=132. Substituting this value of y in (1). x=y−3=132+3=72 ∴x=72 and y=132   Now, to solve  3x−3y=−9−−−(3) and 3x+y=−3−−−(4) (4)−(3)⇒  3x+y  =−3−3x∓3y=∓9–––––––––––––––––                                  4y=6                                     y=32. Substituting this value of y in (1). x=y−3=32−3=−32 ∴x=−32 and y=32. Vertex are (72,132)  &  (−32,32)  Suggest corrections   