<u>Answer</u>:
Equation of a line perpendicular to and passes through the point ( 2 , − 6 ) is
5x – 7y = 4.
<u>Explanation</u>:
Need to write equation of line perpendicular to y = 7/5x + 6 and passes through the point ( 2 , − 6 )
Generic slope intercept form of a line is given by y = mx + c , where m = slope of the line.
On comparing the given slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line , slope m = 7/5
Product of slope of perpendicular line is -1 .
Let say slope of required line perpendicular to y = 7/5x + 6 is represented by m1
And as Product of slope of perpendicular line is -1 .
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so now we need to find the equation of a line whose slope is and passing through (2 , -6).
Equation of line passing through (x1 , y1) and having slope of m is given by
(y – y1) = m (x – x1)
In our case x1 = 2 and y1 = -6 and m = -5/7
Substituting the values in equation of line we get
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=> 7(y +6) = -5x +10
=> 5x – 7y = 10 – 6
=> 5x – 7y = 4
Hence equation of a line perpendicular to and passes through the point ( 2 , − 6 ) is
5x – 7y = 4.