# maths helpWatch

#1
The vertex A if a square ABCD is at the point (6, -4 ) . The diagnol BD has equation 2y-x=6, and the vertex B is nearer to the origin than D.

a)calculate the coordinates of the centre of the square.

b)calculate the coordinates of B and C.

Calculate the coordinates of the centre of the circle that pass through (0,2) , (4,8) and the origin O.
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4 years ago
#2
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4 years ago
#3
(Original post by maths help)
the vertex a if a square abcd is at the point (6, -4 ) . The diagnol bd has equation 2y-x=6, and the vertex b is nearer to the origin than d.

A)calculate the coordinates of the centre of the square.

B)calculate the coordinates of b and c.

Calculate the coordinates of the centre of the circle that pass through (0,2) , (4,8) and the origin o.
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#4
trust me i've tried and tried but kept getting the answer wrong. pls help i have an exam tomorrow, i wanna make sure i understand!!! pls can yu show me all your working, would be well appreciated!!!
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#5
(Original post by SwEaTy AsIaN)
what book?? what are you talking about?? this is a-level stuff :/
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4 years ago
#6
(Original post by maths help)
what book?? What are you talking about?? This is a-level stuff :/
i did further maths at a level and i am currently studying mathematics at university in oxford
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#7
(Original post by SwEaTy AsIaN)
i did further maths at a level and i am currently studying mathematics at university in oxford
can you just help me pls as i don't know where to start...?
i learn by pracitsing and rote-learning the method, and just apply the method to other questions

pls i've tried but i'm struggling pls can you show how you work it out ...
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#8
fyi the first 2 questions are sub-parts for question 1 and the 3rd is just a seperate question
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4 years ago
#9
I would say that since you have the diagonal BD as y=x/2 + 3
The opposite diagonal would be the y =-2/x + 3
Equate and solve and it might get you the Centre of the square, however i'm not too certain this is correct
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4 years ago
#10
The diagonal BD has equation y=1/2x+3.
Then I know that the angLe between the diagonal of the square and one of the sides is 45.
tan(45+a)=(1+tan a)/(1-tan a) is the gradient of the line AD where a= angle between the diagonal and the x-axis.
So the gradient of AD is 3, the line equation is y= 3x+18 and D(0,-6)
For AB the gradient is just -1/3 so I get y=-1/3x -14/3 and B(2,4)
BC has gradient 3, y= 3X-2, CD y=-1/3x -2 and C(0,-2).
To get the centre you need only to find that line AC is y=-2x-2. Then the centre O is (-2,2).
For the centre of the circle, i remember that for three points which aren't on the same line you can always draw a circle, by first drawing the axis of the two segment formed by joining the points. So if you find the equations of the segment OA which is y=0 and has its middle point at x= 1.
And then the segment OB y= 2x and has the middle point at (2,4).
Find the lines which passes trhough these points and are perpendicular to the two segments,
x=1 and y= -1/2x +5 and then find the intersection which is (1,9/2).
I don't know if it is right, I hope it helps.
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#11
C1 is a non calc exam.
It's to do with gradient of perpendicular lines being -1 and t-shirtsy1=m(X-Menx1)
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