If the vertices of a triangle are (1, −3), (4, p) and (−9, 7)

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

$\Rightarrow \frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|=15$

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

$\Rightarrow \frac{1}{2}|p-7+40+27+9 p|=15$

$\Rightarrow|10 p+60|=30$

$\Rightarrow 10 p+60=30$ or $10 p+60=-30$

$\Rightarrow 10 p=-30$ or $10 p=-90$

$\Rightarrow p=-3$ or $p=-9$

Hence, the value of $p$ is $-3$ or $-9$.