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

In $\Delta ABC$, after solving equation of AB, BC and CA

$x-y-4=0$

$3x-7y+8=0$

and $4x-y-31=0$, respectively

We get,

$A=(7,-3)B=(\frac{18}{5},\frac{2}{3})$ and $C=(\frac{209}{25},\frac{61}{25})$

The coordinates of the vertices of the triangle ABC are marked in the following figure.

Point p(a, 2) lie inside on the triangle if

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

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

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

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

$\left\{7\right(3)-7(-3)-8\}\{3a-7(2)-8\}>0$

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

$3a-22>0$

$a>\frac{22}{3}\dots \dots \left(i\right)$

B and P will lie on the same side of AC if

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

$4a-33>0$

$a>\frac{33}{4}\dots \dots \left(ii\right)$

C and P will lie on the same side of BC if

$(\frac{209}{25}+\frac{61}{25}-4)(a+2-4)>0$

$a+2>0$

$a>-2\dots \dots \left(iii\right)$

From (i), (ii), (iii)

$\alpha ϵ(\frac{22}{3},\frac{33}{4})$