Question

find x,if distance between L(x,7) and M(1,15) is 10​

Answer 1

hlo mate ✌️

Distance = √ (x2-x1)^2 + (y2 - y1)^2

points : (x,7) , (1,15)

Distance = 10 units

According to the condition:

--> √ [(1 - x)^2 + ( 15 - 7)^2 ] = 10

--> √ [ (1 + x^2 -2x) - (8)^2] = 10

--> √ 1 + x^2 -2x + 64 = 10

--> √ x^2 -2x + 65 = 10

Squaring both side :

--> x^2 -2x + 65 = 100

--> x^2 -2x - 35 = 0

--> x^2 -7x +5x -35 =0

--> x( x - 7) + 5(x -7) = 0

--> (x + 5)(x - 7) = 0

Case : 1

x+ 5 = 0

x = -5

Case 2 :

x - 7 = 0

x = 7

Hope it helps ❣️

Answer 2

Answer:

LM = $\\ \sf{\sqrt {{( x2 - x1 )}^{2} + {(y2 - y1)}^{2}}}$

$\\ \longrightarrow \sf{10 = \sqrt {{( 1 - x)}^{2} + {(15 - 7)}^{2}}}$

$\\ \longrightarrow \sf{10 = \sqrt {{( 1 - x)}^{2} + {(8)}^{2}}}$

$\\ \longrightarrow \sf{10 = \sqrt {{( 1 - x)}^{2} + 64}}$

$\\ \longrightarrow \sf{100 = {( 1 - x)}^{2} + 64}$

$\\ \longrightarrow \sf{100 - 64 = {( 1 - x)}^{2}}$

$\\ \longrightarrow \sf{36 = {( 1 - x)}^{2}}$

Taking square root on both sides

$\\ \longrightarrow \sf{\sqrt {36} = \sqrt {{( 1 - x)}^{2}}}$

$\\ \longrightarrow \sf{ +\:or\:- 6 = 1 - x }$

$\\ \longrightarrow \sf{ 1 - x = 6\: or 1 - x = -6 }$

$\\ \longrightarrow \sf{ - x = 6 - 1 \: or \:- x = -6 - 1}$

$\\ \longrightarrow \sf{ - x = 5 \: or \:- x = -7}$

$\\ \longrightarrow \sf{ x = -5 \: or \: x = 7}$

hope it helps you