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Find the equation of a straight line which is perpendicular to the line

which
passes through the point (3, -2).

User Kadeshpa
by
5.1k points

1 Answer

3 votes

Answer:

I don't see the equation for the line, but I describe the process.

Explanation:

Let's look for a line that has the form y=mx+b, where m is the slope and b the y-intercept (the value of y when x=0).

The slope of a perpendicular line is the negative inverse of the slope of the reference line. In this case the equation for the reference line is missing. If it is in the form y=mx+b, take the m value, invert it, and then give it a "-" sign. For example, if the line is y = 2x +1, the negative inverse of the slope of 2 would be -(1/2).

The new line will have this new slope [-(1/m)] to make it y = -(1/m)x + b.

To find b, use the given point of (3,-2) in the equation and solve for b:

y = -(1/m)x + b

-2 = -(1/m)*(3) + b

Then solve for b and you'll have the equation for the solution to the problem.

User Piegames
by
5.2k points
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