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Date November 2014 Marks available 4 Reference code 14N.2.sl.TZ0.6
Level SL only Paper 2 Time zone TZ0
Command term Show that and Write down Question number 6 Adapted from N/A

Question

The following diagram shows two triangles, OBC and OBA, on a set of axes. Point C lies on the y-axis, and O is the origin.

The equation of the line BC is y=4.

Write down the coordinates of point C.

[1]
a.

The x-coordinate of point B is a.

(i)     Write down the coordinates of point B;

(ii)     Write down the gradient of the line OB.

[2]
b.

Point A lies on the x-axis and the line AB is perpendicular to line OB.

(i)     Write down the gradient of line AB.

(ii)     Show that the equation of the line AB is 4y+axa216=0.

[4]
c.

The area of triangle AOB is three times the area of triangle OBC.

Find an expression, in terms of a, for

(i)     the area of triangle OBC;

(ii)     the x-coordinate of point A.

[3]
d.

Calculate the value of a.

[2]
e.

Markscheme

(0, 4)     (A1)

Notes: Accept x=0, y=4.

a.

(i)     (a, 4)     (A1)(ft)

Notes: Follow through from part (a).

 

(ii)     4a     (A1)(ft)

Note: Follow through from part (b)(i).

b.

(i)     a4     (A1)(ft)

Note: Follow through from part (b)(ii).

 

(ii)     y=a4x+c     (M1)

Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.

 

4=a4×a+c

c=14×a2+4

y=a4x+14a2+4     (A1)

 

OR

y4=a4(xa)     (M1)

Note: Award (M1) for substitution of their gradient from part (c)(i) in the equation.

 

y=ax4+a24+4     (A1)

4y=ax+a2+16

4y+axa216=0     (AG)

Note: Both the simplified and the not simplified equations must be seen for the final (A1) to be awarded.

c.

(i)     2a     (A1)

 

(ii)     4x2=3×2a     (M1)

Note: Award (M1) for correct equation.

 

x=3a     (A1)(ft)

Note: Follow through from part (d)(i).

 

OR

04=a4(xa)     (M1)

Note: Award (M1) for correct substitution of their gradient and the coordinates of their point into the equation of a line.

 

16a=xa

x=a+16a     (A1)(ft)

Note: Follow through from parts (b)(i) and (c)(i).

 

OR

4×0+axa216=0     (M1)

Note: Award (M1) for correct substitution of the coordinates of A(x, 0) into the equation of line AB.

 

axa216=0

x=a+16aORx=(a2+16)a     (A1)(G1)

d.

4(0)+a(3a)a216=0     (M1)

Note: Award (M1) for correct substitution of their 3a from part (d)(ii) into the equation of line AB.

 

OR

12(a+16a)×4=3(4a2)     (M1)

Note: Award (M1) for area of triangle AOB (with their substituted a+16a and 4) equated to three times their area of triangle AOB.

 

a=2.83(2.82842..., 22, 8)     (A1)(ft)(G1)

Note: Follow through from parts (d)(i) and (d)(ii).

e.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.

Syllabus sections

Topic 5 - Geometry and trigonometry » 5.1 » Equation of a line in two dimensions: the forms y=mx+c and ax+by+d=0 .
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