Question

Perpendicular to the line 4x - 6y + 7 = 0 , passing through (3, 4) .

Answer 1

The equation of line perpendicular to given line passing through (3,4) is:

`y = -(3)/(2)x+(17)/(2)`

Explanation:

Given equation of line is:

`4x - 6y + 7 = 0`

Given equation is in standard form. It has to be converted into slope-intercept form to extract slope from the equation.

So,

`4x+7 = 6y\\6y=4x+7\\(6y)/(6) = (4x+7)/(6)\\y = (4)/(6)x+(7)/(6)\\y=(2)/(3)x+(7)/(6)`

The standard form of slope-intercept form of equation is:

`y=mx+b`

Here, the co-efficient of x is the slope of the line.

So the slope of given line is: 2/3

m = 2/3

The product of slopes of two perpendicular lines is -1

Let m1 be the slope of line perpendicular to given line

Then

`m.m_1 = -1\\(2)/(3) . m_1 = -1\\m_1 = -1*(3)/(2)\\m_1 = -(3)/(2)`

The equation of perpendicular will be:

`y = m_1x+b`

Putting the value of slope

`y = -(3)/(2)x+b`

To find the value of b, putting (3,4) in the equation

`4 = -(3)/(2)(3)+b\\4 = -(9)/(2)+b\\b = 4+(9)/(2)\\b= (8+9)/(2)\\b=(17)/(2)`

So the equation of line perpendicular to given line passing through (3,4) is:

`y = -(3)/(2)x+(17)/(2)`