Question

Find the values of the parameter a so that the point (a, 2) is an interior point of the triangle formed by the lines x + y − 4 = 0, 3x − 7y − 8 = 0 and 4x − y − 31 = 0.

Answer 1

Let ABC be the triangle of sides AB, BC and CA whose equations are x + y − 4 = 0, 3x − 7y − 8 = 0 and 4x − y − 31 = 0, respectively.

On solving them, we get $A\left(7,-3\right)$, $B\left(\frac{18}{5},\frac{2}{5}\right)$ and $C\left(\frac{209}{25},\frac{61}{25}\right)$ as the coordinates of the vertices.
Let P (a, 2) be the given point.



It is given that point P (a,2) lies inside the triangle. So, we have the following:

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

Thus, if A and P lie on the same side of BC, then

$\left(21+21-8\right)\left(3a-14-8\right)>0$

$\Rightarrow a>\frac{22}{3}$ ... (1)

If B and P lie on the same side of AC, then

$\left(\frac{4\times 18}{5}-\frac{2}{5}-31\right)\left(4a-2-31\right)>0$

$\Rightarrow a<\frac{33}{4}$ ... (2)

If C and P lie on the same side of AB, then

$$\begin{gathered} \left(\frac{209}{25}+\frac{61}{25}-4\right)\left(a+2-4\right)>0 \\ \Rightarrow \left(\frac{34}{5}-4\right)\left(a+2-4\right)>0 \end{gathered}$$

$\Rightarrow a>2$ ... (3)

From (1), (2) and (3), we get:

$a\in \left(\frac{22}{3},\frac{33}{4}\right)$