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Let us find the equation of the straight line passing through the point \(A(x_1,y_1)\) and having slope \(m\).
Let \(P(x,y)\) be another point other than \(A\). Then, the slope of the line joining the points \(A\) and \(P\) is given by:
\(m = tan \theta = \frac{y - y_1}{x - x_1}\)
\((x - x_1)m = y - y_1\)
Therefore, the equation of the straight line is \((x - x_1)m = y - y_1\) [Point slope form]
Example:
Find the equation of the line passing through the points \((2,-3)\), and the slope is \(\frac{1}{5}\).
Solution:
Given that \((x_1,y_1) = (2,-3)\) and \(m = \frac{1}{5}\)
Substituting the known values in the formula, \((x - x_1)m = y - y_1\), we get:
\((x - 2)\frac{1}{5} = y + 3\)
\(x - 2 = 5(y + 3)\)
\(x - 2 = 5y + 15\)
\(x - 5y - 15 - 2 = 0\)
\(x - 5y - 17 = 0\)
Therefore, the equation of the straight line is \(x - 5y - 17 = 0\)
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