SOLUTION: Write the equation of a line containing (-1,7) that is perpendiculas to 3x-5y=11. please help I do not get it at all!!! Thank you sooo much. red

Algebra ->  Linear-equations -> SOLUTION: Write the equation of a line containing (-1,7) that is perpendiculas to 3x-5y=11. please help I do not get it at all!!! Thank you sooo much. red      Log On


   

Question 226490: Write the equation of a line containing (-1,7) that is perpendiculas to 3x-5y=11. please help I do not get it at all!!! Thank you sooo much. red
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You have to get the equation in the form y+=+mx+%2B+b
where m is the slope, so
3x+-+5y+=+11
Subtract 3x from both sides
-5y+=+-3x+%2B+11
Divide both sides by -5
y+=+%283%2F5%29%2Ax+-+11%2F5
Now you can see that the slope is m+=+3%2F5
ANY line which is perpendicular to this line
will have a slope which is -%281%2Fm%29, so
-%281%2Fm%29+=+-%281%2F%283%2F5%29%29
-%281%2F%283%2F5%29%29+=+-%285%2F3%29
So now I know that the line I want will look mlike
y+=+-%285%2F3%29x+%2B+b
I'm told the perpendicular goes through (-1,7)
I can use the formula
%28y+-+7%29%2F%28x+-+%28-1%29%29+=+-5%2F3
%28y+-+7%29%2F%28x+%2B+1%29+=+-5%2F3
Multiply both sides by 3%2A%28x%2B1%29
3%2A%28y+-+7%29+=+-5%2A%28x+%2B+1%29
3y+-+21+=+-5x+-+5
3y+=+-5x+%2B+16
y+=+-%285%2F3%29x+%2B+16%2F3 answer
I'll plot both equations

This looks right, because if y+=+0 for both equations,
x=3.67 for one and x+=+3.2 for the other