Any line that is parallel to a line $\displaystyle y=mx+b$ has a slope that is equal to the slope $\displaystyle m$. Given $\displaystyle y=-5x+11$, $\displaystyle m=-5$ and therefore any line parallel to the given line must have a slope of $\displaystyle -5$.

Any line that is parallel to a line $\displaystyle y=mx+b$ has a slope that is equal to the slope $\displaystyle m$. Given $\displaystyle y=\frac{4}{3}x+12$, $\displaystyle m=\frac{4}{3}$ and therefore any line parallel to the given line must have a slope of $\displaystyle \frac{4}{3}$.

Parallel lines have the same slope. Therefore, the slope of the new line is $\displaystyle 2$, as the equation of the original line is $\displaystyle y=2x+10$ $\displaystyle (y=mx+b)$,with slope $\displaystyle m$.

$\displaystyle m=2$     and      $\displaystyle (5,1)$:

$\displaystyle y-y_{1}=m(x-x_{1})$

$\displaystyle y-1=2(x-5)$

$\displaystyle y-1=2x-10$

$\displaystyle y=2x-10+1$

$\displaystyle y=2x-9$

The parallel line has the equation $\displaystyle \dpi{100} \small 4x-2y=5$. We can find the slope by putting the equation into slope-intercept form, y = mx + b, where m is the slope and b is the intercept.  $\displaystyle \dpi{100} \small 4x-2y=5$ becomes $\displaystyle \dpi{100} \small y=2x-\frac{5}{2}$, so the slope is 2.

We know that our line must have an equation that looks like $\displaystyle \dpi{100} \small y=2x+b$. Now we need the intercept. We can solve for b by plugging in the point (4, 1).

1 = 2(4) + b

b = –7

Then the line in question is $\displaystyle \dpi{100} \small y=2x-7$.

Parallel lines have the same slope, so our slope will still be 4. The y-intercept is just the "+b" at the end. In f(x) the y-intercept is 13. In this case, we need to have a y-intercept of -13, so our equation just becomes:

$\displaystyle \small \small f(x)=4x-13$

Two lines are parallel if they have the same slope. The slope of g(x) is 6, so eliminate anything without a slope of 6.

Recall slope intercept form which is $\displaystyle y=mx+b$.

We know that the line must have an m of 6 and an (x,y) of (8,9). Plug everything in and go from there.

$\displaystyle 9=(6\cdot 8)+b$

$\displaystyle \small 9=48+b$

$\displaystyle \small b=9-48=-39$

So we get:

$\displaystyle \small y=6x-39$

Given a line $\displaystyle a$ defined by the equation $\displaystyle f(x)=mx+b$ with slope $\displaystyle m$, any line that is parallel to $\displaystyle a$ also has a slope of $\displaystyle m$. Since $\displaystyle f(x)=7x-12$, the slope $\displaystyle m$ is $\displaystyle 7$ and the slope of any line $\displaystyle g(x)$ parallel to $\displaystyle f(x)$ also has a slope of $\displaystyle m=7$.

Since $\displaystyle g(x)$ also needs to have a $\displaystyle y$-intercept of $\displaystyle 8$, then the equation for $\displaystyle g(x)$ must be $\displaystyle g(x)=7x+8$. 

Given a line $\displaystyle a$ defined by the equation $\displaystyle f(x)=mx+b$ with slope $\displaystyle m$, any line that is parallel to $\displaystyle a$ also has a slope of $\displaystyle m$. Since $\displaystyle f(x)=-\frac{6}{5}x+7$, the slope $\displaystyle m$ is $\displaystyle -\frac{6}{5}$ and the slope of any line $\displaystyle g(x)$ parallel to $\displaystyle f(x)$ also has a slope of $\displaystyle m=-\frac{6}{5}$.

Since $\displaystyle g(x)$ also needs to have a $\displaystyle y$-intercept of $\displaystyle -2$, then the equation for $\displaystyle g(x)$ must be $\displaystyle g(x)=-\frac{6}{5}x-2$. 

Given a line $\displaystyle a$ defined by the equation $\displaystyle f(x)=mx+b$ with slope $\displaystyle m$, any line that is parallel to $\displaystyle a$ also has a slope of $\displaystyle m$. Since $\displaystyle f(x)=2x+5$, the slope $\displaystyle m$ is $\displaystyle 2$ and the slope of any line $\displaystyle g(x)$ parallel to $\displaystyle f(x)$ also has a slope of $\displaystyle m=2$.

Since $\displaystyle g(x)$ also needs to have a $\displaystyle y$-intercept of $\displaystyle -17$, then the equation for $\displaystyle g(x)$ must be $\displaystyle g(x)=2x-17$. 

Perpendicular lines have slopes that are negative reciprocals of each other.