find the coordinates of the point on X axis which is equidistant from the points (-2,5) and (2,-3)
Answers
Answered by
12
Answer:
( -2, 0 )
Step-by-step explanation:
Call the given point A = (-2, 5) and B = (2, -3).
Let X be the point we're looking for. Since it's on the x-axis, X = (x, 0) for some x.
Squares of distances are:
AX² = ( x + 2 )² + ( 0 - 5 )² = x² + 4x + 4 + 25 = x² + 4x + 29
BX² = ( x - 2 )² + ( 0 + 3)² = x² - 4x + 4 + 9 = x² - 4x + 13
Since X is equidistant from A and B:
AX = BX
=> AX² = BX²
=> x² + 4x + 29 = x² - 4x + 13
=> 8x = 13 - 29 = -16
=> x = -2.
So X = ( -2, 0 ).
Answered by
2
Answer:
Step-by-step explanation:
Let the point be (x,0) as it lies on x axis
Now as the point is equidisatant we see that distance will be equal
Or we can say that (x,0) is mid point of the line joining (-2,5) and (2,-3)
So by mid point formula -2+2/2=x
0/2=x
So x=0
So the point is (0,0)
sharmashobha474:
sorry brother your answer is wrong the points are (-2,0)
It doesn't have to be the midpoint. I just lies somewhere on the perpendicular bisector of the line joining (-2,5) and (2,-3).
This question is of RS agarwal the answer which is give in back say the answer of this question is (-2,0)
Yes, and despite the mistake above, it is interesting that samavedamsrinioxk0qw has hit on another to get the answer. The perpendicular bisector goes through the midpoint (0,1) and has slope 1/2. So it has equation y = x/2 + 1. The x-intercept is then where x/2 + 1 = 0, so x = -2. There's the correct answer again!
ok
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