Math, asked by Anonymous, 2 months ago

Mathematics
Class 10
chapter 7
Coordinates Geometry
Exercise 7.1
qué no 9


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if A(0,1) is equidistant from P(5,-3) and R (X,6) , find the value of X. Also find the distance QR and PR also​

Answers

Answered by BadCaption01
2

Given -

  • Q(0 , 1) is equidistant from P(5 , -3) and R(x , 6).

Calculation -

\bf\pink{According\:to\:the\:question,} \\

\longmapsto\:\:\bf\blue{QP\:=\:QR} \\

Squaring both sides,

\longmapsto\:\:\bf{(QP)^2\:=\:(QR)^2} \\

We know that -

\orange\bigstar\:\:\bf\purple{Distance\:formula\:=\:\sqrt{(x_2\:-\:x_1)^2\:+\:(y_2\:-\:y_1)^2}\:} \\

\bf\red{So,} \\

\longmapsto\:\:\bf{\Big(\sqrt{(5\:-\:0)^2\:+\:(-3\:-\:1)^2}\Big)^2\:=\:\Big(\sqrt{(x\:-\:0)^2\:+\:(6\:-\:1)^2}\Big)} \\

\longmapsto\:\:\bf{5^2\:+\:(-4)^2\:=\:x^2\:+\:5^2} \\

\longmapsto\:\:\bf{x^2\:=\:16} \\

\longmapsto\:\:\bf{x\:=\:\sqrt{16}} \\

\longmapsto\:\:\bf\purple{x\:=\:\pm\:4} \\

FOR QR ;-

\red\checkmark\:\:\bf{QR\:=\:\sqrt{(\pm{4}\:-\:0)^2\:+\:(6\:-\:1)^2}\:} \\

:\implies\:\:\bf{QR\:=\:\sqrt{(\pm{4})^2\:+\:5^2}\:} \\

:\implies\:\:\bf{QR\:=\:\sqrt{16\:+\:25}\:} \\

:\implies\:\:\bf\orange{QR\:=\:\sqrt{41}\:} \\

FOR PR ;-

☆ For P(5 , -3) & R(4 , 6),

\red\checkmark\:\:\bf{PR\:=\:\sqrt{(4\:-\:5)^2\:+\:(6\:-\:(-3))^2}\:} \\

:\implies\:\:\bf{PR\:=\:\sqrt{(-1)^2\:+\:9^2}\:} \\

:\implies\:\:\bf{PR\:=\:\sqrt{1\:+\:81}\:} \\

:\implies\:\:\bf\{\color{navy}{PR\:=\:\sqrt{82}\:} \\

☆ For P(5 , -3) & R(-4 , 6),

\red\checkmark\:\:\bf{PR\:=\:\sqrt{(-4\:-\:5)^2\:+\:(6\:-\:(-3))^2}\:} \\

:\implies\:\:\bf{PR\:=\:\sqrt{(-9)^2\:+\:9^2}\:} \\

:\implies\:\:\bf{PR\:=\:\sqrt{81\:+\:81}\:} \\

:\implies\:\:\bf{PR\:=\:\sqrt{162}\:} \\

:\implies\:\:\bf{\color{lime}PR\:=\:9\sqrt{2}\:} \\

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