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Gradient of a line

This topic has made it into the five early topics list as it is one of the fundamental tools required for the ideas of calculus to be introduced.

Gradient of a line

The gradient, often denoted by $m$ of a line segment between two points, $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by

$$ m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} $$

Note here we have used the $\Delta$-notation to mean "change in", so the formula read "change in $y$ divided by the change in $x$", this is significant because we tweak this notation to introduce differentiation.

An additional important result is that if two lines, with gradients, $m$ and $n$, are perpendicular, then $$m\times n = - 1$$

Gradient of a line 1

The points $A(1,2)$, $B(3,7)$ and $C(5,4)$ form a triangle.

Determine the gradients of the line segments $AB$, $AC$ and $BC$

solution - press button to display

Drawing a diagram is almost always helpful!

triangle

The line segment AB has a gradient given by $$ m_{AB} = \frac{\Delta y}{\Delta x} = \frac{7-2}{3-1} = \frac{5}{2} $$

The line segment AC has a gradient given by $$ m_{AC} = \frac{\Delta y}{\Delta x} = \frac{4-2}{5-1} = \frac{1}{2} $$

The line segment AB has a gradient given by $$ m_{BC} = \frac{\Delta y}{\Delta x} = \frac{4-7}{5-3} = -\frac{3}{2} $$

gradient of a line 2

The point $A(4,5)$, $B(10,5)$ and $C(8,k)$ are such that angle $ACB$ is a right angle. Determine the two possible values of $k$

solution - press button to display

If $ACB$ is a right angle, then the line segments $AC$ and $BC$ are perpendicular, so $$ m_{AC} \times m_{BC} = -1 $$

The gradients are: $$m_{AC} = \frac{k-5}{8-4} = \frac{1}{4}(k-5),\; m_{BC} = \frac{k-5}{8-10} = -\frac{1}{2}(k-5)$$ and so $m_{AC}\times m_{BC}$ is $$ m_{AC}\times m_{BC} = -\frac{1}{8}(k-5)^2 $$ It follows that $-1 = -\frac{1}{8}(k-5)^2$, this gives $(k-5)^2 = 8$ and so $k = 5\pm 2\sqrt{2}$