Question

Distance from x axis to point p(-3,-4)

Answer 1

Distance is $\sqrt{ \left ( b+3 \right )^2+ 16}$ where $b$ is constant.

Step-by-step explanation:

Let us assume, coordinate on $x -\text{axis}$ be $(b,0)$ where $b$ is constant.

Formula:

The distance between two points $\left ( x_1,y_1 \right ) \text{and} \left ( x_2,y_2 \right )$ is

$\sqrt{ \left ( x_2-x_1 \right )^2+ \left ( y_2-y_1 \right )^2}$

Here, $\left ( x_1,y_1 \right )=\left ( -3,-4 \right )$ and $\left ( x_2,y_2 \right )=\left ( b,0 \right )$

The distance between given points is $\sqrt{ \left ( b+3 \right )^2+ \left ( 0+4 \right )^2}$

$\Rightarrow \sqrt{ \left ( b+3 \right )^2+ \left ( 0+4 \right )^2}=\sqrt{ \left ( b+3 \right )^2+ 16}$

For example - Let us take $b=4$, The distance between points is

$\Rightarrow \sqrt{ \left ( 4+3 \right )^2+ 16}\\\\\Rightarrow \sqrt{ 49+ 16}$

$\Rightarrow \sqrt{ 65}$                                                                           

For example

Let us take $b=-8$, The distance between points is:

$\Rightarrow \sqrt{ \left ( -8+3 \right )^2+ 16}\\\\\Rightarrow \sqrt{ 25+ 16}\\\\\Rightarrow \sqrt{ 41}$                                                                  

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