Question

line jk passes through points j(-4, -5) and k(-6, 3). if the equation of the line is written in slope-intercept form, y = mx + b, what is the value of b?
-21
-4
11
27

Explanation:

Step1: Calculate the slope \( m \)

The slope formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). For points \( J(-4,-5) \) and \( K(-6,3) \), we have \( x_1=-4,y_1 = - 5,x_2=-6,y_2 = 3 \).

$$ m=\frac{3-(-5)}{-6-(-4)}=\frac{3 + 5}{-6 + 4}=\frac{8}{-2}=-4 $$

Step2: Use point - slope form to find the equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \). We can use point \( J(-4,-5) \) and \( m=-4 \).

$$ y-(-5)=-4(x - (-4)) $$

$$ y + 5=-4(x + 4) $$

Step3: Convert to slope - intercept form (\(y=mx + b\))

Expand the right - hand side:

$$ y+5=-4x-16 $$

Subtract 5 from both sides:

$$ y=-4x-16 - 5 $$

$$ y=-4x-21 $$

In the slope - intercept form \( y = mx + b \), \( m=-4 \) and \( b=-21 \).

Answer:

-21