Point-Slope Form
Another common form of an equation of a line is referred to as the point-slope form, and is written as:
\[y-y_1=m(x-x_1)\]
This equation can be developed from the slope formula.
Recall
Our slope formula is given by
\[\begin{align*} m&=\dfrac{\Delta y}{\Delta x}\\ m&=\dfrac{y_2-y_1}{x_2-x_1} \end{align*}\]
where we use two points \((x_1,y_1)\) and \((x_2,y_2)\) to complete the slope calculation.
We can rearrange this formula into the point-slope form of a linear equation.
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We will keep the point \((x_1,y_1)\) as \((x_1,y_1)\), but we will rename \((x_2,y_2)\) as just \((x,y)\).
- \((x_1,y_1)\) continues to represent a specific known point on the line. \((x,y)\) now represents a general unknown point on the line, and provides the variables \(x\) and \(y\) in our linear equation.
- Multiply both sides by \((x-x_1)\) to get the point-slope form.
\[\begin{align*} m&=\dfrac{\Delta y}{\Delta x}\\ m&=\dfrac{y-y_1}{x-x_1} \\ m(x-x_1) &=y-y_1 \\ y-y_1&=m(x-x_1) \end{align*}\]
Example 4
A line has a slope of \(-2\) and passes through the point \((8,11)\).
- Determine the equation of the line in point-slope form.
- Rearrange the point-slope form into \(y=mx+b\) (slope, \(y\)-intercept) form.
Solution — Part A
We can substitute \(m=-2\), \(x_1=8\), and \(y_1=11\) into the general point-slope form \(y-y_1=m(x-x_1)\):
\[\begin{align*} y-y_1=m(x-x_1)\\ y-11=-2(x-8) \end{align*}\]
Therefore, the equation of this line in point-slope form is \(y-11=-2(x-8)\).
Solution — Part B
We can expand the right side of the equation using the distributive property, and then isolate for \(y\) using inverse operations.
\[\begin{align*} y-11&=-2(x-8)\\ y-11&=-2x+16\\ y&=-2x+16+11\\ y&=-2x+27 \end{align*}\]
Example 5
A line passes through the points \((-6,9)\) and \((10,21)\).
Determine the equation of the line in point-slope form.
Solution
First, we will calculate the slope using the two given points and the slope formula.
\(\begin{align*} m&=\dfrac{\Delta y}{\Delta x}\\ &=\dfrac{y_2-y_1}{x_2-x_1}\\ &=\dfrac{21-9}{10-(-6)}\\ &=\dfrac{12}{16}\\ &=\dfrac{3}{4} \end{align*}\)
We can now substitute the slope and one of the given points into the point-slope form. It does not matter which point we use. We will show both options give the same line.
Method 1: Use the point \((-6,9)\)
\(m=\dfrac{3}{4}\), \(x_1=-6\) and \(y_1=9\)
\[\begin{align*} y-y_1=m(x-x_1)\\ y-9=\dfrac{3}{4}(x-(-6)) \\ y-9=\dfrac{3}{4}(x+6) \tag{1} \end{align*}\]
Method 2: Use the point \((10,21)\)
\(m=\dfrac{3}{4}\), \(x_1=10\) and \(y_1=21\)
\[\begin{align*} y-y_1=m(x-x_1)\\ y-21=\dfrac{3}{4}(x-10) \tag{2} \end{align*}\]
At first glance the slope-point equations of the line appear different. If we rearrange them both into \(y=mx+b\) form, we will see that they represent the same line.
Rearranging Equation \((1)\), we get:
\[\begin{align*} y-9&=\dfrac{3}{4}(x+6)\\ y-9&=\dfrac{3}{4}x+\dfrac{18}{4}\\ y&=\dfrac{3}{4}x+\dfrac{18}{4} +9\\ y&=\dfrac{3}{4}x+\dfrac{9}{2} +\dfrac{18}{2}\\ y&=\dfrac{3}{4}x+\dfrac{27}{2} \end{align*}\]
The line \(y-9=\dfrac{3}{4}(x+6)\) in point-slope form is equivalent to the line \(y=\dfrac{3}{4}x+\dfrac{27}{2}\) in slope y-intercept form.
Rearranging Equation \((2)\), we get:
\[\begin{align*} y-21&=\dfrac{3}{4}(x-10)\\ y-21&=\dfrac{3}{4}x-\dfrac{30}{4}\\ y&=\dfrac{3}{4}x-\dfrac{30}{4} +21\\ y&=\dfrac{3}{4}x-\dfrac{15}{2} +\dfrac{42}{2}\\ y&=\dfrac{3}{4}x+\dfrac{27}{2}\\ \end{align*}\]
The line \(y-21=\dfrac{3}{4}(x-10)\) in point-slope form is equivalent to the line \(y=\dfrac{3}{4}x+\dfrac{27}{2}\) in slope y-intercept form.