Advertisements
Advertisements
प्रश्न
The distance of the point (5, 0) from the origin is ______.
विकल्प
0
5
`sqrt(5)`
52
Advertisements
उत्तर
The distance of the point (5, 0) from the origin is 5.
Explanation:
Distance of the point (5, 0) from the origin is 5 units.
APPEARS IN
संबंधित प्रश्न
Prove that the points A(1, 7), B (4, 2), C(−1, −1) D (−4, 4) are the vertices of a square.
Find the distance between the following pair of point.
P(–5, 7), Q(–1, 3)
Find the distance of the following point from the origin :
(0 , 11)
Find the distance of a point (13 , -9) from another point on the line y = 0 whose abscissa is 1.
Find the relation between x and y if the point M (x,y) is equidistant from R (0,9) and T (14 , 11).
P and Q are two points lying on the x - axis and the y-axis respectively . Find the coordinates of P and Q if the difference between the abscissa of P and the ordinates of Q is 1 and PQ is 5 units.
A(-2, -3), B(-1, 0) and C(7, -6) are the vertices of a triangle. Find the circumcentre and the circumradius of the triangle.
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = – 7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
Find distance CD where C(– 3a, a), D(a, – 2a)
If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
