Question

If the distance between the points M (-1,-1) and N (2,x) is 5 units, then the value of x can be

Answer 1

Solution:

• Here,the distance between the points M(-1,-1) and N(2,x) is 5 units,so we should find the value of x by applying the distance formula,

As we know that,

$\to$Distance=$\sf{\sqrt{{(x_2-x_1)}^{2}+{(y_2-y_1)}^{2} }}$

Given that,

• $\rm{x_1,y_1}$  = (-1,-1)
• $\rm{x_2,y_2}$  = (2,x)

Applying the values,

$\implies\sf{\sqrt{\big(2-(1)\big)^2\: + \:\big(x-(-1)\big)^2}=5}\\ \\ \implies\sf{\sqrt{(2+1)^2\:+\:(x+1)^2}=5}\\ \\ \implies\sf{\sqrt{(3)^2\:+\:(x+1)^2}=5 }\\ \\ \implies\sf{9\:+\:(x+1)^2=5^2}\\ \\ \implies\sf{9\:+\:(x+1)^2\:=25} \\ \\ \implies\sf{(x+1)^2=25 - 9}\\ \\ \implies\sf{(x+1)^2=16}\\ \\ \implies\sf{x+1\:=\: \sqrt{16}}\\ \\ \implies\sf{x +1 = \pm\:4}\\ \\ \implies\sf{x = \pm\:4 -1}\\ \\ \implies\sf{x= 3\:(or)-5}$

• Therefore,the value of x is 3(or)-5.

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Answer 2

$\sf {Ratio = 1.3.Y \tiny3}$

$\sf{Now \: Let \: P ( x ,y )}$

$\sf {X = }\frac{1(6) + 3(2)}{1 + 3} \: = \frac{6 + 6}{4} \: = \frac{12}{4} = 3$

$\sf {X = }\frac{1(5) + 3(1)}{5 + 3} \: = \frac{5 + 3}{4} \: = \frac{8}{4} = 2$

$\sf{So \: P ( x , y ) = ( 3 , 2 ) }$

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