# Perpendicular lines....oohhh fun!Watch

This discussion is closed.
#1
Would someone mind double checking my answer?

Find an equation of the straight line:

a, perpendicular to the line y=3x+5, passing through (1,7)

This is what I did,

M = 3 therefore a line perpendicular to y=3x+5 and has the points (1,7) surely must have the gradient - 3 as 3x-1 = - 3

Next thing I did was to find the equation of the straight line of the line perpendicular to y=3x+5

I got y-7 = -3(x-1)
y-7 = - 3x + 3
3x-y-3-7 = 0

which equals:

3x-y-10

the answer in the back of the book is x+3y-22 = 0 - I have no idea how

Thanks
0
14 years ago
#2
(Original post by DOJO)
Would someone mind double checking my answer?

Find an equation of the straight line:

a, perpendicular to the line y=3x+5, passing through (1,7)

This is what I did,

M = 3 therefore a line perpendicular to y=3x+5 and has the points (1,7) surely must have the gradient - 3 as 3x-1 = - 3

0
14 years ago
#3
m=-1/3

y - 7 = -1/3(x-1)
-3y+21 = x - 1
x+3y-22=0

as required
0
#4
ohhhhhhhhhh, oh ok!

I always thought it was the negative of the gradient as the lines cross each other on the graph left and right...

Thank you v much, that silly mistake would have costed me mega marks!
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