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Question

(a) $y = mx$. It makes an angle of $ \pm {45^o}$ with $2x + 3y = 6$.

 $	an ( \pm {45^o}) = \frac{{m - ( - 2/3)}}{{1 + m( - 2/3)}} = \pm 1$

Vedclass · Accepted Answer

$x - 5y = 0$ ; $5x + y = 0$ ; $\Delta  = \frac{{36}}{{13}}$

Vedclass · Answer

(a) $y = mx$. It makes an angle of $ \pm {45^o}$ with $2x + 3y = 6$.

 $	an ( \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 $(0,\,0)$,

$\left( {\frac{6}{{13}},\frac{{30}}{{13}}} ight)\,,\left( {\frac{{30}}{{13}}, - \frac{6}{{13}}} ight)$.

.$\Delta = \left| {\frac{1}{2}({x_1}{y_2} - {x_2}{y_1})} ight| = \frac{1}{2} 	imes \frac{{936}}{{169}} = \frac{{36}}{{13}}$.

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

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