Question

Find an equation of the line perpendicular to the graph of 14x-7y=8 passing through the point at (-2,5)

Answer 1

Equation of line perpendicular to given graph is:

`y = -(1)/(2)x+4`

Explanation:

Given equation of line is:

14x-7y=8

First of all, we have to convert the given equation into slope-intercept form to find the slope of the line

The slope-intercept form is:

`y = mx+b`

Now

`14x-7y=8\\14x-8 = 7y\\(7y)/(7) = (14x-8)/(7)\\y = (14)/(7)x - (8)/(7)\\y = 2x - (8)/(7)`

The co-efficient of x is 2 so the slope of given line is 2

Let m1 be the slope of line perpendicular to given line

The product of slopes of two perpendicular lines is -1

`m.m_1 = -1\\2.m_1 = -1\\m_1 = -(1)/(2)`

The equation for line perpendicular line to given line will be:

`y = m_1x+b\\y = -(1)/(2)x+b`

To find the value of b, putting (-2,5) in the equation

`5 = -(1)/(2)(-2) + b\\5 = 1+b\\b = 5-1\\b = 4`

The final equation is:

`y = -(1)/(2)x+4`

Hence,

Equation of line perpendicular to given graph is:

`y = -(1)/(2)x+4`