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(A) Let the lines passing through the origin be $y = mx$. These lines make an angle of $\pm 45^\circ$ with the line $2x + 3y = 6$,which has a slope of

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the lines passing through the origin be $y = mx$. These lines make an angle of $\pm 45^\circ$ with the line $2x + 3y = 6$,which has a slope of $m_1 = -2/3$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m m_1} ight|$,we have:$	an(\pm 45^\circ) = \frac{m - (-2/3)}{1 + m(-2/3)} = \pm 1$$\frac{3m + 2}{3 - 2m} = \pm 1$Case $1$: $3m + 2 = 3 - 2m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$.The line is $y = \frac{1}{5}x \Rightarrow x - 5y = 0$.Case $2$: $3m + 2 = -(3 - 2m)$ $\Rightarrow 3m + 2 = -3 + 2m$ $\Rightarrow m = -5$.The line is $y = -5x \Rightarrow 5x + y = 0$.The vertices of the triangle are the intersection points of these lines with $2x + 3y = 6$ and the origin $(0,0)$.Solving $x - 5y = 0$ and $2x + 3y = 6$: $2(5y) + 3y = 6$ $\Rightarrow 13y = 6$ $\Rightarrow y = 6/13, x = 30/13$.Solving $5x + y = 0$ and $2x + 3y = 6$: $y = -5x$ $\Rightarrow 2x + 3(-5x) = 6$ $\Rightarrow -13x = 6$ $\Rightarrow x = -6/13, y = 30/13$.The vertices are $(0,0)$,$A(\frac{30}{13}, \frac{6}{13})$,and $B(-\frac{6}{13}, \frac{30}{13})$.The area $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |0 + \frac{30}{13}(\frac{30}{13} - 0) + (-\frac{6}{13})(0 - \frac{6}{13})| = \frac{1}{2} |\frac{900}{169} + \frac{36}{169}| = \frac{1}{2} 	imes \frac{936}{169} = \frac{468}{169} = \frac{36}{13}$.

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.

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