Gradient and Angle of Inclination

Gradient and Angle of Inclination

Gradient measures vertical change divided by horizontal change: $m=(y_2-y_1)/(x_2-x_1)$. The angle of inclination is measured anticlockwise from the positive x-axis and satisfies $m=\tan\theta$ for a non-vertical line. A positive gradient rises left to right; a negative gradient falls; a vertical line has undefined gradient.

Worked reasoning

For points $(1,2)$ and $(5,10)$, $m=(10-2)/(5-1)=8/4=2$. Then $\theta=\tan^{-1}(2)\approx63.4^\circ$.

Exam method

1. Keep subtraction order consistent top and bottom. 2. Simplify gradient. 3. Use inverse tangent in degree mode for inclination, checking the line’s sign.

Coordinate order protects the sign

Choose one point as point 1 and keep that order throughout: $y_2-y_1$ over $x_2-x_1$. If you reverse both numerator and denominator, the gradient stays the same; reverse only one and it does not. For an inclination angle, a calculator inverse tangent gives a reference angle. Use the sketch and gradient sign to decide whether the line’s inclination lies between 0 and 90 degrees or between 90 and 180 degrees.

One-minute retrieval: Gradient and Angle of Inclination

Close the worked solution. From memory, state its central rule, name one condition that makes it valid, and reconstruct one check that would catch a typical exam error.