The coordinates of two points are A(-2,6) & B(9,3). Find the coordinates of the point C on the X-axis such that AC=BC.
3 Answers
kjcdb8er | Teacher | (Level 1) Associate Educator
Let the point C be at (x,y).
AC--> D^2 = (-2 - x)^2 + (6 - y)^2 = x^2 + 4x + 4 + y^2 - 12y + 36
BC--> D^2 = (x - 9)^2 + (y - 3)^2 = x^2 - 18x + 81 + y^2 - 6y + 9
AC = BC
x^2 + 4x + 4 + y^2 - 12y + 36 = x^2 - 18x + 81 + y^2 - 6y + 9
11x - 25 - 3y = 0
The point C lies on this line.
On the x-axis, y= 0. Therefore:
11x - 25 = 0
x = 25/11
hala718 | High School Teacher | (Level 1) Educator Emeritus
A(-2,6) B (9,3)
We need to find C on x-axis ==> C (x,0)
such that :
AC = BC
AC= sqrt(0-6)^2 + (x+2)^2]= sqrt[(36 + (x+2)^2]
BC= sqrt[(0-3)^2+ (x-9)^2]= sqrt[(9 + (x-9)^2]
==> AC = BC
==> sqrt(36+(x+2)^2]= sqrt[(9+ (x-9)^2]
square both sides:
==> 36 + (x+2)^2 = 9 + (x-9)^2
==> 36 + x^2 +4x + 4 = 9 + x^2 -18x + 81
Now group similars:
==> 22x -50 = 0
==> x= 50/22= 25/11
Then the point C is (25/11, 0)
neela | High School Teacher | (Level 3) Valedictorian
Any point C on X axis has the y coordinate zero. So we take P( h,0) as a point on X axis.
Since CA = CB, by distance formula,
CA^2 = ((h- -2)^2 +(0-6) ^2 = (h+2)^2+36 = h^2+4h+40....(1)
CB^2 = (h-9)^2+(0-3)^2 = = h^2-18h+ 90.....(2).
Right side of (1) and (2) are equal as CA = CB.
h^2+2h+40 = h^2-18h+90
2h+18h = 90 - 40 = 50
20 = 50 Or
h = 50/20 = 2.5
So C has the coordinates (2.5,0).