Since the line passes through the point $(3,17)$ and its slope is $4$, we substitute $x=3$, $y=17,$ and $m=4$ into the equation in slope-intercept form:

\[ \eqalign{ y&=mx+b\\ (17)&=(4)(3)+b\\ 17&=12+b\\ 5 &= b } \]

So, the $y$-intercept is $b=5.$ The coordinates of the $y$-intercept are $(0,5).$

Since the line passes through the point $(7,13)$ and its slope is $-3$, we substitute $x=7$, $y=13,$ and $m=-3$ into the equation in slope-intercept form:

\[ \eqalign{ y&=mx+b\\ (13)&=\left(-3\right)(7)+b\\ 13&= -21+b\\ 34&=b } \]

So, the $y$-intercept is $b=34.$ The coordinates of the $y$-intercept are $(0,34).$

Since the line passes through the point $(12,6)$ and its slope is $\dfrac{1}{4}$, we substitute $x=12$, $y=6,$ and $m=\dfrac{1}{4}$ into the equation in slope-intercept form:

\[ \eqalign{ y &= mx + b \\ (6) &= \left(\dfrac 1 4\right) (12) + b \\ 6 &= 3 + b \\ 3 &= b } \]

So, the $y$-intercept is $b=3.$ The coordinates of the $y$-intercept are $(0,3).$

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

First, we calculate the slope $m$ using the given points:

\[ \begin{align} m &= \frac{y_2 - y_1}{x_2 - x_1}\\ &=\frac{11 -7}{3 - 2}\\ &=\frac{4}{1}\\ &=4\\ \end{align} \]

Substituting the slope $m=4$ into the slope-intercept formula $y=mx+b,$ we reach

\[ y = 4x + b. \]

We now need to find the $y$-intercept $b.$ We can find this by substituting the coordinates of a point on the line. Any point will do, so let's substitute $(2,7)\mathbin{:}$ \[ \begin{align} y &= 4x + b\\ 7 &=4(2) +b\\ 7 &=8+b\\ 7-8 &=b\\ -1 &=b \end{align} \] So the equation of the line is $y=4x-1.$

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

First, we calculate the slope $m$ using the given points:

\[ \begin{align} m &= \frac{y_2 - y_1}{x_2 - x_1}\\[5pt] &=\frac{1 - (-5)}{-2 - 4}\\[5pt] &=\frac{6}{-6}\\[5pt] &=-1\\ \end{align} \]

Substituting the slope $m=-1$ into the slope-intercept formula $y=mx+b,$ we reach

\[ y = - x + b. \]

We now need to find the $y$-intercept $b.$ We can find this by substituting the coordinates of a point on the line. Any point will do, so let's substitute $(-2,1)\mathbin{:}$

\[ \begin{align} y &=-x + b\\[5pt] 1 &=-(-2) +b\\[5pt] 1 &=2+b\\[5pt] 1-2 &=b\\[5pt] -1 &=b \end{align} \]

Therefore, the equation of the line is $y=-x-1.$

We will write the equation of the line in the slope-intercept form $y=mx+b.$ So, we need to work out the slope $m$ and the $y$-intercept $b.$

First, we calculate the slope $m$ using the given points:

\[ \begin{align} m &= \frac{y_2 - y_1}{x_2 - x_1}\\[3pt] &=\frac{2 - 1}{3 - 1}\\[3pt] &=\frac{1}{2} \end{align} \]

Substituting the slope $m=\dfrac 1 2$ into the slope-intercept formula $y=mx+b,$ we reach

\[ y = \dfrac 1 2 x + b. \]

We now need to find the $y$-intercept $b.$ We can find this by substituting the coordinates of a point on the line. Any point will do, so let's substitute $(1,1)\mathbin{:}$

\[ \begin{align} y &= \dfrac{1}{2}x + b\\[5pt] 1 &=\dfrac 1 2(1) +b\\[5pt] 1 &=\dfrac 1 2+b\\[5pt] 1-\dfrac 1 2 &=b\\[5pt] \dfrac 1 2 &=b \end{align} \]

Therefore, the equation of the line is $y=\dfrac {1}{2}x+ \dfrac{1}{2}.$

At the $x$-intercept, the $y$-coordinate is equal to $0.$ So, we can substitute $y=0$ into the equation and then solve for $x\mathbin{:}$ \begin{align*} y&=16x+18\\[5pt] 0 &=16x+18\\[5pt] -18 &= 16x \\[5pt] \dfrac{-18}{16} &= \dfrac{16x}{16} \\[5pt] -\dfrac98 &= x \end{align*}

Therefore, the $x$-intercept is $x=-\dfrac98.$ The coordinates of the $x$-intercept are $\left(-\dfrac98,0 \right).$

At the $x$-intercept, the $y$-coordinate is equal to $0.$ So, we can substitute $y=0$ into the equation and then solve for $x\mathbin{:}$

\[ \eqalign{ y&=-4x-16\\ 0 &=-4x-16\\ 4x &=-16 \\ \dfrac{4x}{4} &= \dfrac{-16}{4} \\ x &= -4 } \]

Therefore, the $x$-intercept is $x=-4.$ The coordinates of the $x$-intercept are $\left(-4,0 \right).$

First, we need to find the equation of the line. We know that $m=3,$ so we can write the equation of the line as $y=3x + b.$

It remains to find the $y$-intercept of the line. We can do that by substituting the point $(1,12)$ into the equation and solving for $b$ as follows:

\begin{align*} y &= 3x+b \\ (12) &= 3(1) + b \\ 12 &= 3+b \\ 9 &= b \end{align*}

So, the equation of the line is $y=3x+9.$

At the $x$-intercept, the $y$-coordinate is equal to $0.$ So, we can substitute $y=0$ into the equation and then solve for $x\mathbin{:}$

\begin{align*} y &= 3x+9 \\ 0 &= 3x+9 \\ -9 &= 3x \\ -3 &= x \end{align*}

Therefore, the $x$-intercept is $x=-3.$ The coordinates of the $x$-intercept are $(-3,0).$

On any horizontal line, all points have the same $y$-coordinate.

The $y$-coordinate of the point $(2,5)$ is $5,$ so the $y$-coordinate of the $y$-intercept is also $5.$

Therefore, the coordinates of the $y$-intercept are $(0,5).$

On any vertical line, all points have the same $x$-coordinate.

The $x$-coordinate of the point $(3,5)$ is $3,$ so the $x$-coordinate of the $x$-intercept is also $3.$

Therefore, the coordinates of the $x$-intercept are $(3,0).$

The line $y=4$ is horizontal. So, there is no $x$-intercept.