Question

find the value of a if the distance between the points (a,2),(3,4) is 2root2

Answer 1

Distance between the points (a,2) and (3,4) is √[(3-a)²+(4-2)²]
=√(9+a²-6a+4)
⇒√(a²-6a+13)=2√2(given)=√8
⇒a²-6a+13=8
⇒a²-6a+5=0
⇒a²-5a-a+5=0
⇒a(a-5)-1(a-5)=0
⇒(a-1)(a-5)=0
⇒a=1 or a=5.

Answer 2

Answer:

Value of a are 1 or 5.

Step-by-step explanation:

Given: Two points $(a,2)$,$(3,4)$ & distance between them is $2\sqrt{2}$.

To find value of a.

We know that distance between two points$(x_{1},y_{1} )$ & $(x_{2},y_{2} )$ is given by

$d=\sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$.

Here, $x_{1}=a,y_{1}=2,x_{2}=3,y_{2}=4$

Substituting values in equation  (1) we have,

$2\sqrt{2} =\sqrt{(3-a)^{2}+(4-2)^{2} }\\ 2\sqrt{2} =\sqrt{(3-a)^{2}+4 }$

Squaring both sides we get,

$8=(3-a)^{2}+4\\ 4=(3-a)^{2}$

Now using, $(a-b)^{2}=a^{2}-2ab+b^{2}$ we have,

$4=9-6a+a^{2}\\a^{2} -6a+5=0\\a^{2} -5a-a+5=0\\a(a-5)-1(a-5)=0\\(a-1)(a-5)=0\\a=1,5$

So, the value of a are 1 or 5.

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