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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

A

(3, 4)

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B

(5/2, 13/2)

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C

(7/2, 13/2)

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D

(1,9)

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Solution

## The correct option is C (7/2, 13/2) Area of the triangle = 12[x1(y2 - y3) + x2(y3 - y1)+ x3(y1 - y2)] = 5 5 = 12[2(-2 - y) + 3(y - 1) + 3x] 5 = 12[-4 - 2y + 3y - 3 + 3x] 3x + y = 17, which is a linear equation in two variable and also a straight line. We are given in question that the third vertex lies on y = x + 3. It means that the point of intersection of both the lines is the vertex. SO, solving y = x + 3 and 3x + y = 17, we get x = 72 and y = 132. Therefore third vertex is (72, 132​)

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