Q46 of 66 Page 7

The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (7/2, y); find the value of y.(CBSE 2017)

Area of triangle = 5 sq. units.

Vertices are (2, 1), (3, –2) and (7/2, y).


Area of a triangle when three points (x1, y1), (x2, y2), (x3, y3) are given:


Area of triangle = 1/2|[x1(y2 – y3) + x2(y3 - y1) + x3(y1 – y2)]|


∴ 5 = 1/2 |[2(-2 - y) + 3(y – 1) + (7/2)(1 + 2)]|


= 1/2|[ -4 -2y + 3y - 3 + (21/2)]|


= 1/2|[y + (7/2)]|


⇒ 5 × 2 = |y + (7/2)|


⇒ 10 = y + (7/2) or 10 = -[y + 7/2]


⇒ y = 10 –(7/2) or 10 = -y – (7/2)


⇒ y = 13/2 or y = -10 – (7/2)


⇒ y = 13/2 or y = -27/2


∴ y = 13/2 or -27/2.

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