Find the equation of a straight line which is perpendicular to the line which
passes through the point (3, -2).

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

Step-by-step 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.

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°