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

Given, vertices of a triangle are (1,-3), (4, p) and (-9, 7).
$x_1=1,y_1=-3$
$x_2=4,y_2=p$
$x_3=-9,y_3=7$
Area of given triangle
$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}\left[1\right(p-7)+4(7+3)+(-9\left)\right(-3-p\left)\right]=\frac{1}{2}[p-7+40+27+9p]=\frac{1}{2}[10p+60]$
=5(p+6)
Here, the obtained expression may be positive or negative.
5(p+6) = 15 or 5(p+6) = -15
p + 6 = 3 or p + 6 = -3
p = or p = -9