I need help please The equation of line XY is (y−3) = negative 2 over 3(x − 4). What is the slope of a line perpendicular to XY?
I think its -2/3
Is the equation of Line XY \[y-3=-2/3(x-4)\]
yes
What is the slope of a line perpindicular to another line?
its reciprocal?
yes, but there is one more thing, it is the negative reciprocal
You find the reciprocal of the slope, but multiply it by -1 as well.
I really dont understand algebra
Okay, how about this. If the slope of XY is -2/3 what happens when you turn it upside down and change the sign?
2/-3
Actually it would be 3/2. The reciprocal would be -3/2 and changing the sign makes it just 3/2
ooh i thought you just turn them
3/2 is the reciprocal of 2/3, 2 is the reciprocal of 1/2, 1/3 would be the reciprocal of what?
You turn the numerator into the denominator and the denominator into the numerator
3/1
exactly :)
thank you how about this one The equation of line QR is y = negative 1 over 2x + 1. Write an equation of a line perpendicular to line QR in slope-intercept form that contains point (5, 6).
What is the slope? It is "rise over run" right? Or the change in y over the change in x.
I believe rise over run i think
If the equation of the line is y=-(1/2)x+1 then the slope is -1/2 right? What is the slope of the perpendicular line? What is the negative reciprocal of -1/2?
2/-1
Yes, you got this down. Now we just need to find a line with that slope and passes through 5,6
ok
Oh wait, I missed the -1... it shoudl actually be 2/1 since it is the negative reciprocal of -1/2, remember to change the sign.
so our new line has a slope of 2, the slope = the change in y over the change in x. m=(y-6)/(x-5)
is it y=2x-3
y-6 is the change of the line from some point y and 6 which is the point it needs to pass through, x-5 is the change in the line where x is some point and 5 is the x value the line has to pass through.
m=2=(y-6)/x-5) y-6=2(x-5) y=2x-4
You can check your equation by sustituting (5,6) into it. If x = 5 then y=2(5)-4=6
ohh lol I was looking at the wrong question thank you though
Join our real-time social learning platform and learn together with your friends!