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)
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-> 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)
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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)
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
