The slope of a line with $\displaystyle x$-intercept $\displaystyle (a, 0)$ and $\displaystyle y$-intercept $\displaystyle (0, b)$ is $\displaystyle m = -\frac {b}{a}$. For Line 1, $\displaystyle b= 8, a= 5$, so Line 1 has slope $\displaystyle -\frac{8}{5}$. The slope of Line 2, which is perpendicular to Line 1, will be the opposite of the reciprocal of this, which is $\displaystyle \frac{5}{8}$. Setting $\displaystyle m$ equal to this and $\displaystyle b=8$, we get

$\displaystyle \frac{5}{8}= -\frac {8}{a}$, or $\displaystyle \frac{5}{8}= \frac {-8}{a}$

Cross-multiplying:

$\displaystyle 5a = -64$

$\displaystyle \frac{5a }{5}= \frac{-64}{5}$

$\displaystyle a = -12 \frac{4}{5}$

The $\displaystyle x$-intercept of Line 2 is $\displaystyle \left ( -12 \frac{4}{5}, 0 \right )$.

All of the equations are given in slope-intercept form $\displaystyle y = mx+b$, so we can answer this question by examining the coefficients of $\displaystyle x$, which are the slopes, and the constants, which are the $\displaystyle y$-intercepts. In each case, since the lines are perpendicular, each $\displaystyle x$-coefficient must be the other's opposite reciprocal, and since the lines have the same $\displaystyle y$-intercept, the constants must be equal.

Of the five pairs, only 

$\displaystyle y = 4x-8$ and $\displaystyle y = - \frac{1}{4}x + \frac{1}{8}$

and 

$\displaystyle y = \frac{2}{5}x + 9$ and $\displaystyle y = - \frac{5}{2}x + 9$

have equations whose $\displaystyle x$-coefficients are the other's opposite reciprocal. Of these, only the latter pair of equations have equal constant terms. 

$\displaystyle y = \frac{2}{5}x + 9$ and $\displaystyle y = - \frac{5}{2}x + 9$

is the correct choice.

The slope of each line can be calculated by putting the equation in slope-intercept form $\displaystyle y = mx+b$ and noting the coefficient of $\displaystyle x$:

Line 1:

$\displaystyle x+2y = 12$

$\displaystyle x+2y - x = 12 - x$

$\displaystyle 2y = -1x + 12$

$\displaystyle \frac{2y }{2}= \frac{-1x + 12}{2}$

$\displaystyle y= -\frac{1}{2}x+6$

Slope of Line 1: $\displaystyle -\frac{1}{2}$

Line 2: 

$\displaystyle 2x+y= 12$

$\displaystyle 2x+y - 2x= 12 - 2x$

$\displaystyle y = -2x+12$

Slope of Line 2: $\displaystyle -2$

Line 3: The equation is already in slope-intercept form; its slope is 2.

Two lines are perpendicular if and only their slopes have product $\displaystyle -1$. The slopes of Lines 1 and 3 have product $\displaystyle -\frac{1}{2} \cdot 2 = -1$; they are perpendicular. The slopes of Lines 1 and 2 have product $\displaystyle -\frac{1}{2} \cdot (- 2 )= 1$; they are not perpendicular. The slopes of Lines 2 and 3 have product $\displaystyle -2 \cdot 2 = -4$; they are not perpendicular. 

Correct response: Line 1 and Line 3 are perpendicular; Line 2 is perpendicular to neither.