The equation of a line passing through the po | vedclass

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(A) Step $1$: Find the point of intersection of the lines $x + 5y + 7 = 0$ and $3x + 2y - 5 = 0$.Multiply the first equation by $3$: $3x + 15y + 21 =

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$2x - 7y - 20 = 0$

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(A) Step $1$: Find the point of intersection of the lines $x + 5y + 7 = 0$ and $3x + 2y - 5 = 0$.Multiply the first equation by $3$: $3x + 15y + 21 = 0$.Subtract the second equation from this: $(3x + 15y + 21) - (3x + 2y - 5) = 0$,which gives $13y + 26 = 0$,so $y = -2$.Substitute $y = -2$ into $x + 5y + 7 = 0$: $x + 5(-2) + 7 = 0 \implies x - 10 + 7 = 0 \implies x = 3$.The point of intersection is $(3, -2)$.Step $2$: Find the slope of the line perpendicular to $7x + 2y - 5 = 0$.The slope of $7x + 2y - 5 = 0$ is $m_1 = -\frac{7}{2}$.The slope of the required line $m_2$ satisfies $m_1 	imes m_2 = -1$,so $m_2 = \frac{2}{7}$.Step $3$: Write the equation of the line using the point-slope form $y - y_1 = m(x - x_1)$.$y - (-2) = \frac{2}{7}(x - 3) \implies 7(y + 2) = 2(x - 3) \implies 7y + 14 = 2x - 6$.Rearranging gives $2x - 7y - 20 = 0$.

Column $I$	Column $II$
$(A)$ $L_1, L_2, L_3$ are concurrent,if	$(p)$ $k=-9$
$(B)$ One of $L_1, L_2, L_3$ is parallel to at least one of the other two,if	$(q)$ $k=-\frac{6}{5}$
$(C)$ $L_1, L_2, L_3$ form a triangle,if	$(r)$ $k=\frac{5}{6}$
$(D)$ $L_1, L_2, L_3$ do not form a triangle,if	$(s)$ $k=5$

The equation of a line passing through the point of intersection of the lines $x + 5y + 7 = 0$ and $3x + 2y - 5 = 0$,and perpendicular to the line $7x + 2y - 5 = 0$,is given by

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