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13 Feb 2023, 22:53
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Difficulty:
55%
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Question Stats:
64% (02:02) correct
36%
(02:16)
wrong
based on 246
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Point A has coordinates (1, 7) and point B has coordinates (13, 1). Point C is placed on the line segment AB such that C is twice as far from A as it is from B. What are the coordinate of point C ?
A. (5, 5)
B. (6, 6)
C. (7, 4)
D. (8, 2)
E. (9, 3)
A. (5, 5)
B. (6, 6)
C. (7, 4)
D. (8, 2)
E. (9, 3)
25 Feb 2023, 03:44
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(P.s this had quite tiresome and lengthy calculation, is this expected?)
We can solve it as follows :
Distance from C to B assume 'x'. Then distance from C to A = '2x'. Total distance = 3x. We know both the endpoints, using distance formula, we can find x.
x= 2 * [square_root]5
Now, let's assume Pt C (p,q). The distance from C to B = 2 * [square_root]5. The distance from C to A = 4 * [square_root]5.
We can form 2 equations using distance formula for the same.
1) (1-q)^2 + (13-p)^2 = 20,
2) (q-7)^2 + (p-1)^2 = 80.
This calc goes on endlessly till we get q= 2p - 15. Substitute this value into one of the eq'ns till we get p=9. And thus, q=3.
Any less calculation intensive alternative will be appreciated.
We can solve it as follows :
Distance from C to B assume 'x'. Then distance from C to A = '2x'. Total distance = 3x. We know both the endpoints, using distance formula, we can find x.
x= 2 * [square_root]5
Now, let's assume Pt C (p,q). The distance from C to B = 2 * [square_root]5. The distance from C to A = 4 * [square_root]5.
We can form 2 equations using distance formula for the same.
1) (1-q)^2 + (13-p)^2 = 20,
2) (q-7)^2 + (p-1)^2 = 80.
This calc goes on endlessly till we get q= 2p - 15. Substitute this value into one of the eq'ns till we get p=9. And thus, q=3.
Any less calculation intensive alternative will be appreciated.
26 Feb 2023, 08:26
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point C is on line AB such that it divided the line in 2:1.
now, there is a direct formula to solve such maths.
C(x,y)={(m*x2+n*x1)/(m+n), (m*y2+n*y1)/(m+n)}
Here, A(x1,y1)=(1,7) and B(x2,y2)=(13,1) and m=2; n=1
therefore, C(x,y)=(2*13+1*1)/(2+1), (2*1+1*7)/(2+1) = (9,3)
now, there is a direct formula to solve such maths.
C(x,y)={(m*x2+n*x1)/(m+n), (m*y2+n*y1)/(m+n)}
Here, A(x1,y1)=(1,7) and B(x2,y2)=(13,1) and m=2; n=1
therefore, C(x,y)=(2*13+1*1)/(2+1), (2*1+1*7)/(2+1) = (9,3)
