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italian18
What is the equation of the line that passes through (12, 4) and is perpendicular to the graph of y = –1/3x – 2?
Look at your slope of that line. The slope is -1/3. In order for your other line to be perpendicular, it needs to be the "negative inverse." That means you have to flip the -1/3 and then multiply it by -1. -1*(-3/1) = 3 So you have your point (12,4) and your slope m=3 Plug these values into y=mx+b to get your b-value (y-intercept) After that, you'll have your m-value and your b-value. Plug those (only those) into y=mx+b and you'll have the eqn of your perpendicular line.
Okay, I'll start from here: So you have your point (12,4) and your slope m=3 Plug these values into y=mx+b to get your b-value (y-intercept) 4 = 3*(12) +b b=4- (3)*12=-32 m=4 b=-32 Plug them into y=mx+b and you have the equation of your line. You're not looking for a point. You're finding an equation.
Do you know how to plug numbers in to formulas?
Read through this whole thing then. You won't be able to do these problems if you don't know how to graph or plug numbers in. http://tutorial.math.lamar.edu/Classes/Alg/Lines.aspx