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प्रश्न
Find the points on the y-axis which is equidistant form the points A(6,5) and B(- 4,3)
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उत्तर
Let P (0, y) be a point on the y-axis. Then as per the question, we have
AP=BP
`⇒ sqrt((0-6)^2 +(y-5)^2) = sqrt((0+4)^2 +(y-3)^2)`
`⇒sqrt((6)^2 +(y-5)^2) = sqrt((4)^2 +(y-3)^2)`
`⇒ (6)^2 +(y-5)^2 = (4)^2 +(y-3)^2` (Squaring both sides)
`⇒ 36+y^2 - 10y +25 = 16+y^2-6y +9`
⇒4y = 36
⇒y = 9
Hence, the point on the y-axis is ( 0,9) .
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संबंधित प्रश्न
On which axis do the following points lie?
Q(0, -2)
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If p(x , y) is point equidistant from the points A(6, -1) and B(2,3) A , show that x – y = 3
If the point `P (1/2,y)` lies on the line segment joining the points A(3, -5) and B(-7, 9) then find the ratio in which P divides AB. Also, find the value of y.
Find the ratio which the line segment joining the pints A(3, -3) and B(-2,7) is divided by x -axis Also, find the point of division.
Points A(-1, y) and B(5,7) lie on the circle with centre O(2, -3y).Find the value of y.
Find the coordinates of the circumcentre of a triangle whose vertices are (–3, 1), (0, –2) and (1, 3).
Show that A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a
rhombus ABCD.
Show that A (−3, 2), B (−5, −5), C (2,−3), and D (4, 4) are the vertices of a rhombus.
Find the value of k, if the points A (8, 1) B(3, −4) and C(2, k) are collinear.
\[A\left( 6, 1 \right) , B(8, 2) \text{ and } C(9, 4)\] are three vertices of a parallelogram ABCD . If E is the mid-point of DC , find the area of \[∆\] ADE.
Write the perimeter of the triangle formed by the points O (0, 0), A (a, 0) and B (0, b).
If the distance between points (x, 0) and (0, 3) is 5, what are the values of x?
If A (2, 2), B (−4, −4) and C (5, −8) are the vertices of a triangle, than the length of the median through vertex C is
If A(4, 9), B(2, 3) and C(6, 5) are the vertices of ∆ABC, then the length of median through C is
In which quadrant does the point (-4, -3) lie?
Write the equations of the x-axis and y-axis.
Find the coordinates of the point whose ordinate is – 4 and which lies on y-axis.
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
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`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.
