Advertisements
Advertisements
प्रश्न
The distance between the points (3, 1) and (0, x) is 5. Find x.
Advertisements
उत्तर
It is given that the distance between the points A (3, 1) and B (0, x) is 5.
∴ AB = 5
AB2 = 25
(0 - 3)2 + (x - 1)2 = 25
9 + x2 + 1 - 2x = 25
x2 - 2x - 15 = 0
x2 - 5x + 3x - 15 = 0
x(x - 5) + 3(x - 5) = 0
(x - 5)(x + 3) = 0
x = 5, -3
APPEARS IN
संबंधित प्रश्न
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and R(–3, 6), find the coordinates of P.
If two vertices of an equilateral triangle be (0, 0), (3, √3 ), find the third vertex
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).
The value of 'a' for which of the following points A(a, 3), B (2, 1) and C(5, a) a collinear. Hence find the equation of the line.
Distance of point (−3, 4) from the origin is ______.
Prove that the points (1 , 1) , (-1 , -1) and (`- sqrt 3 , sqrt 3`) are the vertices of an equilateral triangle.
Use distance formula to show that the points A(-1, 2), B(2, 5) and C(-5, -2) are collinear.
Give the relation that must exist between x and y so that (x, y) is equidistant from (6, -1) and (2, 3).
The distance between point P(2, 2) and Q(5, x) is 5 cm, then the value of x = ______.
The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is ______.

