Advertisements
Advertisements
प्रश्न
Find the distance between the following pairs of point:
`(sqrt(3)+1,1)` and `(0, sqrt(3))`
उत्तर
`(sqrt(3)+1,1)` and `(0, sqrt(3))`
Distance between the given points
= `sqrt((0 -sqrt(3) - 1)^2 + (sqrt(3) - 1)^2)`
= `sqrt(3 + 1 + 2sqrt(3) + 3 + 1 - 2sqrt(3)`
= `sqrt(8)`
= 2`sqrt(2)`
APPEARS IN
संबंधित प्रश्न
If P (2, – 1), Q(3, 4), R(–2, 3) and S(–3, –2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Find the distance between the following pair of points:
(-6, 7) and (-1, -5)
Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of other two
vertices.
If P (x , y ) is equidistant from the points A (7,1) and B (3,5) find the relation between x and y
Find the distance between the following pair of point.
T(–3, 6), R(9, –10)
Show that the points A(1, 2), B(1, 6), C(1 + 2`sqrt3`, 4) are vertices of an equilateral triangle.
Find the distance of the following point from the origin :
(0 , 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.
Prove that the following set of point is collinear :
(4, -5),(1 , 1),(-2 , 7)
ABC is an equilateral triangle . If the coordinates of A and B are (1 , 1) and (- 1 , -1) , find the coordinates of C.
Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.
Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT
Give the relation that must exist between x and y so that (x, y) is equidistant from (6, -1) and (2, 3).
Show that the points (a, a), (-a, -a) and `(-asqrt(3), asqrt(3))` are the vertices of an equilateral 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`
The point which lies on the perpendicular bisector of the line segment joining the points A(–2, –5) and B(2, 5) is ______.
The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, – 4).
If the point A(2, – 4) is equidistant from P(3, 8) and Q(–10, y), find the values of y. Also find distance PQ.
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.