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:

Equation of a line perpendicular to 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 , 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 .

=>

=>

=>

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

=>  

=>

=> 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.

p = 6s

step-by-step explanation: