If poitns `A(3,5)` and B are equidistant from `H(sqrt2,sqrt5)` and B has rational coordinates,then `AB=`

Question

If poitns `A(3,5)` and B are equidistant from `H(sqrt2,sqrt5)` and B has rational coordinates,then `AB=`

If poitns `A(3,5)` and B are equidistant from `H(sqrt2,sqrt5)` and B has rational coordinates,then `AB=`
A. `sqrt(7)`
B. `sqrt((3-sqrt2)^2+(5-sqrt5)^2)`
C. `ssqrt(34)`
D. none of these

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

Find the value of the following: (i) ` log_(10) 2 + log_(10) 5` (ii) ` log_(3) (sqrt(11)-sqrt2) + log_(3) (sqrt11+sqrt2)` (iii) ` log_(7) 35 - log_(7)

2 Answers 5 Views

The locus of a point that is equidistant from the lines `x+y - 2sqrt2 = 0` and `x + y - sqrt2 = 0` is (a) `x+y-5sqrt2=0` (b) `x+y-3sqrt2=0` (c) `2x+2y

2 Answers 6 Views

Convert the following points from polar coordinates to the corresponding Cartesian coordinates. (i) `(2,pi//3)` (ii) `(0.pi//2)` (iii) `(-sqrt(2),pi//

2 Answers 5 Views

Convert the following polar coordinates to its equivalent Cartesian coordinates. (i) `(2,pi)` (ii) `(sqrt(3),pi//6)`

2 Answers 5 Views

Convert the following Cartesian coordinates to the cooresponding polar coordinates using positive r. (i) `(1,-1)` (ii) `(-3,-4)`

2 Answers 5 Views

Le n be the number of points having rational coordinates equidistant from the point `(0,sqrt3)`, the

2 Answers 5 Views

If the vertices of a triangle are `(sqrt5,0)` and `(sqrt3,sqrt2)`, and `(2,1)` then the orthocentre of the triangle is

2 Answers 8 Views

In the `DeltaABC`, the coordinates of B are `(0, 0), AB=2, /_ABC=pi/3` and the middle point of BC has the coordinates (2,0). The centroid of the trian

2 Answers 5 Views

If (4, -3) and (-4, 3) are the coordinates of the ends of the diameter of the circle, then the coordinates of centre of the circle are

2 Answers 5 Views

If (6, -8) and (-6, 8) are the coordinates of the ends points of the diameter of a circle, then the coordinates of the centre of the circle are –

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - D
Let `B-=(alpha,beta)`, where `alpha,beta` are rational.
Given `HA^2=HB^2`
`rArr(3-sqrt2)^2+(5-sqrt5)^2=(alpha-sqrt2)^2+(beta-sqrt5)^2`
`41-6sqrt2-10sqrt5=alpha^2+beta^2+7-2sqrt2alpha-2sqrt5beta`
`rArr alpha^2+beta^2-34++2(3-alpha)sqrt2+2(5-beta)sqrt5=0`
Since `alpha` and `beta` are rational,
`3-alpha=0rArralpha=3`
and `5-beta=0rArrbeta=5`
`therefore AB=0`