5.7.2 The Straight Line, SPM Paper 2


Question 4:


In the diagram above, PQRS is a parallelogram. Find
(a)  the gradient of SR,
(b) the equation of QR,
(c)  the x-intercept of QR.

Solution:
(a)
PQ is parallel to SR, gradient of PQ = gradient of SR.
Gradient of SR= 6 3 =2

(b)
Gradient of QR= 86 50 = 2 5 Substitute m= 2 5  and R (5,8) into y=mx+c 8= 2 5 ( 5 )+c c=6 Therefore equation of QR: y= 2 5 x+6

(c)
For x-intercept, y = 0 0 = 2 5 x + 6 x = 15
Therefore x-intercept of QR = –15.


Question 5:
 
In the diagram above, a straight line 5x +7y + 35 = 0 intersects with the x-axis and y-axis at R and S respectively. Determine
(a) the gradient of the straight line RS.
(b) the x-intercept of the straight line RS.
(c) the distance of RS.
 
Solution:
(a)
5 x + 7 y + 35 = 0 7 y = 5 x 35 y = 5 7 x 5 The gradient of the straight line R S = 5 7 .

(b)
At x-intercept, y = 0 0 = 5 7 x 5 5 7 x = 5 x = 7 x-intercept of the straight line R S = 7.

(c)
Point R = ( 7 , 0 ) and point S = ( 0 , 5 ) Distance of R S = ( 7 0 ) 2 + ( 0 ( 5 ) ) 2 Distance of R S = 49 + 25 Distance of R S = 74 units


Question 6:
 
In the diagram above, O is the origin of the Cartesian plane, AOB is a straight line and OA = AC. Find
(a) the coordinates of C.
(b) the value of h.
(c) the equation of BC.
 
Solution:
(a)
x-coordinate of C = –3 × 2 = –6
Therefore coordinates of C = (–6, 0).

(b)
Gradient of AO= Gradient of OB 0( 4 ) 0( 3 ) = h0 60 4 3 = h 6 h=8

(c)
Gradient of BC= 80 6( 6 ) = 8 12 = 2 3 At point C( 6,0 ), 0= 2 3 ( 6 )+c c=4 The equation of BC is, y= 2 3 x+4

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