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. [CBSE 2012]

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

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|=15 \\ \Rightarrow \frac{1}{2}\left|1\left(p-7\right)+4\left[7-\left(-3\right)\right]+\left(-9\right)\left(-3-p\right)\right|=15 \\ \Rightarrow \frac{1}{2}\left|p-7+40+27+9p\right|=15 \\ \Rightarrow \left|10p+60\right|=30 \end{gathered}$$

$\Rightarrow 10p+60=30$ 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.