5.6 The Straight Line, SPM Practice (Short Questions)


5.6 The Straight Line, SPM Practice (Short Questions)
Question 6:
Diagram below shows a straight line RS with equation 3y = –px – 12, where p is a constant.


It is given that OR : OS = 3 : 2.
Find the value of p.

Solution:
Method 1:
Substitute x = –6 and y = 0 into 3y = –px – 12:
3(0) = –p (–6) – 12
0 = 6p – 12
–6p = –12
p = 2

Method 2:
OR : OS = 3 : 2
OR OS = 3 2 6 OS = 3 2 OS=6× 2 3 =4 units  

Coordinates of S = (0, –4)
Gradient of the straight line RS = 4 6 = 2 3  

Given 3y = –px – 12
Rearrange the equation in the form y = mx + c
y= p 3 x4 Gradient of the straight line RS= P 3 P 3 = 2 3       P=2



Question 7:

The above diagram shows two straight lines, KL and LM, on a Cartesian plane. The distance KL is 10 units and the gradient of LM is 2. Find the x-intercept of LM.

Solution:

Let point N be = (0, 2).
Using Pythagoras’ Theorem,
 LN = √102 – 62 = 8
Point L = (0, 2 + 8) = (0, 10)
y-intercept of LM = 10

Using the gradient formula, m= y-intercept x-intercept 2=( 10 x-intercept ) x-intercept of LM= 10 2 =5