Answer:
Equation of a line perpendicular to[tex]y = \frac{7}{5x} + 6[/tex] and passes through the point ( 2 , − 6 ) is 5x – 7y = 4.
Explanation:
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 [tex]y = \frac{7}{5}x + 6[/tex], 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 .
=> [tex]m \times m1 = -1[/tex]
=>[tex]\frac{7}{5}\times m1 = -1[/tex]
=>[tex]m1 = \frac{-5}{7}[/tex]
so now we need to find the equation of a line whose slope is [tex]\frac{-5}{7}[/tex] 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
=> [tex](y-(-6)) = \frac{-5}{7} (x-2)[/tex]
=>[tex]y +6= \frac{-5}{7} x + \frac{10}{7}[/tex]
=> 7(y +6) = -5x +10
=> 5x – 7y = 10 – 6
=> 5x – 7y = 4
Hence equation of a line perpendicular to[tex]y = \frac{7}{5x} + 6[/tex] and passes through the point ( 2 , − 6 ) is
5x – 7y = 4.
