|
Question 1196468: Find k so that the line through (2, -3) and (k,2) is:
a. parallel to 3x+4y=8
b. perpendicular to 2x-3y= -4
k=?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
Let's solve the equation for y
3x+4y = 8
4y = -3x+8
y = -3x/4+8/4
y = (-3/4)x+2
This equation is in y = mx+b form
m = -3/4 = slope
b = 2 = y intercept
Parallel lines have equal slopes, but different y intercepts.
The line through (2,-3) and (k,2) will also have a slope of -3/4
Let's set up the slope expression for the line through those two points
m = slope
m = rise/run
m = (y2-y1)/(x2-x1)
m = (2-(-3))/(k-2)
m = (2+3)/(k-2)
m = 5/(k-2)
Set this equal to -3/4 and solve for k
m = 5/(k-2)
-3/4 = 5/(k-2)
-3(k-2) = 4*5
-3k+6 = 20
-3k = 20-6
-3k = 14
k = -14/3
======================================================================================
Part (b)
Solve the given equation for y
2x-3y = -4
-3y = -2x-4
y = -2x/(-3)-4/(-3)
y = (2/3)x + 4/3
The slope of this equation is 2/3
Flip the fraction and flip the sign to get -3/2 as the perpendicular slope.
The original slope 2/3 and perpendicular slope -3/2 multiply to -1.
We'll plug -3/2 into the slope formula we set up in part (a) earlier.
m = 5/(k-2)
-3/2 = 5/(k-2)
-3(k-2) = 2*5
-3k+6 = 10
-3k = 10-6
-3k = 4
k = -4/3
|
|
|
| |
