Find an equation

Question

Find an equation of the line that passes through the point (-5,5) and perpendicular to the line y= -(1/3)x+2

Answer 1

y-5 = (-1/3)(x+5)

y = -x/3 -5/3+15/3

y = -x/3+10/3

or

3y =-x +10

x+3y =10 is the parallel line through (-5,5)

to find the perpendicular line through (-5,5)

flip -1/3 upside down and change the sign to +3/1 = 3

then

y-5 = 3(x+5) = 3x+15

y= 3x+15+5

y = 3x +20 is the perpendicular line through (-5,5)

Answer 2

(-5,5) and perpendicular to the line y= -(1/3)x+2

y=-mx+b is slope intercept form of a linear equation with m the slope and b the y intercept.

-1/3 is the slope of the line perpendicular to the line to find. Perpendicular lines have slopes that are negative reciprocals = flip it, change the sign.

-1/3 becomes 3

y = 3x+b

Plug in the given point to find b.

5 = 3(-5)+ b

5 = -15 + b

20 = b

y = 3x + 20

No imagination is needed if the line's equation is found.

Answer 3

Hi Leigh Ann P.,

Perpendicular to == has negative inverse slope of.

So your original line has slope -(1/3) -- (that's the "m" in the standard slope-intercept line format y = mx + b )

so negative inverse = +3. So you now have a general line equation, y = 3x + b.

You plug in the (x, y) of your given point, to get: 5 = 3*(-5) + b . Do I imagine correctly you can solve for b from that?

-- Cheers, --Mr. d.