Question 186620: Wite the slope-intercept form of an equation for a line that passes through (-1,0) and is perpendicular to the graph of 3x-5y=20 Found 2 solutions by nerdybill, uday1435:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Wite the slope-intercept form of an equation for a line that passes through (-1,0) and is perpendicular to the graph of 3x-5y=20
.
First, determine the slope of
3x-5y=20
.
Do this by putting it into the "slope-intercept" form:
y = mx + b
where
m is slope
b is y-intercept
.
3x-5y=20
-5y = 3x + 20
y = (-3/5)x + (-3/5)20
y = (-3/5)x - 12
.
So, if we want to be "perpendicular" to the line above, the NEW slope must be the "negative reciprocal" of -3/5.
Let M = our new slope
M(-3/5) = -1
M = 5/3
.
The NEW line now has
slope of 5/3
passing through (-1,0)
.
Plug the above into the "point-slope" form:
y - y1 = m(x - x1)
y - 0 = (5/3)(x - (-1))
y = (5/3)(x + 1)
y = (5/3)x + 5/3 (this is what they're asking for)
You can put this solution on YOUR website! The given line is 3x – 5y = 20 ; adding 5y – 20 on both sides we get
3x – 5y + 5y – 20 = 20 + 5y – 20
3x – 20 = 5y. interchanging the sides we get
5y = 3x – 20 Dividing both sides by 5 we get
y = (3x – 20)/5 = (3/5) x – 4 comparing with y = mx + b we have (3/5) as m
means the slope of this line is 3/5 . A line perpendicular to this line must have its slops as the negative reciprocal of this line. So the slope of the new line = - (5/3) . Let the new line be
y = (-5/3) x + b ----- equation 1
Given that the new line passes through ( - 1, 0) . so when its x = -1, y =0 . Let us plug-in these values into equation 1
0 = (-5/3) (-1) + b
That is 0 = (5/3) + b
That is b = ( - 5/3)
Pluging in the value of b in equation 1 we get y = - ( 5/3) x - (5/3)
Which is the required equation
