Question

The area of a triangle is 5. If two of its vertices are (2, 1), (3, –2) and the third vertex lies on the line y = x + 3, then the third vertex is –

• A. $\left( - \frac{7}{2}, - \frac{13}{2} \right)$
• B. $\left( - \frac{7}{2},\frac{13}{2} \right)$
• C. $\left( \frac{7}{2}, - \frac{13}{2} \right)$
• D. $\left( \frac{7}{2},\frac{13}{2} \right)$

Answer 1

Answer: (D)

$\left( \frac{7}{2},\frac{13}{2} \right)$

Answer 2

Let the third vertex be (h, k)

Since it lies on line y = x + 3

\ k = h + 3 ........(1)

Also area of triangle is 5

\ $\frac{1}{2}$ [2(–2–k) + 3(k –1) + h (1 + 2) = ± 5

Ž k + h – 7 = ± 10 ........(2)

Solving (1) & (2) we get

h = $\frac{7}{2}$, k =$\frac{13}{2}$

Ž $\left( \frac{7}{2},\frac{13}{2} \right)$

and h = $\frac{- 3}{2}$, k = $\frac{3}{2}$

Ž $\left( - \frac{3}{2},\frac{3}{2} \right)$