# Find the equation of the line containing the points (-2,3) and (2,-3).

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The equation of a line with the points (x1, y1) and (x2, y2) is given by (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1)

We have to find the line with the points (-2, 3) and (2, -3).

The equation of the line is (y + 3)/(x - 2) = (3 + 3)/(-2 - 2)

=> (y + 3)/(x - 2) = 6/(-4)

=> (y + 3)/(x - 2) = -3/2

=> 2*(y + 3) = -3*(x - 2)

=> 2y + 6 = -3x + 6

=> 2y + 3x = 0

**The line with the points (-2, 3) and (2, -3) is 3x + 2y = 0**

Step 1: Find the slope from the two points given to you by using the formula Slope (m) = (y2 –y1)/( x2 – x1)

x1=-2, y1=3, x2=2, y2=-3 so m= (-3 –3)/( 2– -2)=-6/4=**-3/2**

Step2: Find the y-intercept (b) using the slope-intercept form(y=mx+b)

Chose one the point to sub for your x and y (I will choose the first point) so xà -2 and y à 3 and of course you just solved for the slope so you now know m à -3/2

Y = m x + b

3 = (-3/2)( -2) + b Now solve the equation for b

3 = 3 + b

**0 = b **

Step 3:** **Now you know both m(slope) and b(y-intercept) so you can write the equation as

**y=(-3/2) x + 0 OR y=(-3/2) x **