Question

If the vertices of a triangle are (1, –3), (4, p) and (–9, 7) and its area is 15 sq. units, find the value(s) of p.  

Answer 1

Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC.

Here, x₁ = 1, y₁ = −3; x₂ = 4, y₂ = p and x₃ = −9, y₃ = 7

ar(∆ABC) = 15 square units

\[\Rightarrow \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right| = 15\]

\[ \Rightarrow \frac{1}{2}\left| 1\left( p - 7 \right) + 4\left[ 7 - \left( - 3 \right) \right] + \left( - 9 \right)\left( - 3 - p \right) \right| = 15\]

\[ \Rightarrow \frac{1}{2}\left| p - 7 + 40 + 27 + 9p \right| = 15\]

\[ \Rightarrow \left| 10p + 60 \right| = 30\]

\[\Rightarrow 10p + 60 = 30\] or \[10p + 60 = - 30\]

\[\Rightarrow 10p = - 30\] or \[10p = - 90\]

\[\Rightarrow p = - 3\] or \[p = - 9\]

Hence, the value of p is −3 or −9.