Write the equation of a line that is perpendicular to the given line and that passes through the given point. y= 2/3x + 9; (–6, 5)

Hi!

The equation y = 2/3x + 9 is written in slope intercept, or y = mx + b. The m is the slope, and the b is the y intercept.

When a line is perpendicular to another, it meets it at a 90 degree angle. The slope of the line which is perpendicular to one with a given slope is the negative reciprocal of it, for example, if you had a line with slope 1/2, the slope of the line perpendicular to it would be -2.

So for this line, the negative reciprocal of 2/3 is -3/2. You now have one of the slots of the equation filled in, giving you:

y = -3/2x + b

Now, you just need to find b, as slope intercept leaves x and y as variables. Luckily, you are given a point that lays on the line; (-6, 5).

So to solve for b, you can substitute that point in, using -6 as x and 5 as y. That gives you:

5 = -3/2 (-6) + b

Now, just solve for b.

5 = -3/2 (-6) + b
5 = 9 + b
b = -4

And there's your solution for b, which you can fill in, giving you your final answer of:

y = -3/2x - 4

The answer is 10. and if you go by pemdas you would 1. multiplication 2. division 3. addition 4. subtraction