**Question 1**:

In diagram below,

*ABCD*is a trapezium drawn on a Cartesian plane.*BC*is parallel to*AD*and*O*is the origin. The equation of the straight line*BC*is 3*y*=*kx*+ 7 and the equation of the straight line*AD*is $y=\frac{1}{2}x+3$Find

**(a)**the value of

*k*,

**(b)**the

*x*-intercept of the straight line

*BC*.

Solution:Solution:

**(a)**

Equation of

*BC*:3

$\begin{array}{l}y=\frac{k}{3}x+\frac{7}{3}\\ \therefore \text{Gradient of}BC=\frac{k}{3}\\ \text{Equation of}AD:\text{}y=\frac{1}{2}x+3\\ \therefore \text{Gradient of}AD=\frac{1}{2}\end{array}$

*y*=*kx*+ 7$\begin{array}{l}y=\frac{k}{3}x+\frac{7}{3}\\ \therefore \text{Gradient of}BC=\frac{k}{3}\\ \text{Equation of}AD:\text{}y=\frac{1}{2}x+3\\ \therefore \text{Gradient of}AD=\frac{1}{2}\end{array}$

Gradient of

$\begin{array}{l}\frac{k}{3}=\frac{1}{2}\\ \therefore k=\frac{3}{2}\end{array}$
*BC*= gradient of*AD***Equation of**

(b)

(b)

*BC*, $3y=\frac{3}{2}x+7$

For

$\begin{array}{l}3(0)=\frac{3}{2}x+7\\ \frac{3}{2}x=-7\\ x=-\frac{14}{3}\end{array}$
*x-*intercept*, y*= 0Therefore

*x-*intercept of*BC*= $-\frac{14}{3}$**Question 2**:

In diagram below,

*O*is the origin. Straight line*MN*is parallel to a straight line*OK*.**(a)**the equation of the straight line

*MN*,

**(b)**the

*x*-intercept of the straight line

*MN*.

Solution:Solution:

**(a)**Gradient of

*MN*= gradient of

*OK*

*MN*

$\begin{array}{l}=\frac{5-0}{3-0}\\ =\frac{5}{3}\end{array}$

Substitute

5= 5 3 ( −2 )+c *m*= 5/3 and (–2, 5) into*y*=*mx*+*c*15 = – 10 + 3

*c*3

*c*= 25*c*= 25/3

Therefore equation of

*MN*: $y=\frac{5}{3}x+\frac{25}{3}$

**(b)**

For

$\begin{array}{l}0=\frac{5}{3}x+\frac{25}{3}\\ \frac{5}{3}x=-\frac{25}{3}\end{array}$
*x-*intercept*, y*= 05

*x*= –25*x*= –5

Therefore

*x-*intercept of*MN*= –5**Question 3**:

Diagram below shows a straight line

*JK*and a straight line*ST*drawn on a Cartesian plane.*JK*is parallel to*ST.*Find

**(a)**the equation of the straight

*ST*,

**(b)**the

*x*-intercept of the straight line

*ST*.

*Solution:***(a)**

*JK*is parallel to

*ST,*therefore gradient of

*JK*= gradient of

*ST.*

Substitute

*m*= –2 and S(5, 6) into*y*=*mx*+*c*6 = –2 (5) +

*c**c*= 16

Therefore equation of

*ST*:*y*= –2*x*+ 16**For**

(b)

(b)

*x-*intercept

*, y*= 0

0 = –2

*x*+ 162

*x =*16*x*= 8

Therefore

*x-*intercept of*ST*= 8