Question

Area of a triangle is 5. Its two vertices are (2, 1) and (3, –2). Third vertex is on line y = x + 3. That vertex will be-

• A. $\left( \frac{7}{2},\frac{13}{2} \right)$
• B. (8, 14)
• C. $\left( \frac{8}{3},\frac{5}{3} \right)$
• D. $\left( \frac{7}{3},\frac{9}{7} \right)$

Answer 1

Answer: (A)

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

Answer 2

Let vertices are

A (2, 1),B(3, –2), C(a, a + 3)

Area = 5

Ž $\frac{1}{2}$ $\left| \begin{matrix} \alpha & \alpha + 3 & 1 \\ 2 & 1 & 1 \\ 3 & –2 & 1 \end{matrix} \right|$ = ± 5

Ž 3a + a + 3 – 7 = ± 10

Ž 4a – 4 = ± 10

Ž a = $\frac{7}{2}$, a = $\frac{–3}{2}$

$\left( \frac{7}{2},\frac{13}{2} \right)$ or $\left( \frac{–3}{2},\frac{3}{2} \right)$