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astra-53 [7]
3 years ago
11

The endpoints of AB are A(2, 3) and B(8, 1). The perpendicular bisector of AB is CD, and point C lies on AB. The length of CD is

(root of 10) units.
The coordinates of point C are (-6, 2) (5, 2) (6, -2) (10, 4) . The slope of is -3 -1/3 1/3 3 . The possible coordinates of point D are (4, 5) (5, 5) (6, 5) (8, 3) and (2, 1) (4, -1) (5, -1) (6, -1) .
Please pick 1 answer for all the questions
Mathematics
1 answer:
Mrac [35]3 years ago
3 0
<span>The coordinates of point C are ((2 + 8)/2, (3 + 1)/2) = (10/2, 4/2) = (5, 2).
The slope of CD is -1/((1 - 3)/(8 - 2)) = -1/(-2/6) = -1/(-1/3) = 3.
Let the cordinates of D be (a, b) then
sqrt((a - 5)^2 + (b - 2)^2) = sqrt(10)
a^2 - 10a + 25 + b^2 - 4b + 4 = 10
a^2 + b^2 - 10a - 4b = -19 . . . (1)
Also, (b - 2)/(a - 5) = 3
b - 2 = 3a - 15
b = 3a - 13 . . . (2)
Putting (2) into (1) gives:
a^2 + (3a - 13)^2 - 10a - 4(3a - 13) = -19
a^2 + 9a^2 - 78a + 169 - 10a - 12a + 52 = -19
10a^2 - 100a + 240 = 0
a^2 - 10a + 24 = 0
(a - 4)(a - 6) = 0
a = 4 or a = 6
When a = 4, b = 3(4) - 13 = 12 - 13 = -1
When a = 6, b = 3(6) - 13 = 18 - 13 = 5

Therefore, the possible coordinates of point D are (6, 5) and (4, -1)</span>
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Hello,

I note (a,b,c) the result of a quarters, b dimes and c pennies:


2 solutions:
106=( 3, 3, 1)=( 1, 8, 1)



106=( 0, 0, 106) but : 100= 0*25+ 0*10+ 100
106=( 0, 1, 96) but : 100= 0*25+ 1*10+ 90
106=( 0, 2, 86) but : 100= 0*25+ 2*10+ 80
106=( 0, 3, 76) but : 100= 0*25+ 3*10+ 70
106=( 0, 4, 66) but : 100= 0*25+ 4*10+ 60
106=( 0, 5, 56) but : 100= 0*25+ 5*10+ 50
106=( 0, 6, 46) but : 100= 0*25+ 6*10+ 40
106=( 0, 7, 36) but : 100= 0*25+ 7*10+ 30
106=( 0, 8, 26) but : 100= 0*25+ 8*10+ 20
106=( 0, 9, 16) but : 100= 0*25+ 9*10+ 10
106=( 0, 10, 6) but : 100= 0*25+ 10*10+ 0
106=( 1, 0, 81) but : 100= 1*25+ 0*10+ 75
106=( 1, 1, 71) but : 100= 1*25+ 1*10+ 65
106=( 1, 2, 61) but : 100= 1*25+ 2*10+ 55
106=( 1, 3, 51) but : 100= 1*25+ 3*10+ 45
106=( 1, 4, 41) but : 100= 1*25+ 4*10+ 35
106=( 1, 5, 31) but : 100= 1*25+ 5*10+ 25
106=( 1, 6, 21) but : 100= 1*25+ 6*10+ 15
106=( 1, 7, 11) but : 100= 1*25+ 7*10+ 5
106=( 1, 8, 1) is good
106=( 2, 0, 56) but : 100= 2*25+ 0*10+ 50
106=( 2, 1, 46) but : 100= 2*25+ 1*10+ 40
106=( 2, 2, 36) but : 100= 2*25+ 2*10+ 30
106=( 2, 3, 26) but : 100= 2*25+ 3*10+ 20
106=( 2, 4, 16) but : 100= 2*25+ 4*10+ 10
106=( 2, 5, 6) but : 100= 2*25+ 5*10+ 0
106=( 3, 0, 31) but : 100= 3*25+ 0*10+ 25
106=( 3, 1, 21) but : 100= 3*25+ 1*10+ 15
106=( 3, 2, 11) but : 100= 3*25+ 2*10+ 5
106=( 3, 3, 1) is good
106=( 4, 0, 6) but : 100= 4*25+ 0*10+ 0






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