Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

$\displaystyle m=\frac{y'-y}{x'-x}$

Plug in and calculate:

$\displaystyle \small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}$

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

$\displaystyle \small 5=\frac{1}{2}*4+b$

$\displaystyle \small b=5-2=3$

So our answer is: 

$\displaystyle \small y=\frac{x}{2}+3$

The equation for a line in standard form is written as follows:

$\displaystyle y=mx+b$

Where $\displaystyle m$ is the slope and $\displaystyle b$ is the y intercept. We start by calculating the slope between the two given points using the following formula:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{7-3}{6-(-2)}=\frac{4}{8}=\frac{1}{2}$

Now we can plug either of the given points into the formula for a line with the calculated slope and solve for the y intercept:

$\displaystyle y=mx+b$

$\displaystyle 3=(\frac{1}{2})(-2)+b$

$\displaystyle b=4$

We now have the slope and the y intercept of the line, which is all we need to write its equation in standard form:

$\displaystyle y=\frac{1}{2}x+4$

The $\displaystyle y$-intercept of the parabola of the equation can be found by substituting 0 for $\displaystyle x$:

$\displaystyle y = x^{2} + 5x - 6$

$\displaystyle y = 0^{2} + 5 (0) - 6$

$\displaystyle y = -6$

This point is $\displaystyle (0, -6)$.

The vertex of the parabola of the equation $\displaystyle y = a x^{2} + bx+c$ has $\displaystyle x$-coordinate $\displaystyle - \frac{b}{2a}$, and its $\displaystyle y$-coordinate can be found using substitution for $\displaystyle x$. Setting $\displaystyle a = 1$ and $\displaystyle b= 5$:

$\displaystyle x = - \frac{b}{2a} = - \frac{5}{2 \cdot 1} = - \frac{5}{2 }$

$\displaystyle y = \left ( - \frac{5}{2 } \right )^{2} + 5 \left ( - \frac{5}{2 } \right ) - 6$

$\displaystyle y = \frac{25}{4 } - \frac{25}{2 } - 6$

$\displaystyle y = \frac{25}{4 } - \frac{50}{4 } - \frac{24}{4}$

$\displaystyle y = - \frac{49}{4 }$

The vertex is $\displaystyle \left (- \frac{5}{2 },- \frac{49}{4 } \right )$

The line connects the points $\displaystyle (0, -6)$ and $\displaystyle \left (- \frac{5}{2 },- \frac{49}{4 } \right )$. Its slope is

$\displaystyle m= \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

$\displaystyle = \frac{-6-\left ( - \frac{49}{4 } \right )}{0 - \left (- \frac{5}{2 } \right )}$

$\displaystyle = \frac{-\frac{24 }{4 } + \frac{49}{4 } }{ \frac{5}{2 } }$

$\displaystyle = \frac{ \frac{25}{4 } }{ \frac{5}{2 } }$

$\displaystyle = \frac{25}{4 } \cdot \frac{2 } {5}$

$\displaystyle =\frac{5}{2 }$

Since the line has $\displaystyle y$-intercept $\displaystyle (0, -6)$ and slope $\displaystyle m =\frac{5}{2 }$, the equation of the line is $\displaystyle y=\frac{5}{2 } x + (- 6)$, or $\displaystyle y=\frac{5}{2 } x - 6$.

Rewrite this equation in slope-intercept form: $\displaystyle y = mx + b$, where $\displaystyle m$ is the slope.

$\displaystyle Ax + 2Ay = 3A$

$\displaystyle 2Ay = - Ax + 3A$

$\displaystyle \frac{2Ay}{2A} = \frac{- Ax + 3A}{2A}$

$\displaystyle y = -\frac{1}{2}x+ \frac{3}{2}$

The slope is the coefficient of $\displaystyle x$, which is $\displaystyle -\frac{1}{2}$.

Rewrite in the slope-intercept form $\displaystyle y = mx + b$:

$\displaystyle 0.4x - 0.5y = 20$

$\displaystyle \left (0.4x - 0.5y \right )\cdot 10 = 20 \cdot 10$

$\displaystyle \left (0.4x \right )\cdot 10-\left ( 0.5y \right )\cdot 10 = 200$

$\displaystyle 4x-5y = 200$

$\displaystyle 4x-4x-5y = 200-4x$

$\displaystyle -5y = -4x+ 200$

$\displaystyle -5y \div (-5)=\left ( -4x+ 200 \right )\div (-5)$

$\displaystyle y=\left ( -4x \right ) \div (-5)+\left ( 200 \right ) \div (-5)$

$\displaystyle y= \frac{4}{5}x-40$

The slope is the coefficient of $\displaystyle x$, which is $\displaystyle \frac{4}{5}$.

Rewrite in the slope-intercept form $\displaystyle y = mx + b$:

$\displaystyle \frac{2}{5}x + \frac{6}{7}y = 9$

$\displaystyle \frac{2}{5}x - \frac{2}{5}x + \frac{6}{7}y = - \frac{2}{5}x+ 9$

$\displaystyle \frac{6}{7}y = - \frac{2}{5}x+ 9$

$\displaystyle \frac{6}{7}y \cdot \frac{7}{6}=\left ( - \frac{2}{5}x+ 9 \right )\cdot \frac{7}{6}$

$\displaystyle y=- \frac{2}{5}\cdot \frac{7}{6}x+ 9 \cdot \frac{7}{6}$

$\displaystyle y=-\frac{7}{15}x+ \frac{21}{2}$

The slope is the coefficient of $\displaystyle x$, which is $\displaystyle -\frac{7}{15}$

Rewrite in the slope-intercept form $\displaystyle y = mx + b$:

$\displaystyle 5y = x - 7$

$\displaystyle \frac{1}{5} \cdot 5y = \frac{1}{5} \cdot \left ( x - 7 \right )$

$\displaystyle y = \frac{1}{5} x - \frac{7}{5} \right )$

The slope is the coefficient of $\displaystyle x$, or $\displaystyle \frac{1}{5}$.

The slope formula is:

$\displaystyle m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle (1,4)\ and\ (7,10)$

$\displaystyle m= \frac{7-1}{10-4}$

$\displaystyle m= \frac{6}{}6$

$\displaystyle m=1$