Two straight lines 3x + 4y = 5 and 4x - 3y = | vedclass

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(D) Let the given lines be $L_1: 3x + 4y - 5 = 0$ and $L_2: 4x - 3y - 15 = 0$. The slopes are $m_1 = -\frac{3}{4}$ and $m_2 = \frac{4}{3}$.Since $m_1

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$7x + y = 9, x - 7y + 13 = 0$

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(D) Let the given lines be $L_1: 3x + 4y - 5 = 0$ and $L_2: 4x - 3y - 15 = 0$. The slopes are $m_1 = -\frac{3}{4}$ and $m_2 = \frac{4}{3}$.Since $m_1 	imes m_2 = -1$,the lines are perpendicular.Let the line passing through $(1, 2)$ have slope $m$. Its equation is $y - 2 = m(x - 1)$,or $mx - y + (2 - m) = 0$.Since $AB = AC$,the line $BC$ must make equal angles with $L_1$ and $L_2$. Since $L_1 \perp L_2$,the line $BC$ must make an angle of $45^\circ$ with both lines.The angle between the line with slope $m$ and $L_1$ is given by $	an 45^\circ = |\frac{m - (-3/4)}{1 + m(-3/4)}| = 1$.$|\frac{4m + 3}{4 - 3m}| = 1$.Case $1$: $\frac{4m + 3}{4 - 3m} = 1$ $\Rightarrow 4m + 3 = 4 - 3m$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$.The equation is $y - 2 = \frac{1}{7}(x - 1)$ $\Rightarrow 7y - 14 = x - 1$ $\Rightarrow x - 7y + 13 = 0$.Case $2$: $\frac{4m + 3}{4 - 3m} = -1$ $\Rightarrow 4m + 3 = -4 + 3m$ $\Rightarrow m = -7$.The equation is $y - 2 = -7(x - 1)$ $\Rightarrow y - 2 = -7x + 7$ $\Rightarrow 7x + y = 9$.Thus,the lines are $7x + y = 9$ and $x - 7y + 13 = 0$.

Two straight lines $3x + 4y = 5$ and $4x - 3y = 15$ intersect at the point $A$. The equations of the lines passing through $(1, 2)$ and intersecting the given lines at $B$ and $C$ such that $AB = AC$ are

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