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secondary 4 | E Maths
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Please help with part (a)! Thanks!!
Date Posted: 5 years ago
Views: 510
A is (-5, 1) and B is (3, -3).
The reference centre point is O (0, 0). Then, vector OA = (-5 1) and vector OB = (3 -3).
By the addition rule for vectors,
AB
= AO + OB
= -OA + OB
= -(-5 1) + (3 -3)
= (5 -1) + (3 -3)
= (8 -4)
For line AB, gradient
= (-3 - 1) / [3 - (-5)]
= - 4 / 8
= - 1 / 2
(this can also be seen in the direction vector itself, given by the y value of the direction vector divided by the x value of the same direction vector).
Since the line passes B (3, -3),
y = -0.5x + c
-3 = -0.5 (3) + c
Solving for c, c = -1.5
y = -0.5x - 1.5, or 2y = -x - 3.
The reference centre point is O (0, 0). Then, vector OA = (-5 1) and vector OB = (3 -3).
By the addition rule for vectors,
AB
= AO + OB
= -OA + OB
= -(-5 1) + (3 -3)
= (5 -1) + (3 -3)
= (8 -4)
For line AB, gradient
= (-3 - 1) / [3 - (-5)]
= - 4 / 8
= - 1 / 2
(this can also be seen in the direction vector itself, given by the y value of the direction vector divided by the x value of the same direction vector).
Since the line passes B (3, -3),
y = -0.5x + c
-3 = -0.5 (3) + c
Solving for c, c = -1.5
y = -0.5x - 1.5, or 2y = -x - 3.
See 3 Answers
So you have the points
OA=(-5,1)
&
OB=(3,-3)
(a)(i)
To find AB
AB= AO+OB
=-OA+OB
=(5,-1)+(3,-3)
=(8,-4)
(ii)
To find equation of the line AB,
Eqn of line of AB = (-5,1) + λ(8,-4)
Where λ ∈R
OA=(-5,1)
&
OB=(3,-3)
(a)(i)
To find AB
AB= AO+OB
=-OA+OB
=(5,-1)+(3,-3)
=(8,-4)
(ii)
To find equation of the line AB,
Eqn of line of AB = (-5,1) + λ(8,-4)
Where λ ∈R
Thank you!!
done
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Hope this helps. Impt parts in pink !
Date Posted:
5 years ago
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Hope this helps. Impt parts in pink !
Date Posted:
5 years ago
Thanks!!
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