Question

Find p if distance between points (2,3) and (5,p) is 10 units

Answer 1

$\huge{\mathcal{\blue{\underline{Answer}}}}$

Distance formula :

$= \sqrt{(x _{2} - x _{1}) {}^{2} + (y _{2} - y _{1} } ) \\ 10 = \sqrt{(5 - 2) {}^{2} + (p - 3) {}^{2} } \\ (10) {}^{2} = (3) {}^{2} + (p - 3) {}^{2} \\ 100 - 9 = (p - 3) {}^{2} \\ 91 =( p - 3) {}^{2} \\ \sqrt{91} + 3 = p$

Answer 2

$\Huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

$\Large{\sf{Given :}}$

Distance between two points (2 , 3) and (5 , p) is 10 units.

$\rule{200}{2}$

$\LARGE{\sf{To \: Find :}}$

Value of p.

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$\LARGE{\sf{Solution :}}$

We know the distance formula,

$\large{\boxed{\boxed{\green{\tt{Distance \: formula \: = \: \sqrt{(x_{2} - x_{1})^2 (y_{2} - y_{1})^2}}}}}}$

Where,

x2 = 5 , x1 = 2 , y2 = p and y1 = 3

(Putting Values)

$\rule{200}{2}$

$\sf{x = \sqrt{ {( {5 - 2)} }^{2} + ( {p - 3)}^{2} } } \\ \\ \bf{Taking \: square \: root \: to \: LHS} \\ \\ \sf{ {10}^{2} = {3}^{2} + {(p - 3)}^{2} }$

$\sf{100 - 9 = (p - 3)^2 } \\ \\ \sf{91 = (p - 3)^2} \\ \\ \bf{Taking \: square \: to \: LHS}\\ \\ \sf{\sqrt{91} = p - 3} \\ \\ \sf{ \sqrt{91} + 3 =p}$

$\LARGE{\boxed{\boxed{\red{\bf{p = \sqrt{91} + 3}}}}}$