Question

the distance of points p(-3,-4) from the x- axis in units is????? send me answer by solving it ...plz​

Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

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Given:

Coordinates of P(-3,-4)

Origin(0,0)

__________________________________

To find

Distance of P from x-axis

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$\huge{\underline{\underline{\green{\tt{Formula}}}}}$

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

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SoluTion:

Since we need to Find the distance from x-axis we will need to find the coordinates of point on X-axis which has x-coordinate same

So the coordinate on the x-axis must be (-3,0)

Now, by distance formula

$\sqrt{(-3-(-3))^2+(0-(-4))^2} \\ \implies \sqrt{4}^2 \\ \implies \sqrt{16} \\ \implies {\boxed{\boxed{4}}}$

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Answer 2

Answer:

Given:

Coordinates of P(-3,-4)

Origin(0,0)

__________________________________

To find

Distance of P from x-axis

___________________________________________

\huge{\underline{\underline{\green{\tt{Formula}}}}}

Formula

\implies⟹ \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

_________________________________

SoluTion:

Since we need to Find the distance from x-axis we will need to find the coordinates of point on X-axis which has x-coordinate same

So the coordinate on the x-axis must be (-3,0)

Now, by distance formula

\begin{lgathered}\sqrt{(-3-(-3))^2+(0-(-4))^2} \\ \implies \sqrt{4}^2 \\ \implies \sqrt{16} \\ \implies {\boxed{\boxed{4}}}\end{lgathered}

(−3−(−3))

2

+(0−(−4))

2

⟹

4

2

⟹

16

⟹

4

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