Coordinate Geometry

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.

Solution

The area of the triangle, whose vertises are (x₁, y₁), (x₂, y₂) and (x_3, y₃) is

$Area of a ∆ = \frac{1}{2}$ [ x₁(y₂- y₃) + x₂ ( y₃- y₁) + x₃( y₁- y₂) ]

Substituting the given coordinates

$Area of ∆ = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$\Rightarrow \frac{1}{2} [(p - 7) + 40 + 27 + 9 p] = 15$

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

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

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