Question

the equation of line cd is y = 3x − 3. write an equation of a line perpendicular to line cd in slope-intercept form that contains point (3, 1). y = 3x 0 y = −3x − 8 y = negative 1 over 3x 2 y = − 1 over 3x 0

Answer 1

Answer:

The required equation is $y=-\frac{1}{3}x+2$.

Step-by-step explanation:

The equation of line cd is

$y=3x-3$

Slope intercept form of a line is

$y=mx+b$

Where, m is slope and b is y-intercept.

Slope of line cd is 3.

The product of slopes of two perpendicular lines is -1.

$m_1\times m_2=-1$

$3\times m_2=-1$

$m_2=-\frac{1}{3}$

Therefore slope of perpendicular line is $-\frac{1}{3}$.

Point slope form of a line is

$y-y_1=m(x-x_1)$

Slope of perpendicular line is $-\frac{1}{3}$ and line passing through the point (3,1).

$y-1=-\frac{1}{3}(x-3)$

$y=-\frac{1}{3}x+1+1$

$y=-\frac{1}{3}x+2$

Therefore the required equation is $y=-\frac{1}{3}x+2$.

Answer 2

Y=3x-3 => slope=m=3
The new line perpendicular has slop = -1/m=-1/3
y=-1/3x+b, b is intercept point (3,1)
=>b=y+1/3x=1+(1/3)3=2
=> y=-1/3x+2