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(B) The point of intersection of the lines $4x - 3y - 1 = 0$ and $2x - 5y + 3 = 0$ is found by solving them simultaneously.Multiplying the second equa

Vedclass · Accepted Answer

$y - 1 = \pm 1(x - 1)$

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(B) The point of intersection of the lines $4x - 3y - 1 = 0$ and $2x - 5y + 3 = 0$ is found by solving them simultaneously.Multiplying the second equation by $2$,we get $4x - 10y + 6 = 0$.Subtracting this from the first equation: $(4x - 3y - 1) - (4x - 10y + 6) = 0$,which gives $7y - 7 = 0$,so $y = 1$.Substituting $y = 1$ into $2x - 5(1) + 3 = 0$,we get $2x - 2 = 0$,so $x = 1$.The point of intersection is $(1, 1)$.Lines equally inclined to the axes have slopes $m = 	an(45^{\circ}) = 1$ or $m = 	an(135^{\circ}) = -1$.Using the point-slope form $y - y_1 = m(x - x_1)$,the equations are $y - 1 = 1(x - 1)$ and $y - 1 = -1(x - 1)$.These can be written as $y - 1 = \pm 1(x - 1)$.

The equations of the lines which pass through the point of intersection of the lines $4x - 3y - 1 = 0$ and $2x - 5y + 3 = 0$ and are equally inclined to the axes are

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