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Answer 1

Distance between two points P(x1,y1) and Q(x2,y2) is given by: d(P, Q) = √ (x2 − x1)2 + (y2 − y1)2 {Distance formula} 2. Distance of a point P(x, y) from the origin is given by d(0,P) = √ x2 + y2. 3. Equation of the x-axis is y = 0 4.

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Answer 2

Given that , The two points are ( x₁ , y₁ ) and ( x₂ , y₂ ) and this line is parallel to x – axis .

Need To Find : The Distance between them ?

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❍ Let's say that , the two endpoints of line be P( x₁ , y₁ ) & Q( x₂ , y₂ ) .

⠀▪︎ ⠀We know that the Distance between two points ( x₁ , y₁ ) and ( x₂ , y₂ ) is – 

$\qquad \star\:\pmb{\underline {\boxed {\sf \: Distance \:=\: \sqrt{ \Bigg( x_2 - x_1 \Bigg)^2 \: +\:\Bigg( y_2 - y_1\:\Bigg)^2}\:\:}}}\:$

⠀⠀⠀⠀⠀⌬⠀ Finding Distance b/w PQ when this line is parallel to x – axis –

☯︎ Since the Line is Parallel to x – axis .

⠀⠀⠀⠀⠀∴ y₁ = y₂

$\\\qquad \dag{\underline {\frak { Substituting \:known \:Values \:in \:Formula \::\:}}}$

$:\implies \sf AB \:=\: \sqrt{ \Big( x_2 - x_1 \Big)^2 \: +\:\Big( y_2 - y_1\:\Big)^2}\:\:$

$:\implies \sf AB \:=\: \sqrt{ \Big( x_2 - x_1 \Big)^2 \: +\:\Big( 0\:\Big)^2}\:\qquad \because \:\bigg\lgroup \: y_1 = y_2\bigg\rgroup$

$:\implies \sf AB \:=\: \sqrt{ \Big( x_2 - x_1 \Big)^2 \: + \: \Big( 0 \Big) }\:\:$

$:\implies \sf AB \:=\: \sqrt{ \Big( x_2 - x_1 \Big)^2 \: }\:\:$

$:\implies \sf AB \:=\: \pm \Big( x_2 - x_1 \Big) \:\:$

$:\implies \pmb {\underline {\boxed {\purple {\:\frak{ \:AB\:\:=\: |\: x_2 - x_1 |\:}}}}}\:\bigstar \:$

$\:\:\:\therefore \:\underline {\sf \: Hence ,\:The \:Distance \:b/w \:two \: points \:is\: \pmb{\sf |\: x_2 - x_1\: | \:}\:.}$