Question

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

Answer 1

Answer:

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

$y = -\frac{3}{2}x+\frac{17}{2}$

Step-by-step 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,

$4x+7 = 6y\\6y=4x+7\\\frac{6y}{6} = \frac{4x+7}{6}\\y = \frac{4}{6}x+\frac{7}{6}\\y=\frac{2}{3}x+\frac{7}{6}$

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

$m.m_1 = -1\\\frac{2}{3} . m_1 = -1\\m_1 = -1*\frac{3}{2}\\m_1 = -\frac{3}{2}$

The equation of perpendicular will be:

$y = m_1x+b$

Putting the value of slope

$y = -\frac{3}{2}x+b$

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

$4 = -\frac{3}{2}(3)+b\\4 = -\frac{9}{2}+b\\b = 4+\frac{9}{2}\\b= \frac{8+9}{2}\\b=\frac{17}{2}$

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

$y = -\frac{3}{2}x+\frac{17}{2}$