Answer:
The equation of line perpendicular to given line passing through (-15,-4) is:
[tex]y = -\frac{4}{3}x-24[/tex]
Step-by-step explanation:
Given equation is:
[tex]3x-4y = 9[/tex]
The given equation is in standard form. We have to convert it into slope intercept form first.
The slope intercept form is:
[tex]y = mx+b[/tex]
So,
[tex]3x-4y=9\\3x-9 = 4y\\4y = 3x-9\\\frac{4y}{4} = \frac{3x-9}{4}\\y = \frac{3}{4}x - \frac{9}{4}[/tex]
The co-efficient of x is the slope of the line.
m = 3/4
The product of slopes of two perpendicular lines is: -1
Let m1 be the slope of line perpendicular to given line then
[tex]m.m_1 =-1\\\frac{3}{4} . m_1 = -1\\m_1 = -1 * \frac{4}{3} \\m_1 = -\frac{4}{3}[/tex]
Putting in slope-intercept form
[tex]y = m_1x+b\\y = -\frac{4}{3}x+b[/tex]
To find the value of b, putting (-15,-4) in equation
[tex]-4 = -\frac{4}{3}(-15) +b\\-4 = 20+b\\-4-20 = b\\b = -24[/tex]
The equation will be:
[tex]y = -\frac{4}{3}x-24[/tex]
Hence,
The equation of line perpendicular to given line passing through (-15,-4) is:
[tex]y = -\frac{4}{3}x-24[/tex]
