For a given line $\displaystyle a$ defined by the equation $\displaystyle y=mx+b$, any line perpendicular to $\displaystyle a$ must have a slope that is the negative reciprocal of $\displaystyle a$'s slope $\displaystyle m$, $\displaystyle -\frac{1}{m}$.

In this instance, the slope of line $\displaystyle a$ is $\displaystyle m=\frac{9}{8}$, so $\displaystyle -\frac{1}{m}=-\frac{1}{\frac{9}{8}}=-\frac{8}{9}$. The only line provided with an equation that has this slope is $\displaystyle y=-\frac{8}{9}x+4$. 

For a given line $\displaystyle a$ defined by the equation $\displaystyle y=mx+b$, any line perpendicular to $\displaystyle a$ must have a slope that is the negative reciprocal of $\displaystyle a$'s slope $\displaystyle m$, $\displaystyle -\frac{1}{m}$.

Since in this case $\displaystyle m=\frac{5}{6}$, 

$\displaystyle -\frac{1}{m}=-\frac{1}{\frac{5}{6}}=\frac{6}{5}$.

In order to solve this problem, we need first to transform the equation from standard form to slope-intercept form:

$\displaystyle y = mx + b$

Transform the original equation to find its slope.

$\displaystyle 3x + 6y = 12$ 

First, subtract $\displaystyle 3x$ from both sides of the equation.

$\displaystyle 3x-3x + 6y = 12-3x$

Simplify and rearrange.

$\displaystyle 6y = -3x + 12$ 

Next, divide both sides of the equation by 6.

$\displaystyle \frac{6y}{6} = -\frac{3x}{6} + \frac{12}{6}$

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

The slope of our first line is equal to $\displaystyle -\frac{1}{2}$. Perpendicular lines have slopes that are opposite reciprocals of each other; therefore, if the slope of one is x, then the slope of the other is equal to the following: 

$\displaystyle -\frac{1}{x}$

Let's calculate the opposite reciprocal of our slope:

$\displaystyle -\frac{1}{2}\rightarrow -\left (-\frac{2}{1} \right )=2$

The slope of our line is equal to 2. We now have the following partial equation:

$\displaystyle y = 2x + b$

We are missing the y-intercept, $\displaystyle b$. Substitute the x- and y-values in the given point $\displaystyle (-2, 6)$ to solve for the missing y-intercept. 

$\displaystyle 6 = 2(-2) + b$

$\displaystyle 6 = -4 + b$ 

Add 4 to both sides of the equation.

$\displaystyle 6+4 = -4+4 + b$

$\displaystyle 10 = b$

Substitute this value into our partial equation to construct the equation of our line:

$\displaystyle y = 2x + 10$

Use the following formula to find the distance between two points:

$\displaystyle \text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Plug in the points that are given.

$\displaystyle \text{Distance}=\sqrt{(4-(-1))^2+(-2-3)^2}$

$\displaystyle \text{Distance}=\sqrt{(5)^2+(-5)^2}$

$\displaystyle \text{Distance}=\sqrt{25+25}=\sqrt{50}=\sqrt{5*5*2}=5\sqrt2$

The distance between two points is given by the following equation:

$\displaystyle \text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Now, using the two given points, plug them in to find the distance.

$\displaystyle \text{Distance}=\sqrt{(-2-8)^2+(-3-1)^2}=\sqrt{100+16}=\sqrt{116}=10.77$

Use the distance formula to find the length of this line segment:

$\displaystyle \text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute in the values provided:

$\displaystyle \text{Distance}=\sqrt{(-2-8)^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}$

At this point, break down the square root:

$\displaystyle \sqrt{104}=\sqrt{2\cdot52}=\sqrt{2\cdot2\cdot26}=\sqrt{2\cdot2\cdot2\cdot13}$

You can remove two of the twos and put them on the outside of the square root symbol, and multiply the two and the thirteen that remain underneath the square root symbol:

$\displaystyle \sqrt{2\cdot2\cdot2\cdot13}=2\sqrt{2\cdot13}=2\sqrt{26}$

Use the distance formula to find the length of the line segment.

$\displaystyle D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Where,

$\displaystyle (x_1,y_1)=(-2,-1)$ and $\displaystyle (x_2,y_2)=(8,1)$.

$\displaystyle \text{Distance}=\sqrt{(8-(-2))^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}=2\sqrt{26}$

Use the distance formula to find the length of the line segment.

$\displaystyle D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

$\displaystyle (x_1,y_1)=(0,4)$ and $\displaystyle (x_2,y_2)=(12,6)$.

$\displaystyle \text{Distance}=\sqrt{(12-0)^2+(6-4)^2}=\sqrt{144+4}=\sqrt{148}=12.166$

Use the distance formula to find the length of the line.

$\displaystyle D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

$\displaystyle (x_1,y_1)=(2,4)$ and $\displaystyle (x_2,y_2)=(7,8)$.

$\displaystyle \text{Distance}=\sqrt{(7-2)^2+(8-4)^2}=\sqrt{25+16}=\sqrt{41}$

Use the distance formula to find the length of the leg.

$\displaystyle D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

$\displaystyle (x_1,y_1)=(-1,-2)$ and $\displaystyle (x_2,y_2)=(-4,-1)$.

$\displaystyle \text{Length}=\sqrt{(-4-(-1))^2+(-1-(-2))^2}=\sqrt{9+1}=\sqrt{10}$