The equation of a straight line passing through a point \((x_1,y_1)\) with slope (gradient) \(m\) is:
\[
y - y_1 = m(x - x_1)
\]
This is known as the point–slope form of a line.
Why it’s true
Slope \(m\) is defined as the change in \(y\) over the change in \(x\): \(\frac{y-y_1}{x-x_1} = m\).
Rearranging gives \(y-y_1=m(x-x_1)\).
So this equation captures all points \((x,y)\) that make the slope equal to \(m\) when connected to \((x_1,y_1)\).
Recipe (how to use it)
Identify the slope \(m\) and a point \((x_1,y_1)\) on the line.
Substitute into \(y-y_1=m(x-x_1)\).
Simplify into slope–intercept form \(y=mx+c\) if needed.
Spotting it
This form is useful when you know a point on the line and the slope, but not the intercept.
Common pairings
Slope–intercept form \(y=mx+c\).
General linear form \(ax+by+c=0\).
Parallel and perpendicular line problems.
Mini examples
Given: Point (2,3), slope 4. Equation: \(y-3=4(x-2)\) → \(y=4x-5\).
Given: Point (-1,2), slope -½. Equation: \(y-2=-\tfrac{1}{2}(x+1)\) → \(y=-\tfrac{1}{2}x+1.5\).
Given: Point (0,5), slope 3. Equation: \(y-5=3(x-0)\) → \(y=3x+5\).
Pitfalls
Mixing up \((x,y)\) (variable coordinates) with \((x_1,y_1)\) (fixed point).
For vertical lines slope is undefined, so this form does not apply.
Exam strategy
Always substitute the given point into \((x_1,y_1)\) carefully.
Convert into slope–intercept form if asked for the equation in \(y=mx+c\).
For perpendicular lines, remember slope is negative reciprocal.
Summary
The point–slope form of a line \(y-y_1=m(x-x_1)\) is powerful when you know one point and the slope. It can easily be transformed into other linear equation forms.
Worked examples
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Find the equation of the line through (1,2) with slope 3.
\( Use y-y1=m(x-x1). \)
\( y-2=3(x-1) \)
\( Simplify: y=3x-1 \)
Answer:
\( y=3x-1 \)
Equation of line with slope -2 through (0,5).
\( y-5=-2(x-0) \)
\( y=-2x+5 \)
Answer:
\( y=-2x+5 \)
Find line through (2,4) with slope ½.
\( y-4=½(x-2) \)
\( y=½x+3 \)
Answer:
\( y=½x+3 \)
Equation of line with slope 4 passing through (-1,2).