Write the equation of the line containing (-3,4) and perpendicular to y = 2/3x + 2.
Write your answer in slope-intercept form.
Not sure how to do this one.
m = -3/2
y = mx + b
y - 4 = -3/2 (x-4)
am i on the right track?
would I add 4 to the y side to make the 4 cancel out so is by itself?
where do i put +2?
No.
Y=mx+b
4=-3/2 (-3)+b
solve for b. then put it into this:
y=-3/2 x + b
You are on the right track! To find the equation of a line perpendicular to the given line y = (2/3)x + 2, the first step is to determine the slope of the perpendicular line.
The given line has a slope of 2/3, so the slope of the perpendicular line is the negative reciprocal of 2/3, which is -3/2.
Now, using the point-slope form of the equation of a line (y - y1 = m(x - x1)), where (x1, y1) is a point on the line and m is the slope, we can substitute the given point (-3,4) and the slope -3/2 into the equation.
y - 4 = (-3/2)(x - (-3))
To simplify further and convert the equation into slope-intercept form (y = mx + b), distribute -3/2 to the x - (-3) inside the parentheses:
y - 4 = (-3/2)(x + 3)
y - 4 = -3/2x - 9/2
To isolate y on one side, add 4 to both sides:
y = -3/2x - 9/2 + 4
Combining like terms:
y = -3/2x - 9/2 + 8/2
y = -3/2x - 1/2
Therefore, the equation of the line perpendicular to y = (2/3)x + 2 and passing through the point (-3,4) in slope-intercept form is y = -3/2x - 1/2.