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Question 504669: Write an equation of a line passing through point(-1,9) and perpendicular to the line x+3y=2
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! line x+3y=2
3y=-x+2
y=(-x+2)/3
y=(-1/3)x+2/3
passing through the point (-1,9)
Equation of the line slope-intercept form
y=mx+b
Given line
y=(-1/3)x+2/3
Slope of given line m = -1/3
Since the required line is perpendicular to the given line the sign of this line will be opposite to sign of given line's slope and it will be reciprocal to the slope of the given line
Thus
slope (m) of required line will be = 3
Now we have a slope and we have a point (-1, 9)
We can easily find the equation of the line either by slope-intercept form or by Point slope-form.
Use slope-intercept
y=mx+b ..........(1)
here
y=9, x=-1, m= 3 and we have to find b
Plug in the value in the equation of line
y=mx+b
9=3(-1)+b
9=-3 + b
9+3= b
12=b
b=12
Put the value m and b in (1)
y=3x+12
Equation of the line passing through (-1, 9) and perpendicular to line x+3y=2 is
y=3x+12
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