Write an equation of a line that passes through (2, -3) and is perpendicular to the line y=2x-3

Question

I need to know y=mx+b and I also need different ways to solve the problem

Answer 1

The slope for the line y=2x-3 is simply 2 (whatever coefficient is in front of the x). We first need to determine the slope of the perpendicular line, we'll just call it y2 to distinguish it from the first line y.

Since y2 is also a straight line, it MUST satisfy the equation y2 = mx + b, otherwise it would no longer be a line. (x is the variable. m and b are constants. We're here to determine the values of m and b).

To solve for the slope m, notice that the slope of the perpendicular line y2 is simply the negative reciprocal of the original line (True for all perpendicular lines). Therefore, m = -(1/2).

To solve for b, we use the fact that the perpendicular line y2 needs to pass through (2,-3).

-> We already know from above that m= -(1/2) . So, y2 = -(1/2)x + b. Since, the line passes through (2,-3)

-> we can plug x=2, y2=-3 into the equation for y2. Thus, we get -3 = -(1/2)*2 + b.

-> Solving for b, we get b=-2.

So, the final answer is y2 = -(1/2)x + (-2).

*In these situations, you almost always want to solve for m (the slope) before solving for b (also called the y-intercept).*