Advertisements
Advertisements
प्रश्न
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
Advertisements
उत्तर
We know that any point on x-axis has coordinates of the form (x, 0).
Abscissa of point B = 11
Since, B lies of x-axis, so its co-ordinates are (11, 0).
AB = `sqrt((11 -7)^2 + (0 -3)^2)`
= `sqrt(16 + 9)`
= `sqrt(25)`
= 5 units
APPEARS IN
संबंधित प्रश्न
Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle
If a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
Find the distance of a point P(x, y) from the origin.
Determine whether the points are collinear.
P(–2, 3), Q(1, 2), R(4, 1)
Find the distance between the following point :
(sec θ , tan θ) and (- tan θ , sec θ)
Prove that the points (1 , 1) , (-1 , -1) and (`- sqrt 3 , sqrt 3`) are the vertices of an equilateral triangle.
Find the distance between the following pair of points:
`(sqrt(3)+1,1)` and `(0, sqrt(3))`
Point P (2, -7) is the centre of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of AB.

By using the distance formula prove that each of the following sets of points are the vertices of a right angled triangle.
(i) (6, 2), (3, -1) and (- 2, 4)
(ii) (-2, 2), (8, -2) and (-4, -3).
Find the distance between the points O(0, 0) and P(3, 4).
