5.6 SPM Practice (Short Questions)

Posted on April 23, 2020 by user

5.6.2 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

\begin{array}{l} \frac{O R}{O S} = \frac{3}{2} \\ \frac{6}{O S} = \frac{3}{2} \\ O S = 6 \times \frac{2}{3} = 4 units \end{array}

Coordinates of S = (0, –4)

Gradient of the straight line RS =

- \frac{- 4}{- 6} = - \frac{2}{3}

Given 3y = –px – 12
Rearrange the equation in the form y = mx + c

\begin{array}{l} y = - \frac{p}{3} x - 4 \\ Gradient of the straight line R S = - \frac{P}{3} \\ ∴ - \frac{P}{3} = - \frac{2}{3} \\ P = 2 \end{array}

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

\begin{array}{l} Using the gradient formula, m = - \frac{y-intercept}{x-intercept} \\ 2 = - (\frac{10}{x-intercept}) \\ x-intercept of L M = - \frac{10}{2} = - 5 \end{array}