The line L has equation y=5-2x. Find an equation of the line perpendicular to L, which passes through the point P (3,-1).

The gradient of the line L is -2 so the gradient of the line perpendicular to L will be the negative reciprocal of -2 which is 1/2.The equation of the line perpendicular to L will be y=mx+c where m is the gradient and c is the y-intercept. When we put in the gradient we've found we get y=1/2x+c. We can find c by substituting in our x and y values from the point P which the line passes through. We get -1=3/2+c so c=-5/2 and therefore the line perpendicular to L has equation y=1/2x-5/2.

KN

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