What point on the X-axis is equidistant from (7, 6) and (-3, 4) ? – InterviewSolution

InterviewSolution

1.	What point on the X-axis is equidistant from (7, 6) and (-3, 4) ?
Answer» We know that `y`-co-ordinate of a point on X-axis is always 0. So, let a point on X-axis be `P(x, 0)` and let two given points be `A(7, 6) and B(-3, 4).` According to the condition, `" "PA=PB` `rArr" "sqrt((x-7)^(2)+(0-6)^(2))=sqrt((x+3)^(2)+(0-4)^(2))` Squaring both sides, we have `" "x^(2)-14x+49 +36=x^(2)+6x+9 +16` `rArr" "20x=60" "rArr" "x=3` `therefore` Required point is (3, 0)

Discussion

No Comment Found

Related InterviewSolutions

Student of a school are standing in rows and columns in their playground for a drill practice. A, B, C, D are the positions of four students as shown in the figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from eachl of the four students A, B, C and D? If so, what should be his position ?
Wrtie the slope of X-axis and Y-axis .
Seg AB is parallel to Y-axis and Co-ordinates of point A are `(1,3)` then co-ordinates of point B can be ………….. A)`(3,1)` B)`(5,3)` C)`(3,0)` D)`(1,-3)`A. `(3,1)`B. `(5,3)`C. `(3,0)`D. `(1,-3)`
Out of the following, Point _________ lies to the right of the origin on X- axis.A. `(-2,0)`B. `(0,2)`C. `(2,3)`D. `(2,0)`
The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.
Prove that the points (5, -2), (-4, 3) and (10, 7) are the vertices of an isosceles right-angled triangle.
Find the area of the triangle, whose vertices are (a,c+a), (a,c) and (-a,c-a).
Find the area of triangle, whose vertices are (2,3), (7,5) and (-7,-5).
Find the co-ordinates of a point which divides the line joining the points `A(5, -2) and B(4, 6)` in the ratio 1 : 2 externally.
If (x,y) be any point on the line segment joinijng the points (a,0)and (0,b) then prove that `x/b+y/b=1.`