# 5.6 The Straight Line, SPM Practice, Paper 1 (Short Questions)

Question 1:
Diagram below shows a straight line RS on a Cartesian plane.

Find the gradient of RS.

Solution:
$\begin{array}{l}\text{Using gradient formula}\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ \text{Gradient of}RS=\frac{3-1}{5-\left(-1\right)}=\frac{2}{6}=\frac{1}{3}\end{array}$

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

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:
$\begin{array}{l}RS=10\text{units,}OS=\text{6 units}\\ OR=\sqrt{{10}^{2}-{\left(-6\right)}^{2}}=8\text{units}\\ \\ \text{y-intercept of}RS=-6\\ \text{x-intercept of}RS=-8\\ \text{Using the gradient formula,}m=-\frac{\text{y-intercept}}{\text{x-intercept}}\\ \therefore \text{Gradient of}RS=-\left(\frac{-6}{-8}\right)=-\frac{3}{4}\end{array}$

Question 4:
The gradient of the straight line 3x – 4= 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
$\begin{array}{l}\text{}y=\frac{5}{2}\\ \therefore y\text{-intercept}=\frac{5}{2}.\end{array}$