WMA11 January 2020

6. WMA11/01​​ Edexcel IAL P1 January 2020,​​ Q6 (Straight Line Graphs)

The line l1​​ has equation 3x – 4y + 20 = 0

The line l2​​ cuts the x-axis at R(8,0) and is parallel to l1

(a) Find the​​ equation of l2, writing your answer in the form ax + by + c = 0, where a, b and c are integers to be found.

(3)

The line l1​​ ​​ cuts the x-axis at P and the y-axis at Q.

Given that PQRS is a parallelogram, find

(b) the area of PQRS,

(3)

(c) the coordinates of​​ S.

(2)

SOLUTION​​ 

a-​​ 

Since line l2​​ is parallel to l1, the gradient of both lines must be same. So, to find the equation of line l2, apply the point-slope formula.​​ 

First, let us find the gradient of l1.

3x – 4y + 20 = 0

-4y=-3x-20

y=-3-4x-20-4

y=34x+204

On comparing with y=mx+c, the gradient of line l1​​ & l2​​ is​​ 

m1=m2=34

Now, using point slope formula, to find the equationof line l2​​ where point is (8, 0) & gradient ¾.​​ 

y-y1=m2(x-x1)

y-0=34(x-8)

4y=3x-24

0=3x-4y-24

3x-4y-24=0

b-​​ 

3x – 4y + 20 = 0

When y=0,​​ 

3x – 4(0)+20 = 0

3x+20=0

3x=-20

x=-203

x=-623

When x=0,​​ 

3x – 4y + 20=y

3(0) – 4y + 20=y

-4y+20=0

20=4y

y=5

Hence,​​ Q0,5

Area of ∆PQR=128--623 x 5

Area of ∆PQR=1103

∆QR≡∆RSP

Area PQRS=2 x1103

A=2203 sq unit

c-​​ 

Q→R=P→S

Q→R=8-5

P→S=8-5

-623+80-5=113-5

S113,-5