Question

01. Find the value of 'a' if the distance between the points (a 2), (3, 4) is 2✓2
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Answer 1

$\huge \boxed{ \colorbox{red}{ \blue{a = 1 \& 5}}}$

Step-by-step explanation:

Q.

Find the value of ‘a’, of the distance between the points (a, 2), (3, 4) is 2√2

.

Solution -

$let \: point \\ \\ A( x_{1}, y_{1}) = a,2 \\ = > x_{1} = a, y_{1} = 2 \\ \\ B = ( x_{2}, y_{2} )= 3,4 \\ = > x_{2} = 3, y_{2} = 4$

So,

$distance \: AB =2 \sqrt{2} \\ \\ = > \sqrt{ { (x_{1} - x_{2}) }^{2} + ( y_{1} - { y_{2}) }^{2} } = 2 \sqrt{2} \\ \\ = > \sqrt{(a - {3)}^{2} + ( 2 - {4)}^{2} } = 2 \sqrt{2} \\ \\ = > \sqrt{ {a}^{2} - 2(a)(3) + {3}^{2} + ( { - 2)}^{2} } = 2 \sqrt{2} \\ \\ = > \sqrt{ {a}^{2} - 6a + 9 + 4 } = 2 \sqrt{2} \\ \\ = > \sqrt{ {a}^{2} - 6a + 13 } = 2 \sqrt{2} \\ \\ = > {( \sqrt{ {a}^{2} - 6a + 13} )}^{2} = ( {2 \sqrt{2}) }^{2} \\ \\ = > {a}^{2} - 6a + 13 = 8 \\ \\ = > {a}^{2} - 6a + 13 - 8 = 0 \\ \\ = > {a}^{2} - 6a + 5 = 0 \\ \\ = > {a}^{2} - a - 5a + 5 = 0 \\ \\ = > a(a - 1) - 5(a - 1) = 0 \\ \\ = > (a - 1)(a - 5) = 0$

The value of a,

a - 1 = 0

a = 1

.

a - 5 = 0

=> a = 5

.

Hence the value of a = 1 or 5.

both will be correct.

hope it helps.