Question

in module mxo2011 you found the gradients and lengths of lines using coordinate geometry. you need to refer to this module if you need a reminder.

to find the gradients of the sides of quadrilateral abcd

click through the slideshow for an example.

gradient of dc

ab is parallel to dc because the gradients are equal.

Explanation:

Identify the coordinates of points D and C

Find coordinates from the graph.
The coordinates are:
\(D(3, -7)\)
\(C(9, -4)\)

State the gradient formula

Recall the slope formula.
The gradient \(m\) is:

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

Substitute the coordinates into the formula

Calculate the gradient of \(DC\).
Using \(D(3, -7)\) as \((x_1, y_1)\) and \(C(9, -4)\) as \((x_2, y_2)\):

$$m = \frac{-4 - (-7)}{9 - 3}$$

Simplify the expression

Compute the final value.
Simplify the numerator and denominator:

$$m = \frac{-4 + 7}{6} = \frac{3}{6} = \frac{1}{2}$$

Answer:

The gradient of \(DC\) is \(\frac{1}{2}\).