document.write( "Question 40449: I posted a question just a few minutes ago and Nate answered it but he didn't exactly answer what I am looking for. My slope is 2/3 but how do I get this into y=mx+b format. The question is Line 2 is perpendicular to a line that passes through the points (4,-3) and (-2,6). I know that the slope is 2/3 but how do I get this into y=mx+b? \n" ); document.write( "

Algebra.Com's Answer #25837 by smik(40) $\"\"$ $\"About$

You can put this solution on YOUR website!
Well, you can put that slope value into y=mx+b, but I don't believe you can get the value for b. Let me explain. First of all, the equation would be y=(2/3)x+b. Now, what you'd need is one single point that Line 2 passes through. We know that there's a line with the points (4,-3) and (-2,6) but we can't get the value for b without a point that Line 2 passes through.
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\n" ); document.write( "I mean, you can imagine this graphically as well. Plot the two given points and draw the first line. Now, Line 2 has to have a slope of 2/3, but that can be anywhere on the first line. I could shift line 2 up and down and still maintain its slope.
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\n" ); document.write( "However, if you do find a point that Line 2 passes through then you just plug the point (x,y) into the equation y = (2/3)x + b, and solve for b. \n" ); document.write( "

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