Let A = (a_1, a_2) and B = (b_1, b_2) be two | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The distance $d$ between two points $A(a_1, a_2)$ and $B(b_1, b_2)$ is given by $d = \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2}$.Since $a_1, a_2, b_1, b

Vedclass · Accepted Answer

$\sqrt{83}$

Vedclass · Answer

(C) The distance $d$ between two points $A(a_1, a_2)$ and $B(b_1, b_2)$ is given by $d = \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2}$.Since $a_1, a_2, b_1, b_2$ are integers,let $x = |b_1 - a_1|$ and $y = |b_2 - a_2|$,where $x$ and $y$ are non-negative integers.Thus,$d^2 = x^2 + y^2$.We check the given options:$A: \sqrt{65} = \sqrt{8^2 + 1^2}$,which is possible.$B: \sqrt{74} = \sqrt{7^2 + 5^2}$,which is possible.$C: \sqrt{83}$. We check if $83$ can be written as the sum of two squares. The squares less than $83$ are $0, 1, 4, 9, 16, 25, 36, 49, 64, 81$. No two of these sum to $83$.$D: \sqrt{97} = \sqrt{9^2 + 4^2}$,which is possible.Therefore,$\sqrt{83}$ is not a possible value.

Let $A = (a_1, a_2)$ and $B = (b_1, b_2)$ be two points in the plane with integer coordinates. Which one of the following is not a possible value of the distance between $A$ and $B$?

Similar Questions

Find the points of trisection of the line segment joining the points $A(2, 1)$ and $B(5, 3)$.

Three vertices of a parallelogram taken in order are $(-1, -6)$,$(2, -5)$ and $(7, 2)$. The fourth vertex is

If the line $2x - 3y + 4 = 0$ divides the line segment joining the points $A(-2, 3)$ and $B(3, -2)$ in the ratio $m:n$,then the point which divides $AB$ in the ratio $-4m:3n$ is

The straight line $3x + y = 9$ divides the line segment joining the points $(1, 3)$ and $(2, 7)$ in the ratio

If the line $2x - y - 4 = 0$ divides the line segment joining the points $(2, -1)$ and $(1, -4)$ at the point $(a, b)$ in the ratio $m:n$,then $4(a - b(\frac{m}{n})^2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module