Carcass wrote:
Line q on a coordinate plane is defined by the equation \(-2x + 3y = 6.\)
Quantity A |
Quantity B |
Slope of a line perpendicular to line q |
The slope of a line parallel to line q |
A quick solution is to sketch the given line, and then sketch a perpendicular line.
To sketch the line given by the equation \(-2x + 3y = 6\), we need only find two points that satisfy the equation.
Let's plug in some EASY values.
If x = 0, then we get: \(-2(0) + 3y = 6\)
When we solve this for y, we get: y = 2
So, one point on the line is (0, 2)
Let's now find a second point.
If y = 0, then we get: \(-2x + 3(0) = 6\)
When we solve this for x, we get: x = -3
So, another point on the line is (-3, 0)
We can now plot both points and draw our line:
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Notice that the line has a POSITIVE slope.
This means that any line that's PARALLEL with the line will also have a POSITIVE slope.
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Conversely, any line that's PERPENDICULAR to the line will also have a NEGATIVE slope.
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So, we get:
QUANTITY A: some NEGATIVE slope
QUANTITY B: some POSITIVE slope
Answer: B
Cheers,
Brent