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secondary 3 | A Maths
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Pls help. Need it urgently
Date Posted: 3 years ago
Views: 204
It is wrong. For question 1a) As gradient of QS should follow y=x + 3, the gradient of QS is also 1 as the lines are parellel. Hence the equation of QS is y-1=1(x-2) > y=x-1. So the c = y intercept is -1.
Thus we can now find equation of the line PR where using the coordinates of P, 0=m(-1) - 1 giving m = -1. Thus equation of PR is y=-x-1.
Alternatively , the fomula of two gradients for perpendicular line is m1 x m2 =-1. Thus if QS gradient is 1, PR gradient is -1 ( perpendicalr lines intersect in rhombas)
Thus we can now find equation of the line PR where using the coordinates of P, 0=m(-1) - 1 giving m = -1. Thus equation of PR is y=-x-1.
Alternatively , the fomula of two gradients for perpendicular line is m1 x m2 =-1. Thus if QS gradient is 1, PR gradient is -1 ( perpendicalr lines intersect in rhombas)
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done
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clear
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Please check. I think area can use some formula but i not sure which.
Date Posted:
3 years ago
It is wrong. For question 1a) As gradient of QS should follow y=x + 3, the gradient of QS is also 1 as the lines are parellel. Hence the equation of QS is y-1=1(x-2) > y=x-1. So the c = y intercept is -1.
Thus we can now find equation of the line PR where using the coordinates of P, 0=m(-1) - 1 giving m = -1. Thus equation of PR is y=-x-1.
Alternatively , the fomula of two gradients for perpendicular line is m1 x m2 =-1. Thus if QS gradient is 1, PR gradient is -1 ( perpendicalr lines intersect in rhombas)
Thus we can now find equation of the line PR where using the coordinates of P, 0=m(-1) - 1 giving m = -1. Thus equation of PR is y=-x-1.
Alternatively , the fomula of two gradients for perpendicular line is m1 x m2 =-1. Thus if QS gradient is 1, PR gradient is -1 ( perpendicalr lines intersect in rhombas)
Ok noted! Thanks for explaining:)
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