line jk passes through points j(-4, -5) and k(-6, 3). if the equation of the line is written in slope…

Explanation:

Step1: Calculate the slope $m$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Given $J(-4,-5)$ and $K(-6,3)$, we have $x_1=-4,y_1 = - 5,x_2=-6,y_2 = 3$. Then $m=\frac{3-(-5)}{-6-(-4)}=\frac{3 + 5}{-6 + 4}=\frac{8}{-2}=-4$.

Step2: Substitute a point and the slope into the equation $y=mx + b$

Substitute the point $J(-4,-5)$ and $m=-4$ into $y=mx + b$. We get $-5=-4\times(-4)+b$.

Step3: Solve for $b$

First, simplify the right - hand side: $-5 = 16 + b$. Then subtract 16 from both sides: $b=-5 - 16=-21$.

Answer:

-21