Find the distance of the point $P( -3,\ -4)$ from the $x-axis$.

Given: Point $( -3,\ -4)$.

To do: To find the distance of point $( -3,\ -4)$ from $x-axis$.

Solution: 

As given, point $A=( -3,\ -4)$

Co-ordinates of $x-axis\ B=( -3,\ 0)$

Distance of the point $( -3,\ -4)$ from $x-axis\ AB=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$

$\Rightarrow AB=\sqrt{( -3-( -3))^2+( 0-( -4))^2}$

$\Rightarrow AB=\sqrt{0+16}$

$\Rightarrow AB=\pm4$

$\because$ Distance can't be negative, therefore we reject the value $-4$.

Therefore, distance of the point $( -3,\ -4)$ from the $x-axis$ is $4$ units.

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Updated on: 10-Oct-2022

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