(D) Given equations are $y - x + 7 = 0$ $(1)$ and $y + 2x - 2 = 0$ $(2)$.Subtracting $(1)$ from $(2)$,we get $(y + 2x - 2) - (y - x + 7) = 0$,which si

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(D) Given equations are $y - x + 7 = 0$ $(1)$ and $y + 2x - 2 = 0$ $(2)$.Subtracting $(1)$ from $(2)$,we get $(y + 2x - 2) - (y - x + 7) = 0$,which simplifies to $3x - 9 = 0$,so $x = 3$.Substituting $x = 3$ in $(1)$,we get $y - 3 + 7 = 0$,so $y = -4$.The point of intersection is $(3, -4)$.The equation of a line passing through the origin $(0, 0)$ and $(x_1, y_1)$ is given by $y = \frac{y_1}{x_1}x$.Substituting $(3, -4)$,we get $y = \frac{-4}{3}x$,which implies $3y = -4x$ or $4x + 3y = 0$.

Class 11 Mathematics Pair of straight lines Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

Easy

The equation of the straight line joining the origin to the point of intersection of $y - x + 7 = 0$ and $y + 2x - 2 = 0$ is

A
$3x + 4y = 0$
B
$3x - 4y = 0$
C
$4x - 3y = 0$
D
$4x + 3y = 0$

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Pair of straight lines

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Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines

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MediumAP EAMCET 2012

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The equation of the straight line joining the origin to the point of intersection of $y - x + 7 = 0$ and $y + 2x - 2 = 0$ is

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