Please help with this.

The line that passes through (1,5) and (4,-4) 

Point 1: $$x_1 = 1, y_1=5$$

Point 2: $$x_2=4, y_2=-4$$

$$m=\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}$$

$$\begin{array}{rcl} (x-x_1)(y_2-y_1) & = & (y-y_1)(x_2-x_1)\\ x(y_2-y_1)-x_(y_2-y_1) & = & y(x_2-x_1) - y_1(x_2-x_1)\\ y(x_2-x_1) &=& x(y_2-y_1)-x_1(y_2-y_1) + y_1(x_2-x_1) \quad | \quad : (x_2-x_1) \\\\ y &=& \left( \frac{y_2-y_1}{x_2-x_1} \right)x -x_1 \left( \frac{y_2-y_1}{x_2-x_1}\right) + y_1\\\\ y &=& \underbrace{ \left( \frac{y_2-y_1}{x_2-x_1} \right)}_{m} x + \underbrace{ \frac{y_1x_2-x_1y_2}{x_2-x_1} }_{b} \end{array}$$

$$y =\left( \frac{y_2-y_1}{x_2-x_1} \right) x + \frac{y_1x_2-x_1y_2}{x_2-x_1}$$

$$y =\left( \frac{-4-5}{4-1} \right) x + \frac{5*4-1(-4)}{4-1}$$

$$y =\left( \frac{-9}{3} \right) x + \frac{20+4 }{3}$$

$$y =-3x +8$$

$$slope=\frac{y_2-y_1}{x_2-x_1}$$

Slope-intercept form:

$$y=(slope)x+b$$

To find b, take one point, and put that in for y and x. Then, solve for b!

Let me know if you need any more help.

thank you guys