5.6 SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

Diagram below shows a straight line RS on a Cartesian plane.

Find the gradient of RS.

Solution:

$\begin{array}{l} Using gradient formula \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ Gradient of R S = \frac{3 - 1}{5 - (- 1)} = \frac{2}{6} = \frac{1}{3} \end{array}$

Question 2:

In diagram below, PQ is a straight line with gradient

$- \frac{1}{2}$ .

Find the x-intercept of the straight line PQ.

Solution:

$\begin{array}{l} Using the gradient formula, m = - \frac{y-intercept}{x-intercept} \\ - \frac{1}{2} = - (\frac{- 3}{x-intercept}) \\ x-intercept = 3 \times (- 2) = - 6 \end{array}$

Question 3:

Diagram below shows a straight line RS drawn on a Cartesian plane.

It is given that the distance of RS is 10 units.

Find the gradient of RS.

Solution:

$\begin{array}{l} R S = 10 units, O S = 6 units \\ O R = \sqrt{10^{2} - {(- 6)}^{2}} = 8 units \\ y-intercept of R S = - 6 \\ x-intercept of R S = - 8 \\ Using the gradient formula, m = - \frac{y-intercept}{x-intercept} \\ ∴ Gradient of R S = - (\frac{- 6}{- 8}) = - \frac{3}{4} \end{array}$

Question 4:

The gradient of the straight line 3x – 4y = 24 is

Solution:

Rearrange the equation in the form y = mx+ c

3x – 4y = 24

4y = 3x – 24

$y = \frac{3}{4} x - 6$

Therefore, gradient of the straight line =

$\frac{3}{4} .$

Question 5:

Determine the y-intercept of the straight line 3x + 2y = 5

Solution:

For y-intercept, x = 0

3(0) + 2y = 5

$\begin{array}{l} y = \frac{5}{2} \\ ∴ y -intercept = \frac{5}{2} . \end{array}$