**5.6 The Straight Line, SPM Practice (Short Questions)**

**Question 6:**

Diagram below shows a straight line

*RS*with equation 3*y*= –*px*– 12, where*p*is a constant.It is given that

*OR*:

*OS*= 3 : 2.

Find the value of

*p*.

Solution:Solution:

*Method 1:*Substitute

*x*= –6 and*y*= 0 into 3*y*= –*px*– 12:3(0) = –

*p*(–6) – 120 = 6

*p*– 12–6

*p*= –12

*p***= 2**

Method 2:Method 2:

*OR*:

*OS*= 3 : 2

$\begin{array}{l}\frac{OR}{OS}=\frac{3}{2}\\ \frac{6}{OS}=\frac{3}{2}\\ OS=6\times \frac{2}{3}=4\text{units}\end{array}$

Coordinates of

*S*= (0, –4)

Gradient of the straight line

*RS*= $-\frac{-4}{-6}=-\frac{2}{3}$Given 3

*y*= –

*px*– 12

Rearrange the equation in the form

*y*=

*mx*+

*c*

**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:Solution:

Let point

*N*be = (0, 2).Using Pythagoras’ Theorem,

*LN*= √10

^{2}– 6

^{2}= 8

Point

*L*= (0, 2 + 8) = (0, 10)*y*-intercept of

*LM*= 10

$\begin{array}{l}\text{Using the gradient formula,}m=-\frac{\text{y-intercept}}{\text{x-intercept}}\\ 2=-\left(\frac{10}{\text{x-intercept}}\right)\\ \text{x-intercept of}LM=-\frac{10}{2}=-5\end{array}$