WMA11 Jan 2021

4. WMA11/01​​ Edexcel​​ IAL P1 January 2021 IAL Q4 (Straight Line Graphs & Quadratics)

The points P and Q, as shown in Figure 2, have coordinates (–2, 13) and (4, –5) respectively.​​ 

The straight line l passes through P and Q.​​ 

• Find an equation for l, writing your answer in the form y = mx + c, where m and c are integers to be found.​​ 

(3)

The quadratic curve C passes through P and has a minimum point at Q.​​ 

• Find an equation for C.​​ 

(3)

The region R, shown shaded in Figure 2, lies in the second quadrant and is bounded by C and l only.​​ 

• Use inequalities to define region R.​​ 

(2)

SOLUTION

a-​​ 

The points P and Q are given as​​ P-2, 13​​ &​​ Q4, -5. So, to find the equation of a line, we need to find the gradient of the line.

ml=13--5-2-4

ml=18-6

ml= -3

Now, using point-slope formula to find the equation of a line PQ.​​ 

y-y1=mlx-x1

y--5=-3x-4

y+5= -3x+12

y=-3x+7

b-​​ 

Since this is a quadratic curve, the equation of a curve can be represented in the form of a completed square/vertex form, y=a(x-p)2+q. Since point Q4, -5​​ is a vertex of​​ the quadratic curve, the equation of a curve becomes.​​ 

y=a x-42-5

Now, subtituting the value of P as​​ P-2, 13.

13=a -2-42-5

13=a36-5

18=36a

a=12

Therefore, the equation​​ y=a x-42-5​​ becomes​​ 

y=12x-42-5

y=12x2-8x+16-5

y=12x2-4x+8-5

y=12x2-4x+3

c-​​ 

Since the region is enclosed between the solid lines, the inequalities must be either​​ ≤​​ or​​ ≥.​​ 

Keeping in​​ view the above flash card.​​ 

Region is​​ 

• below/outside the quadratic curve; therefore,

y≤12x2-4x+3

• above the straight line PQ; therefore,​​ 

y≥-3x+7

• left to the line​​ x=2, so​​ 

x≤-2