Sarthaks Test
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The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). Third vertex lies on y =x + 3. Find the third vertex.

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Let A(2, 1) and B(3, -2) be the vertices of Δ

And C(x,y) be the third vertex

Area of ΔABC

But it is given that area of ΔABC=5

But it is given that third vertices lies on y = x +3

Hence subsisting value of y in (i)

3x + x +3 =17 or 3x + x +3 = -3

Hence coordinates of c will be (7/2,13/2) or (-3/2,3/2)

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