Discussion

Question:

The area of a triangle is 6, two of its vertices are (1, 1) and (4, –1), the third vertex lies on y = x + 5. Find the third vertex.

Explanation:

If the coordinates of the vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃) then its area is given by

A = $\frac{1}{2}$ $|\begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix}|$

∴ If the third vertex of the given triangle is (x, y)

6 = $\frac{1}{2}$ $|\begin{matrix} 1 & 4 & x \\ 1 & - 1 & y \\ 1 & 1 & 1 \end{matrix}|$

∴ 12 = 1(−1 − y) − 4(1 − y) + x(1 + 1)

∴ 12 = −1 – y − 4 + 4y + 2x

∴ 2x + 3y = 17 …(i)

But (x, y) also lies on y = x + 5 …(ii)

Solving (i) and (ii)

x = $\frac{2}{5}$ , y = $\frac{27}{5}$

Hence, option (a).

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