To solve for the x-intercept, substitute 0 for \(\displaystyle y\) and solve for \(\displaystyle x\): 

\(\displaystyle x + 2Ay = 4A\)

\(\displaystyle x+ 2A \cdot 0= 4A\)

\(\displaystyle x= 4A\)

The \(\displaystyle x\)-intercept is \(\displaystyle (4A,0)\).

Substitute 0 for \(\displaystyle x\) and solve for \(\displaystyle y\):

\(\displaystyle Ax + 2Ay = 3A\)

\(\displaystyle A \cdot 0 + 2Ay = 3A\)

\(\displaystyle 2Ay = 3A\)

\(\displaystyle 2Ay \div 2A = 3A\div 2A\)

\(\displaystyle y = \frac{3A}{2A} = \frac{3}{2}\)

The \(\displaystyle y\)-intercept is \(\displaystyle \left (0, \frac{3}{2} \right )\)

The slope of the line is 

\(\displaystyle m = \frac{y _{2}- y _{1}}{x _{2}- x _{1}}=\frac{1- 5}{7- 2} = \frac{-4}{5} = - \frac{4}{5}\)

Use the point slope form to find the equation of the line.

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

\(\displaystyle y - 5= - \frac{4}{5} \left ( x - 2\right )\)

Now substitute \(\displaystyle x=0\) and solve for \(\displaystyle y\).

\(\displaystyle y - 5= - \frac{4}{5} \left ( 0 - 2\right )\)

\(\displaystyle y - 5= \frac{8}{5}\)

\(\displaystyle y = \frac{8}{5} + 5 = \frac{8}{5} + \frac{25}{5} = \frac{33}{5} = 6\frac{3}{5}\)

The \(\displaystyle y\)-intercept is \(\displaystyle \left (0, 6\frac{3}{5} \right )\)

This can best be solved using a diagram and noting the intercepts of the line of the equation \(\displaystyle 3x + 4y = 15\), which are calculated by substituting 0 for \(\displaystyle x\) and \(\displaystyle y\) separately and solving for the other variable.

\(\displaystyle x\)-intercept:

\(\displaystyle 3x + 4y = 15\)

\(\displaystyle 3x + 4 \cdot 0 = 15\)

\(\displaystyle 3x = 15\)

\(\displaystyle x = 5\)

\(\displaystyle y\)-intercept:

\(\displaystyle 3x + 4y = 15\)

\(\displaystyle 3 \cdot 0 + 4y = 15\)

\(\displaystyle 4y = 15\)

\(\displaystyle y = 3 \frac{3}{4}\)

Now, we can make and examine the diagram below - the red line is the graph of the equation \(\displaystyle 3x + 4y = 15\):

The pink triangle is the one whose area we want; it is a right triangle whose legs, which can serve as base and height, are of length \(\displaystyle 5, 3\frac{3}{4}\). We can compute its area:

\(\displaystyle \frac{1}{2} \cdot 5 \cdot 3\frac{3}{4}= \frac{1}{2} \cdot \frac{5}{1} \cdot \frac{15}{4} = \frac{75}{8} = 9 \frac{3}{8}\)

To solve for the \(\displaystyle x\)-intercept, you have to set \(\displaystyle y\) to zero and solve for \(\displaystyle x\):

\(\displaystyle 3x+2y=2\)

\(\displaystyle 3x+2(0)=2\)

\(\displaystyle 3x=2\)

\(\displaystyle x=\frac{2}{3}\)

\(\displaystyle (\frac{2}{3},0)\)

To solve for the \(\displaystyle y\)-intercept, you have to set \(\displaystyle x\) to zero and solve for \(\displaystyle y\):

\(\displaystyle 3x+2y=2\)

\(\displaystyle 3(0)+2y=2\)

\(\displaystyle 2y=2\)

\(\displaystyle y=1\)

\(\displaystyle (0,1)\)

For some real number \(\displaystyle b\), the \(\displaystyle y\)-intercept of the line will be some point \(\displaystyle (0,b)\). We can set up the slope equation and solve for \(\displaystyle b\) as follows:

\(\displaystyle \frac{y_2-y_1}{x_2-x_1} = m\)

\(\displaystyle \frac{b-N}{0-6} =-\frac{4}{5}\)

\(\displaystyle \frac{b-N}{-6} =-\frac{4}{5}\)

\(\displaystyle \frac{b-N}{-6} \cdot (-6)=-\frac{4}{5}\cdot (-6)\)

\(\displaystyle b-N = \frac{24}{5}\)

\(\displaystyle b-N+N =N+ \frac{24}{5}\)

\(\displaystyle b =N+ \frac{24}{5}\)

Substitute 0 for \(\displaystyle x\):

\(\displaystyle y = 3x^{2} -2x - 16\)

\(\displaystyle y = 3 \cdot 0^{2} -2 \cdot 0 - 16 = 0 - 0 -16 = -16\)

The \(\displaystyle y \;\)-intercept is \(\displaystyle (0,-16)\)

Set \(\displaystyle y = 0\)

\(\displaystyle y = 3x^{2} -2x - 16\)

\(\displaystyle 3x^{2} -2x - 16 = 0\)

Using the \(\displaystyle ac\)-method, we look to split the middle term of the quadratic expression into two terms. We are looking for two integers whose sum is \(\displaystyle -2\) and whose product is \(\displaystyle 3 \left ( -16 \right ) = -48\); these numbers are \(\displaystyle -8,6\).

\(\displaystyle 3x^{2} -8x + 6x - 16 = 0\)

\(\displaystyle \left (3x^{2} -8x \right )+ \left ( 6x - 16 \right ) = 0\)

\(\displaystyle x \left (3x -8 \right )+ 2 \left (3 x - 8 \right ) = 0\)

\(\displaystyle \left ( x + 2 \right )\left (3x -8 \right ) = 0\)

Set each linear binomial to 0 and solve:

\(\displaystyle x + 2 = 0\)

\(\displaystyle x + 2 - 2 = 0 - 2\)

\(\displaystyle x = -2\)

or 

\(\displaystyle 3x - 8 =0\)

\(\displaystyle 3x - 8+ 8 =0 + 8\)

\(\displaystyle 3x = 8\)

\(\displaystyle 3x\div 3 = 8 \div 3\)

\(\displaystyle x = \frac{8}{3}\)

There are two \(\displaystyle x\)-intercepts - \(\displaystyle (-2,0), \left ( \frac{8}{3}, 0 \right )\)

The two points have the same \(\displaystyle x\) coordinate, which is 5; the line is therefore vertical. This makes the line parallel to the \(\displaystyle y\)-axis, meaning that it does not intersect it. Therefore, the line has no \(\displaystyle y\)-intercept.