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


5.6 The Straight Line, SPM Paper 1 (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
O R O S = 3 2 6 O S = 3 2 O S = 6 × 2 3 = 4 units  

Coordinates of = (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 x 4 Gradient of the straight line R S = 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 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 L M = 10 2 = 5  

2 Comments to “5.6 The Straight Line, SPM Paper 1 (Short Questions)”

  1. I would like to know Question 6) OR:OS= 3:2, thus if OR is 6, isn’t OS should be 9?
    Using the equation 3OR=2OS, so how can OS be less (4 in your answer) than OR if you need 3 OR to just get 2 OS???!

    1. Dear Gabriel,
      Please refer to method 2,
      When OR:OS = 3:2
      OR/OS = 3/2
      2OR = 3OS (using cross multiplication)
      OS = 2/3 OR
      OS = 2/3 (6 units)
      OS = 4 units

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