Find the equation of a line that is perpendicular to the line y=1/8x +8 and contains the point (-2,0).
Perpendicular lines have their slopes as a flipped fraction of the original slope and opposite signed. A perpendicular line to an equation with a slope of (2/5) would have a slope of (-5/2)
What is the slope of the equation you posted? Flip the fraction and make it negative to find the slope of the perpendicular equation.
*the slope is always the number next to x
The slope of the equation is (1/8) so a perpendicular slope would be -8
-8x +8
Ignore the +8, that's the y=intercept of the original equation.
ok
Now we use point-intercept form. y-(y1)=m(x-x1) Using the point (-2,0). 0 is y1, m is perpendicular slope (-8), and x is -2
y-0=-8(x-(-2) solve the equation for y
Can you do that?
trying to solve it.... a bit slow but getting there
y=-8x +16
y-0=-8(x-(-2)) y-0=-8(x+2) x-(-2) = x+2
You are close, it should be y=-8x-16 (double checking)
y=-8(x+2) y=-8x-16
sure negative 8 times positive 2 is negative 16. got it. thx
i always miss that
People get confused with x-negative number, Minus a negative is a positive.
Will remember that. thx.
Thanks larrybox for simplifying this for me. you sure made it easy to understand.
In the eqation y=5x+5, what would be the perpendicular slope?
so what happens to the y intercept
meant -5x
Be careful, in this equation, the fraction is (5/1)
So the perpendicular slope would be (-1/5)
Nothing happens to the y-intercept, it just is not needed in calculation.
you know what? I thought about that but changed my mind. silly me!
Anything over 1 is itself. (5/1) 5, (-1/1) = -1
maths maths maths! not my strongest point at all
if the perpendicular slope for 1/8x = -8, then 5x = 5/1 would be -1/5
i should start saying math is interesting and gradually will get there! Thx lbp
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