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 the line y = x + 3.

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

Also area of triangle is s

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

Ž k + 3h – 7 = ± 10 ....... (ii)

Solving (i) & (ii) we get

h = $\frac{7}{2}$, $\frac{- 3}{2}$ and k = $\frac{13}{2}$, $\frac{3}{2}$

\ third vertex is either $\left( \frac{7}{2},\frac{13}{2} \right)$ or $\left( \frac{- 3}{2},\frac{3}{2} \right)$