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

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

Answer 2

The required equation is
`y=-(1)/(3)x+2`.

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* m_2=-1`

`3* m_2=-1`

`m_2=-(1)/(3)`

Therefore slope of perpendicular line is
`-(1)/(3)`.

Point slope form of a line is

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

Slope of perpendicular line is
`-(1)/(3)` and line passing through the point (3,1).

`y-1=-(1)/(3)(x-3)`

`y=-(1)/(3)x+1+1`

`y=-(1)/(3)x+2`

Therefore the required equation is
`y=-(1)/(3)x+2`.