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secondary 3 | A Maths
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Hello can someone help me with question 14 part (iv) and (v)? I solved the rest and I got y=1/2x+15 for (ii) and Q(0,15) for (iii). Thanks!
Date Posted: 4 years ago
Views: 189
(ii) from the graph, can see that PQ has a negative gradient.
correct answer should be y = -(1/2)x+15
(iii) your answer for Q is correct.
(iv)
for 2 perpendicular lines, product of their gradients is -1, so perpendicular bisector of PQ has gradient 2.
midpoint of PQ is
((0+6)/2,(12+15)/2) = (3,13.5)
equation of perpendicular bisector of PQ is
(y-13.5)/(x-3) = 2, giving y = 2x+7.5 …. (1)
equation of OR is y= -(11/2)x …. (2)
solve (1) & (2) together gives
coordinates of R as (-1, (11/2))
(v)
use "shoelace" method to calculate area,
OR
area OPQR
= area OPQ + area OQR
= (1/2)(15)(6) + (1/2)(15)(1)
= 52.5 sq. units
correct answer should be y = -(1/2)x+15
(iii) your answer for Q is correct.
(iv)
for 2 perpendicular lines, product of their gradients is -1, so perpendicular bisector of PQ has gradient 2.
midpoint of PQ is
((0+6)/2,(12+15)/2) = (3,13.5)
equation of perpendicular bisector of PQ is
(y-13.5)/(x-3) = 2, giving y = 2x+7.5 …. (1)
equation of OR is y= -(11/2)x …. (2)
solve (1) & (2) together gives
coordinates of R as (-1, (11/2))
(v)
use "shoelace" method to calculate area,
OR
area OPQR
= area OPQ + area OQR
= (1/2)(15)(6) + (1/2)(15)(1)
= 52.5 sq. units
Okay thank uu
See 1 Answer
done
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Saw your answer to the previous part. Gradient of the line should be -1/2 and not 1/2, think it’s a typo but just in case!
Date Posted:
4 years ago
Yep its a typo
Thank you though!
you’re welcome!
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