Here you will learn what is the point slope form of a line equation with proof and examples.
Letโs begin โย
What is the Point Slope Form of a Line ?
The equation of a line which passes through the point \(P(x_1, y_1)\) and has the slope โmโ is
\(y โ y_1\) = m\((x โ x_1)\)
Proof :
Let \(Q(x_1, y_1)\) be the point through which the line passes and letย P(x, y) be any point on the line.
Then the slope of the line is \(y โ y_1\over x โ x_1\)
but, m is the slope of lineย
\(\therefore\) m = \(y โ y_1\over x โ x_1\) \(\implies\) \(y โ y_1\) = m(\(x โ x_1\))
Hence, \(y โ y_1\) = m\(x โ x_1\) is the required equation of the line.
Example : Find the equation of a line passing through (2, -3) and inclined at an angle of 135 with the positive direction of x-axis.
Solution : Here, m = slope of the line = tan 135 = tan(90 + 45) = -cot 45 = -1
\(x_1\) = 2, \(y_1\) = -3
So, the equation of the line is
\(y โ y_1\) = m(\(x โ x_1\))
i.e. y โ (-3) = -1(x โ 2)
\(\implies\) y + 3 = -x + 2 \(\implies\) x + y + 1 = 0.
which is the required equation of line.
Example : Determine the equation of line through the point (-4, -3) and parallel to x-axis
Solution : Here, m = slope of the line = 0,
\(x_1\) = -4, \(y_1\) = -3
So, the equation of the line is
\(y โ y_1\) = m(\(x โ x_1\))
i.e. y + 3 = 0(x + 4)
\(\implies\) y + 3 = 0.
which is the required equation of line.
