Question

Find the number of straight line passing through (2, 4) & formed a triangle with coordinate ones whose area is 16 square.

Answer 1

Given,

$\Rightarrow$ Area $\left(△AOB\right)=16sq.units$

From the equation of straight lines -

$\Rightarrow y-4=m(x-2)$

$y-4=mx-2m$

$mx-y=2m-4$

$\Rightarrow A=(\frac{2m-4}{m},0)$ $B=(0,4-2m)$

given,

$\Rightarrow \frac{1}{2}\left|\left|\begin{array}{ccc}0& 0& 1\\ 0& 4-2m& 1\\ \frac{2m-4}{m}& 0& 1\end{array}\right|\right|=16$

$\Rightarrow \frac{{(2m-4)}^2}{m}=\pm 32$ [ removing modulus sign ]

$\Rightarrow 4m^2-16m+16-32m=0$ [ by taking $+32$ value ]

$\Rightarrow 4m^2-48m+16=0$

here, $D={(-48)}^2-4\times 4\times 16>0$

So, $m$ has $2$ values

if second case has taken -

$\Rightarrow \frac{{(2m-4)}^2}{m}=-32$

$\Rightarrow 4m^2-16m+16+32m=0$

$4m^2+16m+16=0$

$m^2+4m+4=0$

${(m+2)}^2=0$

$\Rightarrow m=-2$

$\therefore$ $m$ has total $3$ values.

$\therefore$ Total 3 straight lines is possible.

Hence, the answer is $3.$