Straight lines 2x + y = 5 and x - 2y = 3 inte | vedclass

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(D) The given lines are $L_1: 2x + y - 5 = 0$ and $L_2: x - 2y - 3 = 0$. Since the product of their slopes is $( -2 ) 	imes ( 1/2 ) = -1$,the lines a

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$A$ or $B$ both

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(D) The given lines are $L_1: 2x + y - 5 = 0$ and $L_2: x - 2y - 3 = 0$. Since the product of their slopes is $( -2 ) 	imes ( 1/2 ) = -1$,the lines are perpendicular. Let $A$ be the intersection point. Since $AB = AC$,$	riangle ABC$ is an isosceles right-angled triangle with the right angle at $A$. The line $BC$ must make an angle of $45^\circ$ or $135^\circ$ with the given lines. The slopes of the angle bisectors of the lines $L_1$ and $L_2$ are given by $\frac{2x+y-5}{\sqrt{5}} = \pm \frac{x-2y-3}{\sqrt{5}}$,which simplifies to $x + 3y - 2 = 0$ and $3x - y - 8 = 0$. The slopes of these bisectors are $-1/3$ and $3$. Since $BC$ is perpendicular to the angle bisector,the slope of $BC$ is either $3$ or $-1/3$. For slope $m = 3$ passing through $(2, 3)$: $y - 3 = 3(x - 2) \implies 3x - y - 3 = 0$. For slope $m = -1/3$ passing through $(2, 3)$: $y - 3 = -1/3(x - 2) \implies x + 3y - 11 = 0$. Thus,both lines are possible.

Straight lines $2x + y = 5$ and $x - 2y = 3$ intersect at the point $A$. Points $B$ and $C$ are chosen on these two lines such that $AB = AC$. Then the equation of a line $BC$ passing through the point $(2, 3)$ is

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