6.8.4 Coordinate Geometry Long Question (Question 7 & 8)


Question 7:
Solutions by scale drawing will not be accepted.
Diagram below shows a triangle OPQ. Point S lies on the line PQ.

(a) A point Y moves such that its distance from point S is always 5 uints.
Find the equation of the locus of Y.  
(b) It is given that point Pand point Q lie on the locus of Y    .
 Calculate
 (i) the value of k,
 (ii) the coordinates of Q.
(c) Hence, find the area, in uint2, of triangle OPQ.

Solution:
(a)
The equation of the locus  Y   ( x , y )  is given by  Y S = 5  units ( x 5 ) 2 + ( y 3 ) 2 = 5 x 2 10 x + 25 + y 2 6 y + 9 = 25 x 2 + y 2 10 x 6 y + 9 = 0

(b)(i)
Given P (2, k) lies on the locus of Y.
(2)2 + (k)2– 10(2) – 6(k) + 9 = 0  
4 + k2– 20 – 6k + 9 = 0
k2 – 6k – 7 = 0
(k – 7) (k + 1) = 0
k = 7   or   k = – 1
Based on the diagram, k = 7. 
 
(b)(ii) 
As P and Q lie on the locus of Y, is the midpoint of PQ. P = (2, 7), S = (5, 3).
Let the coordinates of Q = (x, y),
( 2+x 2 , 7+y 2 )=( 5,3 ) 2+x 2 =5    and     7+y 2 =3 2+x=10  and    7+y=6 x=8 and    y=1
Coordinates of point Q = (8, –1).

(c)
Area of  OPQ = 1 2 | 0  8  2    0  1  7   0 0 | = 1 2 |0+( 8 )( 7 )+00( 1 )( 2 )0| = 1 2 | 58| =29  units 2


Question 8:
Diagram below shows a quadrilateral ABCD. Point C lies on the y-axis.

The equation of a straight line AD is 2y = 5x – 21
(a) Find
(i) the equation of the straight line AB,
(ii) the coordinates of A,
(b) A point P moves such that its distance from point D is always 5 units.
Find the equation of the locus of P.

Solution:
(a)(i)
2y=5x21 y= 5 2 x 21 2 m AD = 5 2 m AB × m AD =1 m AB × 5 2 =1 m AB = 2 5 Equation of AB y y 1 = m AB ( x x 1 ) y+1= 2 5 ( x+2 ) 5y+5=2x4 5y=2x9

(a)(ii)
2y=5x21 ………. ( 1 ) 5y=2x9 ………. ( 2 ) ( 1 )×5:10y=25x105 ………. ( 3 ) ( 2 )×2:10y=4x18 ………. ( 4 ) ( 2 )( 4 ):0=29x87 x=3 From ( 1 ), 2y=1521 2y=6 y=3 A=( 3 , 3 )

(b)
y=2, 4=5x21 5x=25 x=5 Point D=( 5, 2 ) PD=5 ( x5 ) 2 + ( y2 ) 2 =5 ( x5 ) 2 + ( y2 ) 2 =25 x 2 10x+25+( y 2 4y+4 )=25 x 2 + y 2 10x4y+4=0

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