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(A) Step $1$: Find the intersection point of $y = x + 7$ and $x + 2y + 1 = 0$. Substitute $y = x + 7$ into $x + 2y + 1 = 0$: $x + 2(x + 7) + 1 = 0$ $x

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$2x + 5y = 0$

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(A) Step $1$: Find the intersection point of $y = x + 7$ and $x + 2y + 1 = 0$. Substitute $y = x + 7$ into $x + 2y + 1 = 0$: $x + 2(x + 7) + 1 = 0$ $x + 2x + 14 + 1 = 0$ $3x = -15$ $x = -5$. Then $y = -5 + 7 = 2$. The intersection point is $(-5, 2)$. Step $2$: Find the slope of the line perpendicular to $5x - 2y + 7 = 0$. The slope of $5x - 2y + 7 = 0$ is $m_1 = -\frac{5}{-2} = \frac{5}{2}$. The slope of the perpendicular line is $m_2 = -\frac{1}{m_1} = -\frac{2}{5}$. Step $3$: Use the point-slope form $y - y_1 = m(x - x_1)$ with point $(-5, 2)$ and slope $m = -\frac{2}{5}$. $y - 2 = -\frac{2}{5}(x + 5)$ $5(y - 2) = -2(x + 5)$ $5y - 10 = -2x - 10$ $2x + 5y = 0$.

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	$(s)$ $k=5$

Find the equation of the line perpendicular to $5x - 2y + 7 = 0$ and passing through the intersection of the lines $y = x + 7$ and $x + 2y + 1 = 0$.

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