Question

which of the A(1,3) ,B(-3,2) ,C(3,4) and D(4,2) is farthest to the origin​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{A(1,3), B(-3,2), C(3,4), D(4,2)}$

$\underline{\textbf{To find:}}$

$\textsf{which of the given points is farthest to the origin}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Formula used:}}$

$\mathsf{The\;distance\;between\;the\;points\;(x_1,y_1)\;and\;(x_2,y_2)\;is}$

$\boxed{\mathsf{d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}}$

$\mathsf{Here,;O\;is\;(0,0)}$

$\mathsf{OA=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\mathsf{OA=\sqrt{(0-1)^2+(0-3)^2}}$

$\mathsf{OA=\sqrt{1+9}}$

$\mathsf{OA=\sqrt{10}}$

$\mathsf{OB=\sqrt{(0+3)^2+(0-2)^2}}$

$\mathsf{OB=\sqrt{9+4}}$

$\mathsf{OB=\sqrt{13}}$

$\mathsf{OC=\sqrt{(0-3)^2+(0-4)^2}}$

$\mathsf{OC=\sqrt{9+16}}$

$\mathsf{OC=\sqrt{25}}$

$\mathsf{OD=\sqrt{(0-4)^2+(0-2)^2}}$

$\mathsf{OD=\sqrt{16+4}}$

$\mathsf{OD=\sqrt{20}}$

$\implies\mathsf{\sqrt{10}<\sqrt{13}<\sqrt{20}<\sqrt{25}}$

$\implies\mathsf{OA<OB<OD<OC}$

$\therefore\textsf{The point C(3,4) is fathest to the origin}$

Answer 2

Answer:

The farthest from origin is point D

Step-by-step explanation:

Given:

$A(1,3) ,B(-3,2) ,C(3,4), D(4,2)$

we need to find the farthest distance from the origin

So by applying the distance formula we get

$OA=\sqrt{(1)^2+(3)^2}\\\\OA=\sqrt{10}$

$OB=\sqrt{(-3)^2+(2)^2}\\\\OB=\sqrt{13}$

$OC=\sqrt{(3)^2+(4)^2} \\\\OC=\sqrt{25}$

$OD=\sqrt{(4)^2+(2)^2} \\\\OD=\sqrt{20}$

So farthest is point D