Question

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


❀✿ᏅᏬᎬᏕᎿᎨᎾᏁ᭄✯♡࿐
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​

Answer 1

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),

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

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

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

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

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