WMA11 October 2021

5. WMA11/01 (Edexcel) IAL P1 Oct 2021 Q5 Straight Line Graphs

The line l1​​ has equation 3y – 2x = 30​​ 

The line l2​​ passes through the point A(24, 0) and is perpendicular to l1​​ 

Lines l1​​ and l2​​ meet at the point P​​ 

(a) Find, using algebra and showing your working, the coordinates of P.​​ 

(5)

Given that l1​​ meets the x‑axis at the point B,​​ 

(b) find the area of triangle BPA.​​ 

(3)

SOLUTION​​ 

a-​​ 

Consider the line l1, to its gradient.​​ 

3y – 2x = 30 

3y= 2x+30

y=23x+303

y=23x+10

So​​ the gradient of line l1​​ is​​ 

m1=23

Consider the line l2, which is perpendicular to line l1​​ and it passes through the point​​ A(24, 0).​​ 

m1×m2=-1

23×m2=-1

m2=-32

Using the point-slope formula.​​ 

y-y1=mx-x1

y-0= -32x-24

2y=-3x+72

y=-32x+722

Point P is the point of intersection of both the lines. Hence, solving them​​ simultaneously.

On sibtracting both equations, we get

23x+10=-32x+722

23x+32x=-10+36

136x=26

x=12

Now, substituting the value of x in either equations of line.​​ 

y=23x+10

y=23(12)+10

y=18

Hence, the point P is (12, 18)

b-​​ 

The point at which line l1​​ cuts x-axis, y=0.​​ 

3y-2x=30

-2x=30

x=-15

So the coordinate of B is (-15, 0)

So the area​​ of a triangle is​​ 

Area =12x b x h

Where​​ 

Base=15+24=39

Height= y-coordinate of P=18

Thus,​​ 

Area =12x 39 x 18

Area =39 x 9

Area =351 sq unit