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(b) The equation of lines are $y - {y_1} = \frac{{m \pm 	an \alpha }}{{1 \mp m	an \alpha }}(x - {x_1})$

==> $y - 4 = \frac{{1 \pm 	an 45^\cir

Vedclass · Accepted Answer

$9\over2$

Vedclass · Answer

(b) The equation of lines are $y - {y_1} = \frac{{m \pm 	an \alpha }}{{1 \mp m	an \alpha }}(x - {x_1})$

==> $y - 4 = \frac{{1 \pm 	an 45^\circ }}{{1 \mp 	an 45^\circ }}(x - {x_1})$

==> $y - 4 = \frac{{1 \pm 1}}{{1 \mp 1}}(x - 3)$ ==> $y = 4$ or $x = 3$

Hence, the lines which make the triangle are $x - y = 2,$ $x = 3$ and $y = 4$. The intersection points of these lines are $(6,\,4),\,(3,\,1)$ and $(3,\,4)$

 $\Delta = \frac{1}{2}[6( - 3) + 3(0) + 3(3)]$ $ = \frac{9}{2}$.

Two lines are drawn through $(3, 4)$, each of which makes angle of $45^\circ$ with the line $x - y = 2$, then area of the triangle formed by these lines is

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