document.write( "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
\n" ); document.write( "m = y2-y1 / x2-x1 slope Formula
\n" ); document.write( "4/3 = -3-r /2-5 Let (5,r)= (x1,y1) and (2,-3)= x2,y2)
\n" ); document.write( "4/3 = -3-r / 3 Subtact
\n" ); document.write( "4(-3-r) = 3(3) Find the cross products
\n" ); document.write( "10+4r = 9 Simplify
\n" ); document.write( "35r = 39 Add 30 to each side and simplify
\n" ); document.write( "11.6 = 13 Divide each side by 3 and simplify
\n" ); document.write( "So the line goes threw
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #315597 by Alan3354(69443)\"\" \"About 
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
\n" ); document.write( "Slope m = diffy/diffx
\n" ); document.write( "m = (-3-r)/(2-5) = 4/3
\n" ); document.write( "(-3-r)/(-3) = 4/3
\n" ); document.write( "(3+r) = 4
\n" ); document.write( "r = 1
\n" ); document.write( "--------------
\n" ); document.write( "This is what I did
\n" ); document.write( "m = y2-y1 / x2-x1 slope Formula
\n" ); document.write( "4/3 = -3-r /2-5 Let (5,r)= (x1,y1) and (2,-3)= x2,y2)
\n" ); document.write( "4/3 = -3-r / 3 Subtact ***** DEN is -3
\n" ); document.write( "4(-3-r) = 3(3) Find the cross products
\n" ); document.write( "*********** should be 4*-3 = 3*(-3-r)
\n" ); document.write( "-12 = -9-3r
\n" ); document.write( "etc
\n" ); document.write( "10+4r = 9 Simplify
\n" ); document.write( "35r = 39 Add 30 to each side and simplify
\n" ); document.write( "11.6 = 13 Divide each side by 3 and simplify\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );