I do AS level Maths and im doing a past exam paper (OCR) to prepare for my exams, im stuck on a question, i have the answer booklet but i dont get how 2√10=40.
For context the question is:
A,B and C have coordinates (5,1), (p,7) and (8,2) respectively.
Given the distance between points A and B is twice the distance between points A and C, calculate the possible values of p.
Thanks.
Can anyone tell me how 2√10 = 40?
2008-12-13 3:02 pm
回答 (13)
2008-12-13 3:08 pm
✔ 最佳答案
2√10 does not equal 40, it is a misprint, it should read:2√10=√40
---------------------
Because:
2√10 =
√4 x √10 =
√[4 x 10] =
√40
2008-12-13 3:06 pm
The equation is wrong. 2â10=â40
2008-12-13 3:06 pm
2â10 = â4 x â10 = â40
2008-12-13 3:15 pm
Distance between A and B is sqrt[(p-5)^2 + (7-1)^2]
dAB = sqrt[(p - 5)^2 + 36]
Distance between A and C is sqrt[(8-5)^2 + (2-1)^2]
dAC = sqrt[9 + 1] = sqrt (10)
dAb = 2dAC
so sqrt[(p - 5)^2 + 36] = 2sqrt (10)
square both sides
(p - 5)^2 + 36 = 40 (2^2 = 4 and sqrt(10)^2 = 10)
(p - 5)^2 = 4
Square root each side
p - 5 = 2 or p - 5 = -2
p = 7 or p = 3
dAB = sqrt[(p - 5)^2 + 36]
Distance between A and C is sqrt[(8-5)^2 + (2-1)^2]
dAC = sqrt[9 + 1] = sqrt (10)
dAb = 2dAC
so sqrt[(p - 5)^2 + 36] = 2sqrt (10)
square both sides
(p - 5)^2 + 36 = 40 (2^2 = 4 and sqrt(10)^2 = 10)
(p - 5)^2 = 4
Square root each side
p - 5 = 2 or p - 5 = -2
p = 7 or p = 3
2008-12-13 3:09 pm
AB = 2AC
=> AB^2 = 4AC^2
=> (p-5)^2 + (7-1)^2 = 4 [ (8-5)^2 + (2-1)^2 ]
=> (p-5)^2 = 4(9+1) - 36 = 4
=> p-5 = 2 or -2
=> p = 7 or 3.
=> AB^2 = 4AC^2
=> (p-5)^2 + (7-1)^2 = 4 [ (8-5)^2 + (2-1)^2 ]
=> (p-5)^2 = 4(9+1) - 36 = 4
=> p-5 = 2 or -2
=> p = 7 or 3.
2008-12-15 10:41 pm
It is very astute of you to give the context of the question. Too many people don't even give the full question, never mind the context.
But I think it is possible to see where the confusion arises.
From the information given, we get that AB² = (5 - p)² + 36
And AC² = 3² + 1² = 10
Now, instead of complicating things by taking square roots to find the actual lengths of AB and AC, it is simpler to work with the squares.
We know that AB = 2AC,
therefore AB² = 4AC²
And therefore (5 - p)² + 36 = 4 x 10 (it would be 2 x â10 if we were dealing with the lengths of the lines. But we are considering the squares, so it is 4 x 10. I suspect you have mixed up the two)
So the equation to solve is (5 - p)² + 36 = 40
But I think it is possible to see where the confusion arises.
From the information given, we get that AB² = (5 - p)² + 36
And AC² = 3² + 1² = 10
Now, instead of complicating things by taking square roots to find the actual lengths of AB and AC, it is simpler to work with the squares.
We know that AB = 2AC,
therefore AB² = 4AC²
And therefore (5 - p)² + 36 = 4 x 10 (it would be 2 x â10 if we were dealing with the lengths of the lines. But we are considering the squares, so it is 4 x 10. I suspect you have mixed up the two)
So the equation to solve is (5 - p)² + 36 = 40
2008-12-13 5:56 pm
-Y----------------------- B2
--|------B1 \ -----------/\
--|----------|-\---------/-|-\
--|----------|---\----- /--|---\
--2----------|-----\--/---|-----| C (8,2)
--1---------T1--- A \/--T2-|
X|----------|------/------|-----|
------------3------5-----7----8
=> AC^2 = ( 8 -5 )^2 + ( 2 - 1 )^2
=> AC^2 = 3^2 + 1^2
=> AC^2 = 10
=> AC = â10
=> AB = 2AC
=> AB = 2â10
See the diagram above
Let T ( T1 ; T2 ) be the value of B from Y axis
=> AB^2 = AT^2 + TB^2
=> (2â10)^2 = AT^2 + ( 7 - 1)^2
=> 40 = AT^2 + 36
=> AT^2 = 40 - 36
=> AT^2 = 4
=> AT = +/- â4
=> AT = +/- 2
=>P = A - AT
=>P = 5 - ( +/- 2 )
=>P = 3 or 7
Answer: P be 3 or 7; therefore B can be be points B1 ( 3,7) or B2 ( 7, 7)
Your question about 2â10 = 40 should actually be (2â10)^2= 40
as seen from above :
=> (2â10)^2 = AT^2 + ( 7 - 1)^2
=> 40 = AT^2 + 36
=> (2â10)^2 = 40
--|------B1 \ -----------/\
--|----------|-\---------/-|-\
--|----------|---\----- /--|---\
--2----------|-----\--/---|-----| C (8,2)
--1---------T1--- A \/--T2-|
X|----------|------/------|-----|
------------3------5-----7----8
=> AC^2 = ( 8 -5 )^2 + ( 2 - 1 )^2
=> AC^2 = 3^2 + 1^2
=> AC^2 = 10
=> AC = â10
=> AB = 2AC
=> AB = 2â10
See the diagram above
Let T ( T1 ; T2 ) be the value of B from Y axis
=> AB^2 = AT^2 + TB^2
=> (2â10)^2 = AT^2 + ( 7 - 1)^2
=> 40 = AT^2 + 36
=> AT^2 = 40 - 36
=> AT^2 = 4
=> AT = +/- â4
=> AT = +/- 2
=>P = A - AT
=>P = 5 - ( +/- 2 )
=>P = 3 or 7
Answer: P be 3 or 7; therefore B can be be points B1 ( 3,7) or B2 ( 7, 7)
Your question about 2â10 = 40 should actually be (2â10)^2= 40
as seen from above :
=> (2â10)^2 = AT^2 + ( 7 - 1)^2
=> 40 = AT^2 + 36
=> (2â10)^2 = 40
2008-12-13 3:23 pm
as far as i can see p has to be either 3 or 7. if you use graph paper you see those points are the only two that work that twice as far ac than ab and still on 7 on the y axis
now if 2 is also equal to root 4
so if you multiply the root 4 and the root 10 you get 40 well root 40 anyway
now if 2 is also equal to root 4
so if you multiply the root 4 and the root 10 you get 40 well root 40 anyway
2008-12-13 3:10 pm
2â10
= â4 x â10
= â(4 x 10)
= â40
= â4 x â10
= â(4 x 10)
= â40
2008-12-13 3:09 pm
well this needs the understanding of surds. when a number is in front of the â it means it has been square rooted, there for you square it. as in your example 2â10, 2^2 is 4 and so 4x10 is 40.
heres another example:
3â20 = 3^2 is 9, 9x20 is 180.
hope that was helpful
heres another example:
3â20 = 3^2 is 9, 9x20 is 180.
hope that was helpful
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