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Question 460129: Find the value of r so that the line through (5,r)and (2,-3)has a slope of 4/3 I know I did some thing wrong I just don't know what please help me out thanks This is what I did
m = y2-y1 / x2-x1 slope Formula
4/3 = -3-r /2-5 Let (5,r)= (x1,y1) and (2,-3)= x2,y2)
4/3 = -3-r / 3 Subtact
4(-3-r) = 3(3) Find the cross products
10+4r = 9 Simplify
35r = 39 Add 30 to each side and simplify
11.6 = 13 Divide each side by 3 and simplify
So the line goes threw
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the value of r so that the line through (5,r)and (2,-3)has a slope of 4/3
Slope m = diffy/diffx
m = (-3-r)/(2-5) = 4/3
(-3-r)/(-3) = 4/3
(3+r) = 4
r = 1
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This is what I did
m = y2-y1 / x2-x1 slope Formula
4/3 = -3-r /2-5 Let (5,r)= (x1,y1) and (2,-3)= x2,y2)
4/3 = -3-r / 3 Subtact ***** DEN is -3
4(-3-r) = 3(3) Find the cross products
*********** should be 4*-3 = 3*(-3-r)
-12 = -9-3r
etc
10+4r = 9 Simplify
35r = 39 Add 30 to each side and simplify
11.6 = 13 Divide each side by 3 and simplify
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