Find the equation of the line that passes through (-2,5) and (-6,-3)
y=mx+b
5=m(-2)+b
-3=m(-6)+b
divide the second equation by -3
then add the equations. That gives you b.
then go back to either equation to solve for m.
can you show me them
thats part of algebra 2 that's easy were doing that now in pre algebra
To find the equation of a line that passes through two points, you can use the slope-intercept form of a linear equation, which is given by:
y = mx + b
where m is the slope of the line, and b is the y-intercept.
Step 1: Find the slope (m) of the line using the formula:
m = (y2 - y1) / (x2 - x1)
Let's label point 1 as (x1, y1) = (-2, 5) and point 2 as (x2, y2) = (-6, -3).
m = (-3 - 5) / (-6 - -2)
m = -8 / (-6 + 2)
m = -8 / -4
m = 2
Step 2: Use one of the given points and the slope (m) to find the y-intercept (b).
Using point 1 (-2, 5):
y = mx + b
5 = 2(-2) + b
5 = -4 + b
b = 9
Step 3: Substitute the slope (m) and y-intercept (b) into the slope-intercept form equation.
y = mx + b
y = 2x + 9
Therefore, the equation of the line that passes through (-2, 5) and (-6, -3) is y = 2x + 9.