Asked by Anonymous
                Passing through point (1,4); perpendicular to 
2x+y=5
How do I solve this? I think it might be this but not sure please let me know what step I left out
Thanks:-)
2x+y=5
y=5-2x
y=2x-5
M1=2
4=-1/2(1)+c=1+c
C=4
Y=-1/4x-1
            
        2x+y=5
How do I solve this? I think it might be this but not sure please let me know what step I left out
Thanks:-)
2x+y=5
y=5-2x
y=2x-5
M1=2
4=-1/2(1)+c=1+c
C=4
Y=-1/4x-1
Answers
                    Answered by
            Reiny
            
    from y = 5 - 2x
y = -2x + 5 , (why did you switch the signs in yours ???)
the slope is -2
so the slope of the new line is 1/2
y = (1/2)x + b
sub in the point (4,1)
4 = (1/2)(1) + b
4 - 1/2 = b
b = 7/2
new equation is
y = (1/2)x + 7/2
    
y = -2x + 5 , (why did you switch the signs in yours ???)
the slope is -2
so the slope of the new line is 1/2
y = (1/2)x + b
sub in the point (4,1)
4 = (1/2)(1) + b
4 - 1/2 = b
b = 7/2
new equation is
y = (1/2)x + 7/2
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