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A pair of straight lines drawn through the origin form with the line $2x + 3y = 6$ an isosceles right angled triangle, then the lines and the area of the triangle thus formed is
$x - 5y = 0$ ; $5x + y = 0$ ; $\Delta = \frac{{36}}{{13}}$
$3x - y = 0$ ; $5x + y = 0$ ; $x + 3y = 0$; ;$\Delta = \frac{{36}}{{13}}$$\Delta = \frac{{12}}{{17}}$
$5x - y = 0$ ; $x + 5y = 0$ ; $\Delta = \frac{{13}}{5}$
None of these
Solution
(a) $y = mx$. It makes an angle of $ \pm {45^o}$ with $2x + 3y = 6$.
$\tan ( \pm {45^o}) = \frac{{m – ( – 2/3)}}{{1 + m( – 2/3)}} = \pm 1$
or $3m + 2 = \pm (3 – 2m)$$ \Rightarrow m = \frac{1}{5}, – 5$
Hence sides are
$x – 5y = 0,$
$5x + y = 0$
and $2x + 3y = 6$.
Solving in pairs, vertices are $(0,\,0)$,
$\left( {\frac{6}{{13}},\frac{{30}}{{13}}} \right)\,,\left( {\frac{{30}}{{13}}, – \frac{6}{{13}}} \right)$.
.$\Delta = \left| {\frac{1}{2}({x_1}{y_2} – {x_2}{y_1})} \right| = \frac{1}{2} \times \frac{{936}}{{169}} = \frac{{36}}{{13}}$.
