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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 $(x_{1}, y_{1})$ is

y - y_{1} = a (x - x_{1})

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

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 $y$ : $\begin{array}{l} y - 5 & = 3 (x - 2) \\ y & = 3 x - 6 + 5 = 3 x - 1 \end{array}$

Example 2

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

First, you find the slope:

a = \frac{- 9 - 9}{3 - (- 3)} = - \frac{18}{6} = - 3

Then, select one of the points in the exercise and put it together with the slope into the point-slope equation:

\begin{array}{l} y - (- 9) & = - 3 (x - 3) \\ y + 9 & = - 3 x + 9 \\ y & = - 3 x \end{array}

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

y = - 3 x

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 - y_{1} = f^{'} (x_{1}) (x - x_{1}),

where $(x_{1}, y_{1})$ is a point on the tangent (often the point of tangency) and $f^{'} (x_{1})$ 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 $f (x) = x^{2} + 3 x - 2$ , find the equation for the tangent at $x = 3$

To fill in the equation, you need values for $y_{1}$ and $f^{'} (x_{1})$ . You know that $x_{1} = 3$ , so $y_{1} = f (3)$ and $f^{'} (x_{1}) = f^{'} (3)$ . We begin by computing $f^{'} (x)$ : $\begin{array}{l} f^{'} (x) & = 2 x + 3 \\ f^{'} (x_{1}) & = f^{'} (3) = 2 (3) + 3 = 9 \\ y_{1} & = f (x_{1}) = f (3) \\ = {(3)}^{2} + 3 (3) - 2 = 16 \end{array}$

Now you put this into the equation and get

\begin{array}{l} y - 16 & = 9 (x - 3) \\ y & = 9 x - 27 + 16 = 9 x - 11 \end{array}

Example 4

Let $g (x) = e^{\sin x}$ . Find the equation for the tangent at $x = 0$ .

You need the values of $f^{'} (x_{1})$ and $y_{1}$ . First, differentiate the function:

f^{'} (x) = \cos x \cdot e^{\sin x}

Now you can calculate $f^{'} (x_{1})$ . Since $x_{1} = 0$ , you get

\begin{array}{l} f^{'} (x_{1}) & = f^{'} (0) \\ = \cos 0 \cdot e^{\sin 0} = 1 \cdot e^{0} = 1 \end{array}

You find $y_{1}$ by putting $x_{1}$ into $f (x)$ :

y_{1} = f (x_{1}) = f (0) = e^{\sin 0} = e^{0} = 1

You put all this into the equation and get

\begin{array}{l} y - 1 & = 1 (x - 0) \\ y & = x + 1 \end{array}

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