Solve the first equation, then inverse the slope with a negative. Add in the points with your (y - y')=m(x-x')

usually with a pencil

Given the equation of a line and a point P1 = (x1,y1) If the equation of the line is not in the form: Y = m1(X) + b1 where "m1" may be either a positive or a negative number and "b1" may be either a positive or a negative number, then use algebra to rewrite the equation of the line in that form. The slope of a line that is perpendicular to the original line, m2, is the negative reciprical of the slope of the original line -- that is to say: m2 = -1/m1 You now have part of the equation for the second line that is perpendicular to the first line: Y = m2(X) + b2 but you do not know the value of "b2"; here is how you find the value of "b2": Substitute "y1" given in the Point P1 for "Y" and "x1" given in the Point P1 for "X" into the equation for the line and then solve for "b2" as shown below: y1 = m2(x1) + b2 b2 = y1 - m2(x1) You now know the values of m2 and b2; therefore, you can write the equation of the line that is perpendicular to the given line and passes through the given point: Y = m2(X) + b2 When you set X = 0, you see that Y = b2. Therefore, Point P2 = (0, b2) is a point on the perpendicular line. Put down a straight edge along the given point P1 = (x1,y1) and the point P2 = (0, b2) and draw the line.

In general, if Y = mx+b, then y - b = mx, or (x, y-b) = (t, mt) or (x, y-b) = t(1, m) where (1, m) is a vector pointing parallel to the line (as x increases 1, y -b increases by m) To find a vector perpendicular to the line, it must be perpendicular to all parallel vectors including (1,m). The dot product of any two vectors A*B = mag(A)mag(B)Cos(angleAOB), and if A is perpendicular to B, then Cos(AOB) = 0. so (p, q) * (1, m) = 0, p + mq = 0, p = -mq and this vector is q(-m, 1) where q is an arbitrary scalar. Given point (a,b), the line perpendicular to y = mx + b through (a,b) is the set of all (x, y) such that (x, y) = (a,b) + q(-m, 1), for some q contained in the real numbers.