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What Is the Point-Slope Equation?

A straight line is defined by two points, or by one point and the slope of the line. There is an ingenious formula that allows you to find the formula for a straight line using one point and the slope of the line:

Formula

Point-Slope Equation

The formula that defines the line with slope a through the point (x1,y1) is

y y1 = a(x x1)

Solve the equation with respect to y and the expression looks like the function for a straight line y = ax + b.

Example 1

Find the function for the line through (2, 5) with a slope of 3

Put the numbers into the point-slope equation and solve for y: y 5 = 3(x 2) y = 3x 6 + 5 = 3x 1

Example 2

Find the function for the straight line through the points (3, 9) and (3,9)

First, you find the slope:

a = 9 9 3 (3) = 18 6 = 3

Then, select one of the points in the exercise and put it together with the slope into the point-slope equation: y (9) = 3(x 3) y + 9 = 3x + 9 y = 3x

Since b = 0, you know that the line passes through the origin. The function for the straight line is

y = 3x

If you know the function f(x), you can use the point-slope equation to find the equation of the tangent line at a point on the graph of f(x). This is because the slope of the tangent is equal to the value of the derivative of the function f(x) at the same point.

Formula

The Equation for an Arbitrary Tangent

y y1 = f(x 1)(x x1),

where (x1,y1) is a point on the tangent (often the point of tangency) and f(x1) is the slope of the point. When using the formula, you must always solve for y— that is, get y alone on one side.

Example 3

Given the function f(x) = x2 + 3x 2, find the equation for the tangent at x = 3

To fill in the equation, you need values for y1 and f(x1). You know that x1 = 3, so y1 = f(3) and f(x1) = f(3). We begin by computing f(x): f(x) = 2x + 3 f(x 1) = f(3) = 2(3) + 3 = 9 y1 = f(x1) = f(3) = (3)2 + 3(3) 2 = 16

Now you put this into the equation and get y 16 = 9(x 3) y = 9x 27 + 16 = 9x 11

Example 4

Let g(x) = esin x. Find the equation for the tangent at x = 0.

You need the values of f(x1) and y1. First, differentiate the function:

f(x) = cos x esin x

Now you can calculate f(x1). Since x1 = 0, you get f(x 1) = f(0) = cos 0 esin 0 = 1 e0 = 1

You find y1 by putting x1 into f(x):

y1 = f(x1) = f(0) = esin 0 = e0 = 1

You put all this into the equation and get y 1 = 1(x 0) y = x + 1

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