# Two-Point Form

Two points lying on a line will also uniquely determine the line. The following figure shows two points \(A\left( {{x_1},{y_1}} \right)\) and \(B\left( {{x_2},{y_2}} \right)\) lying on a line:

How do we find the equation of this line? Easy – we first find the slope of this line using the two points, and then use the point-slope form.

Observe the following figure, which shows that \(\angle BAC = \theta \) (the inclination angle of the line):

Clearly, the slope of the line is

\[m = \frac{{BC}}{{AC}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]

Thus, using the point-slope form, the equation of the line is

\[\begin{align}&y - {y_1} = m\left( {x - {x_1}} \right)\\&\Rightarrow \,\,\,y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)\\&\Rightarrow \,\,\,\frac{{y - {y_1}}}{{x - {x_1}}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\end{align}\]

This is known as the **two-point form** of a line’s equation, as we have used two points lying on the line to specify it.

**Example 1:** Find the equation of the line passing through \(A\left( {1,2} \right)\) and \(B\left( { - 1,3} \right)\) .

**Solution:** The following figure depicts this line:

Using the two-point form, we have:

\[\begin{align}&\frac{{y - {y_1}}}{{x - {x_1}}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\\&\Rightarrow \,\,\,\frac{{y - 2}}{{x - 1}} = \frac{{3 - 2}}{{ - 1 - 1}}\\&\Rightarrow \,\,\,\frac{{y - 2}}{{x - 1}} = \frac{1}{{ - 2}}\\&\Rightarrow \,\,\, - 2y + 4 = x - 1\\&\Rightarrow \,\,\,x + 2y - 5 = 0\end{align}\]

**Example 2:** Find the *y*-intercept of the line passing through the points \(A\left( {3, - 2} \right)\) and \(B\left( { - 1,3} \right)\) .

**Solution:** The following figure shows the specified line:

Using the two-point form, the equation of the line is:

\[\begin{align}&\frac{{y - {y_1}}}{{x - {x_1}}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\\&\Rightarrow \,\,\,\frac{{y - \left( { - 2} \right)}}{{x - 3}} = \frac{{3 - \left( { - 2} \right)}}{{ - 1 - 3}}\\&\Rightarrow \,\,\,\frac{{y + 2}}{{x - 3}} = \frac{5}{{ - 4}}\\&\Rightarrow \,\,\, - 4y - 8 = 5x - 15\\&\Rightarrow \,\,\,5x + 4y - 7 = 0\end{align}\]

Now, we rearrange this equation into the slope-intercept form:

\[\begin{align}&4y = - 5x + 7\\&\Rightarrow \,\,\,y = \left( { - \frac{5}{4}} \right)x + \frac{7}{4}\end{align}\]

It is evident that the *y*-intercept is

\[c = \frac{7}{4}\]

**Example 3:** Write the equation of the line whose *x*-intercept is *a* and *y*-intercept is *b*, as shown in the following figure:

**Solution:** The line passes through the points \(A\left( {a,0} \right)\) and \(B\left( {0,b} \right)\). Using the two-point form, the equation of the line is

\[\begin{align}&\frac{{y - {y_1}}}{{x - {x_1}}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\,\,\, \Rightarrow \,\,\,\frac{{y - 0}}{{x - a}} = \frac{{b - 0}}{{0 - a}}\\&\Rightarrow \,\,\,\frac{y}{{x - a}} = \frac{b}{{ - a}}\,\,\, \Rightarrow \,\,\, - ay = bx - ab\\&\Rightarrow \,\,\,bx + ay = ab\,\,\, \Rightarrow \,\,\,\frac{x}{a} + \frac{y}{b} = 1\end{align}\]

This form of a line’s equation is known as the **intercept form** – when the intercepts of the lines on the two axes are known, we can use this equation directly. For example, consider a line whose intercepts on the *x* and *y* axes are 3 and \( - 5\) respectively:

The equation of this line, using the intercept form, will be

\[\begin{align}&\frac{x}{3} + \frac{y}{{ - 5}} = 1\\&\Rightarrow \,\,\, - 5x + 3y = - 15\\&\Rightarrow \,\,\,5x - 3y - 15 = 0\end{align}\]