Question

If the distance between the points (-3,p) (3,5) is 3 root5 then find the value of p​

Answer 1

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

$\sf \bullet Two\ points \ (-3,p) \& \ (3,5)$

$\sf \bullet Distance\ between\ them \ is\ 3\sqrt{5} .$

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

$\sf \bullet The \ value \ of \ p .$

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

$\underline{\purple{\tt Given\ two \ points\ are , }}$

• A ( -3 , p )
• B ( 3 , 5 )

$\underline{\purple{\tt We\ know \ Distance\ Formula \ as : }}$

$\large\boxed{\pink {\bf\purple{\dag} Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}$

$\tt:\implies Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\tt:\implies Distance=\sqrt{[3-(-3)]^2+(p-5)^2}\\\\\tt:\implies 3\sqrt5 = \sqrt{(3+3)^2+p^2+25 - 10p } \\\\\tt:\implies (3\sqrt5)^2= 36 + p^2 + 25 - 10p \\\\\tt:\implies 45 = p^2-10p+61\\\\\tt:\implies p^2-10p+61-45=0\\\\\tt:\implies p^2-10p+16=0\\\\\tt:\implies p^2-2p-8p+16=0\\\\\tt:\implies p(p-2)-8(p-2)=0\\\\\tt:\implies (p-2)(p-8)=0\\\\\underline{\boxed{\red{\tt\longmapsto x \qquad=\qquad 2,8 }}}$

$\boxed{\green{\bf\pink{\dag}\:\:Hence\:the \ points \ can \ be \ 2 \ or \ 8 .}}$