The end points 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 10 unites. The coordinates of point C are ?. The slope of CD is ?. The possible coordinates of point D are ? and ?.
Answers
Answer:
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The end points 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 √10 unites .
C( x , y ) = ((8 +2) / 2 , ( 1 + 3 )/ 2 )
= (10/2, 4/2) = (5, 2).
The slope of AB = (1 - 3)/(8 - 2) = (-2/6) = (-1/3)
The slope of CD is -1/(The slope of AB) = -1/(-1/3) = 3.
Let the coordinate of D be (a, b) then
√{ (b - 2)^2 +(a - 5)^2 } = √10
a^2 - 10a + 25 + b^2 - 4b + 4 = 10
a^2 + b^2 - 10a - 4b = - 19 . ... ... (1)
Also, slope of CD = (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( a ,b ) are (6, 5) and (4, -1) ( answer )
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