Question

-
Find the value of 'a', when the distance between the points (3. a) and (4, 1) is √10
units.​

Answer 1

☯ Let the given points be A(3, a) and B(4, 1).

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» Using Distance formula,

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$\star\:\boxed{\sf{\pink{D = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}}}}$

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• Distance formula is used to find the distance b/w two given points. Here, distance is given as √10.

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Therefore,

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$:\implies\sf\sqrt{10} = \sqrt{(1 - a)^2 + (4 - 3)^2} \\\\\\:\implies\sf 10 = (1 - a)^2 + (4 - 3)^2 \qquad\bigg\lgroup\bf Squaring \ both \ sides \bigg\rgroup \\\\\\:\implies\sf 10 = 1 + a^2 - 2a + 1 = 0 \\\\\\:\implies\sf a^2 - 2a - 8 = 0 \\\\\\:\implies\sf a^2 - 4a + 2a - 8 = 0 \\\\\\:\implies\sf a(a - 4) + 2(a - 4) = 0 \\\\\\:\implies\sf (a - 4) (a + 2) = 0 \\\\\\:\implies\sf a = 4 \: or \: a = -2$

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$\therefore\:{\underline{\sf{Hence,\ The \ value \ of \ 'a' \ are \ {\textsf{\textbf{ 4 or - 2}}}.}}}$

Answer 2

Given :-

➪ Points ( 3 , a ) and ( 4, 1 )

Let them be :

A( 3, a ) and B( 4 , 1 )

➪ Distance between these points = √10

To Find :-

➪ Value of 'a'

Solution :-

➪ As we know that

Distance Formula = √( x₂ - x₁ )² + ( y₂ - y₁ )²

➪ Let's find out the value of a  !

$\sf \implies \sqrt{10} = \sqrt{(1-a)^{2} + ( 4 -3 )^{2}}$

$\sf \{ Squaring \; both \; sides \}$

$\sf \implies 10 = (1-a)^{2} + (4-3)^{2}$

$\sf \implies 10 = 1 + a^{2} - 2a + 1 = 0$

$\sf \implies a^{2} - 2a - 8 = 0$

$\sf \implies a^{2} - 4a + 2a - 8 = 0$

$\sf \implies a(a-4) + 2(a-4) = 0$

$\sf \implies (a-4)(a-2) = 0$

∴ The value of 'a' =  -2  or  4

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