# What is the equation of a line perpendicular to 8y-2x=-6 and also passing through the point (2,3).

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### 2 Answers

We are asked to find the equation of a line perpendicular to

8y-2x = -6 which passes through the point (2,3).

We begin by finding the slope of the given line. We will convert the given equation to slope-intercept form.

=>8y - 2x = -6

=> 8y = 2x - 6

=> y = 2/8 x - 6/8

=> y = 1/4 x - 3/4

The slope of a line perpendicular to the above equation would be -4.

We substitute -4 as the perpendicular slope and the given point (2,3) into the slope intercept form to find "b."

=>y = mx + b

=> 3 = -4(2) + b

=> 3 = -8 + b

=> 11 = b

We now have the slope and the y intercept for the equation of the perpendicular line.

Substitute our values into the slope intercept form.

y = mx + b

y = -4x + 11.

**The answer is y = -4x + 11 (slope-intercept form). The standard form of the answer is 4x + y = 11.**

** **

The product of the slope of perpendicular lines is -1. First find the slope of the given line.

8y - 2x = -6

=> 8y = 2x - 6

=> y = x/4 - 3/4

This is in the slope-intercept form y = mx + c where m is the slope.

The slope is 1/4. As a line perpendicular to this line has a slope which is the negative reciprocal, the slope of the required line is -4. The line also passes through (2,3)

The equation of the line is : (y - 3)/(x - 2) = -4

=> y - 3 = -4x + 8

=> 4x + y - 11 = 0

**The equation of the required line is 4x + y - 11 = 0**