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प्रश्न
Find a point on the x-axis which is equidistant from the points (7, 6) and (−3, 4).
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उत्तर
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`
Here we are to find out a point on the x−axis which is equidistant from both the points A(7,6) and B(−3,4).
Let this point be denoted as C(x, y).
Since the point lies on the x-axis the value of its ordinate will be 0. Or in other words, we have y = 0.
Now let us find out the distances from ‘A’ and ‘B’ to ‘C’
`AC = sqrt((7 - x)^2 + (6 - y)^2)`
`= sqrt((7 - x)^2 + (6 - 0)^2)`
`AC = sqrt((7-x)^2 + (6)^2)`
`BC= sqrt((-3-x)^2 + (4- y)^2)`
`= sqrt((-3-x)^2 + (4 - 0)^2)`
`BC = sqrt((-3-x)^2 + (4)^2)`
We know that both these distances are the same. So equating both these we get,
AC = BC
`sqrt((7 - x)^2 + (6)^2) = sqrt((-3-x)^2 + (4)^2)`
Squaring on both sides we have,
`(7 -x)^2 + (6)^2 = (-3-x)^2+ (4)^2`
`49 + x^2 -14x + 36 = 9 + x^2 + 6x + 16`
20x = 60
x = 3
Hence the point on the x-axis which lies at equal distances from the mentioned points is (3,0)
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संबंधित प्रश्न
Find the distance between the following pair of points:
(a, 0) and (0, b)
Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6,2) are the vertices of a square.
Prove that the points (4, 5) (7, 6), (6, 3) (3, 2) are the vertices of a parallelogram. Is it a rectangle.
Determine the ratio in which the point (-6, a) divides the join of A (-3, 1) and B (-8, 9). Also, find the value of a.
If p(x , y) is point equidistant from the points A(6, -1) and B(2,3) A , show that x – y = 3
Find the co-ordinates of the point equidistant from three given points A(5,3), B(5, -5) and C(1,- 5).
Points P, Q, and R in that order are dividing line segment joining A (1,6) and B(5, -2) in four equal parts. Find the coordinates of P, Q and R.
The line segment joining A( 2,9) and B(6,3) is a diameter of a circle with center C. Find the coordinates of C
Find the centroid of ΔABC whose vertices are A(2,2) , B (-4,-4) and C (5,-8).
Find the ratio in which the line segment joining the points A (3, 8) and B (–9, 3) is divided by the Y– axis.
In \[∆\] ABC , the coordinates of vertex A are (0, - 1) and D (1,0) and E(0,10) respectively the mid-points of the sides AB and AC . If F is the mid-points of the side BC , find the area of \[∆\] DEF.
What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4)?
If the distance between the points (3, 0) and (0, y) is 5 units and y is positive. then what is the value of y?
What is the distance between the points \[A\left( \sin\theta - \cos\theta, 0 \right)\] and \[B\left( 0, \sin\theta + \cos\theta \right)\] ?
The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (−3, 4) are
The point R divides the line segment AB, where A(−4, 0) and B(0, 6) such that AR=34AB.">AR = `3/4`AB. Find the coordinates of R.
If the points P(1, 2), Q(0, 0) and R(x, y) are collinear, then find the relation between x and y.
Given points are P(1, 2), Q(0, 0) and R(x, y).
The given points are collinear, so the area of the triangle formed by them is `square`.
∴ `1/2 |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = square`
`1/2 |1(square) + 0(square) + x(square)| = square`
`square + square + square` = 0
`square + square` = 0
`square = square`
Hence, the relation between x and y is `square`.
Statement A (Assertion): If the coordinates of the mid-points of the sides AB and AC of ∆ABC are D(3, 5) and E(–3, –3) respectively, then BC = 20 units.
Statement R (Reason): The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
The distance of the point (–6, 8) from x-axis is ______.
The distance of the point (–1, 7) from x-axis is ______.
