Question

if the distance between the origin and the point p (x,15) is 17 units , the value of x is​

Answer 1

Answer:

The value of x is – 8 or 8 if the distance between the origin and the point p (x, 15) is 17 units.

Step-by-step explanation:

The distance between the origin and the point p (x, 15) is 17 units.

The coordinates of the origin are (0, 0).

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So we have :

$\quad ~$★ x₁ = 0

$\quad~$★ x₂ = x

$\quad~$★ y₁ = 0

$\quad~$★ y₂ = 15

$~$

Substituting these values in the distance formula and solving for x :

$\sf \quad : \implies \: \sqrt{ {( x_{2} - x_{1}) }^{2} + {( y_{2} - y_{1}) }^{2} }$

$\sf\quad : \implies \: \sqrt{ {(x - 0)}^{2} + {(15 - 0)}^{2} } = 17$

$\sf\quad : \implies \: \sqrt{ {x}^{2} + {( 15)}^{2} } = 17$

$\sf \quad : \implies \: \sqrt{ {x}^{2} + 225} = 17$

$\sf\quad : \implies \: {x}^{2} + 225 = 289$

$\sf \quad : \implies \: {x}^{2} + 225 - 289 = 0$

$\sf \quad : \implies \: {x}^{2} - 64 = 0$

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Factorising x² – 64 using the identity :

$\sf\quad : \implies \: (a - b)(a + b) = {a}^{2} - {b}^{2}$

$\sf \quad: \implies \: {a}^{2} = {x}^{2} \: \: \: \: \: {b}^{2} = 64$

$\sf \quad : \implies \: {x}^{2} - 64 = (x + 8)(x - 8)$

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Finding the zeroes of (x + 8) and (x – 8) :

$\sf \quad : \implies \: (x + 8) : x = - 8$

$\sf \quad : \implies \: (x - 8) : x = 8$

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The value of x is either – 8 or 8.