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.
please help.
To find the equation of a line perpendicular to another line, you need to find the negative reciprocal of the slope of the original line.
The equation y = 2/3x + 2 is in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
The slope of the original line is 2/3. To find the negative reciprocal, you take the reciprocal, which means flipping the fraction upside down, and then take the negative sign:
Negative reciprocal = -3/2
So, the slope of the line perpendicular to y = 2/3x + 2 is -3/2.
Now, to find the equation of the line containing the point (-3,4) and having the slope -3/2, you can use the point-slope form of a line:
y - y1 = m(x - x1)
where (x1, y1) represents the coordinates of the given point, and m represents the slope.
Plugging in the values (-3,4) for (x1,y1) and -3/2 for m, we get:
y - 4 = -3/2(x - (-3))
Simplifying further:
y - 4 = -3/2(x + 3)
Expanding the brackets:
y - 4 = -3/2x - 9/2
Now, rearranging the equation to slope-intercept form, we isolate y on one side:
y = -3/2x - 9/2 + 4
Simplifying:
y = -3/2x - 9/2 + 8/2
y = -3/2x - 1/2
Therefore, the equation of the line perpendicular to y = 2/3x + 2 and containing the point (-3,4) is y = -3/2x - 1/2, in slope-intercept form.