Simple maths help....Whocan help?

Question

If the distance between the points A(k, 0) and B(0, k) is 10. find the possible values of k.
Q2) The coordinates of two points are A(-2, 6) and B(9, 3). Find the coordinates of the points C on the x-axis such that AC=BC

Answer 1

1.distance between points = ((x2-x1)^2 + (y2-y1)^2)^(1/2)

that is square root of sum of squares of the difference betwen x and y co-ordinates respectively.

now, here distance = ((0-k)^2+(k-0)^2)^(1/2)
= (k^2+k^2)^(1/2)
= (2k^2)^(1/2)
which is equal to 10 (given)
=> 10 = (2k^2)^(1/2)
squaring both sides
=> 100 = 2k^2
=> 50 = k^2
=> k = (+or - )(50)^(1/2) i.e. 5*(2)^(1/2) i.e. five multiplied to square root of 2. Ans.



2. It means you have to find the mid point betwen A and B
which is C = (A+B)/2

A=(x1,y1)=(-2,6) here
B=(x2,y2)=(9,3) here
for x co-ordinate :- x=(x1+x2)/2 = (-2+9)/2 = 7/2 = 3.5
for y co-ordinate :- y=(y1+y2)/2 = (6+3)/2 = 9/2 = 4.5

hence C = (3.5,4.5) Ans.

Answer 2

d^2 = (k-0)^2+(0-k)^2 = 2 k^2 = 100
k^2 = 50 and k = +- sqrt50

2 C(x,0) (x+2)^2+36 = (x-9)^2 +9 ===> 4x+40= -18x +90

22x=50 x= 25/11