Question

If the distance between P (4, 0) and Q(0, x) is 5 units, the value of x will be * 1 point

Answer 1

$\sf \: Distance \: formula \: = \sqrt{ {({x}_{1} - {x}_{2})}^{2}+ {( {y}_{1} - {y}_{2} )}^{2}}$

Substituting values,

$\sf \: \rightarrow \sqrt{ {({x}_{1} - {x}_{2})}^{2}+ {( {y}_{1} - {y}_{2} )}^{2}} \\ \: \rightarrow \sf \: \sqrt{ {(4 - 0)}^{2}+ {( 0 - x)}^{2}} = 5\\ \: \rightarrow \sf \: \sqrt{ {4 }^{2}+ {x}^{2}} = 5 \\ \: \rightarrow \sf \: \sqrt{ 16+ {x}^{2}} = 5 \\ \\ \bigstar \: \sf \underline{ \: squaring\: both \: sides} \\ \\\sf \: \rightarrow 16 +{x}^{2}} = 25 \\ \sf \: \rightarrow {x}^{2} \: = 25 - 16 \\ \sf \: \rightarrow {x}^{2} = 9 \\ \sf \: \rightarrow x = 3$

Answer 2

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QUESTION :

If the distance between P (4, 0) and Q(0, x) is 5 units, the value of x will be * 1 point.

SOLUTION :

Distance Formulae :

$\sf{ \sqrt { { x_{2} - x_{1} } ^ 2 + {y_{2} - y_{1} } ^ 2 } }$

Substituting the given Values ,

$\sf{ \sqrt { { 0 - 4 } ^ 2 + {x - 0 } ^ 2 } }$

=> $\sf{ \sqrt { 16 + {x} ^ 2 } }$

=> Squareing :

X^2 + 16 = 25

=> X ^ 2 = 9

=> X = 3 and -3