SOLUTION: write an equation of a line that passes through the point (5,-3) and is perpendicular to the line 4x-7y=11 4x -7y=11 slope 4over7 y-y1=m(x-x1)

Question 406571: write an equation of a line that passes through the point (5,-3) and is perpendicular to the line 4x-7y=11
4x -7y=11 slope 4over7 y-y1=m(x-x1)

7y=4x+11
7 7 7
y=4+11
7 7
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
slope is 4/7 which is right.
The perpendicular line will have a slope which is negative reciprocal of the slope of the given line.
Since m1*m2=-1
so slope of the required line will be -7/4
..
line passes through (5,-3)
plug the value of slope , and the point to find b the y intercept
-3=(-7/4)*5+b
-3+35/4=b
+23/4 = b
...
the equation will be y = (-x/4)+23/4
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