Question

the distance of the point P( - 5 - 12) from the origin is​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

13 units is your answer.

$\huge\star{\green{\underline{\mathfrak{Explaination :-}}}}$

The point given is P (-5,-12)

So, x = -5 and y = -12.

The formula to find the distance of a point from origin is $\sqrt {{x}^{2}+{y}^{2}}$.

Putting the values of x and y in the formula :-

$\leadsto {\sqrt {{-5}^{2}+{-12}^{2}}}$

$\leadsto {\sqrt {25 + 144}}$

$\leadsto {\sqrt {169}}$

$\leadsto{ 13}$

So, the distance between the origin and the point given is 13 units.

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Additional important formulas :-

• To find the distance between two points say P ($x_1,y_1$) and Q($x_2,y_2$) is : $\sqrt {{(x_2-x_1)}^{2}+{(y_2-y_1)}^{2}}$

• To find the distance of a point say P (x,y) from origin is : $\sqrt {{x}^{2}+{y}^{2}}$

• Coordinates of the point P (x,y) which device the line segment joining the points A ($x_1,y_1$) and B ($x_2,y_2$) internally in the ratio $m_1 : m_2$ is : $\dfrac {m_1 x_2 + m_2 x_1}{m_1+m_2},\dfrac {m_1 y_2 + m_2 y_1}{m_1+m_2}$

• The midpoint of the line segment joining the points P ($x_1,y_1$) and Q($x_2,y_2$) is : $\dfrac{x_1+x_2}{2} , \dfrac {y_1+y_2}{2}$

• The area of the triangle formed by the points ($x_1,y_1$),($x_2,y_2$) and ($x_3,y_3$) is the numerical value of the expression:$\dfrac{1}{2}[x_1 (y_2-y_3)+x_2 (y_3-y_1)+x_3 (y_1-y_2)]$

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Answer 2

$\huge{\underline{\underline{\tt{\red{Answer\leadsto 13}}}}}$

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$\huge{\underline{\underline{\blue{\tt{\mid{GiveN}}}}}}$

•Coordinates of Point P(-5,-12)

•Distance from origin

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$\huge{\underline{\blue{\displaystyle{Formula}}}}$

Distance from origin

$\implies OP = \sqrt{x_1 ^2 + y_1 ^2}$

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SoluTion:

Applying the formula

OP= $\sqrt{x_1 ^2 + y_1 ^2}$

$\implies\sqrt{(-5) ^2 + (-12) ^2} \\ \implies \sqrt{25+144} \\ \implies \sqrt{169} \\ \\ \huge{\boxed{\leadsto{13 \:units}}}$

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