Question

What is the horizontal distance of the point (3,-5)from origin ?

Answer 1

Given :-

A point ( 3 , - 5 )

To Find :-

Horizontal Distance from the origin

Solution :-

Let's Understand some basic information ;

• X - axis is also known as abscissa

• Y - axis is also known as ordinate

• Coordinates of Origin = ( 0 , 0 )

Distance between any two points $\bf ( x_1 , y_1 )$ and $\bf ( x_2 , y_2 )$ is given by ;

$\quad \qquad { \bigstar { \underline { \boxed { \pmb { \bf { \red { \underbrace { \sqrt{(x_2 - x_ 1)² + ( y_2 - y_1)²} }}}}}}}}{\bigstar}\quad \qquad$

See the attachment ( Rough graph ) !!!

From here we can find the distance easily , if they lie on X - axis or Y - axis or vertical distance , But for horizontal Distance , we have to understand from where it is being measured , so it is being measured between O and let us name the point ( 3 , - 5 ) as " A " . Now , The distance we have to find is OA . Now , let's start !!!!!!

__________________________

By Distance formula ;

$\quad \leadsto \quad \bf OA = \sf \sqrt{(3-0)² + ( -5-0)²}$

${ : \implies \quad \bf OA = \sf \sqrt{3² + ( - 5 )²}}$

${ : \implies \quad \bf OA = \sf \sqrt{9 + 25}}$

${ : \implies \quad \bf \therefore \quad OA = \sf \sqrt{34} \: units}$

Henceforth , The Required Answer is $\bf \sqrt{34}$ units :)