What is the equation of the line that passes through the point (-2,3) and that is perpendicular to the line represented by 3x-2y=-2?
A. 2x+3y=5
B. 2x+3y=7
C. -2x=3y=5
D. 3x-2y=-12
I think that the answer is D, but I am not too sure about it.
Little tricks that can come in handy ...
For a linear equation ax^2 + bx + c = 0
the slope is -a/b
so the slope of the given line is -3/-2 = 3/2
thus the slope of the new line must be
-2/3
But your choice of D has a slope of 3/2, so it can't be D
so write the given choices into the general form and decide which one has a slope of -2/3
Ok, thanks so much for the help! =)
btw, I have a typo, the equation should have said:
ax + by + c = 0
To find the equation of a line that is perpendicular to the given line, we first need to find the slope of the given line.
The given line is in the form of `Ax + By = C`, where A, B, and C are constants. To find the slope of this line, we need to rearrange the equation in the form `y = mx + b`, where m is the slope.
For the given line, 3x - 2y = -2, we can rearrange it to solve for y:
-2y = -3x - 2
Dividing both sides by -2:
y = (3/2)x + 1
Now that we have the slope of the given line, which is 3/2, we can find the slope of the perpendicular line. The slope of a line that is perpendicular to another line is the negative reciprocal of the slope of that line. So, the slope of the perpendicular line is -2/3.
We also know that the line passes through the point (-2, 3). Using the point-slope form of a line, which is y - y1 = m(x - x1), we can substitute the values and solve for y:
y - 3 = (-2/3)(x - (-2))
y - 3 = (-2/3)(x + 2)
y - 3 = (-2/3)x - 4/3
Now, rearranging the equation to the standard form Ax + By = C, we get:
(2/3)x + y = 4/3 + 3
(2/3)x + y = 4/3 + 9/3
(2/3)x + y = 13/3
Multiplying every term by 3 to eliminate fractions, we get:
2x + 3y = 13
Therefore, the equation of the line that passes through (-2, 3) and is perpendicular to the line 3x - 2y = -2 is B. 2x + 3y = 7.