Question

Write the equation of a line that is perpendicular to y = 7/5 x + 6 y= 5/7and that passes through the point ( 2 , − 6 )

Answer 1

Answer:

Equation of a line perpendicular to$y = \frac{7}{5x} + 6$ 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 $y = \frac{7}{5}x + 6$, 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 .

=> $m \times m1 = -1$

=>$\frac{7}{5}\times m1 = -1$

=>$m1 = \frac{-5}{7}$

so now we  need to find the equation of a line whose slope is $\frac{-5}{7}$ 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

=>  $(y-(-6)) = \frac{-5}{7} (x-2)$

=>$y +6= \frac{-5}{7} x + \frac{10}{7}$

=> 7(y +6) = -5x +10

=> 5x – 7y = 10 – 6

=> 5x – 7y = 4  

Hence equation of a line perpendicular to$y = \frac{7}{5x} + 6$  and passes through the point ( 2 , − 6 ) is  

5x – 7y = 4.