Question

Find the value of x for which the distance between points A(x, 7) and B (–2 , 3 ) is 4√5 units.​

Answer 1

$\sqrt{(-2-x) ^{2} + (3-7) ^2} = 4\sqrt{5} \\\\4 + x^{2} + 4x + 4 ^2 = 16 \times 5\\\\x^{2} +4x + -60 = 0\\\\x(x+10) - 6(x + 10)\\\\(x-6)(x+10)\\\\x = 6 \ or -10$

Answer 2

Answer:

Required value of x is (-10) or 6.

Step-by-step explanation:

Given, two points are A(x, 7) and B (–2 , 3 ).

It is also given that distance between A and B is equal to 4√5 units.

We know, if (a,b) and (p,q) are two points then distance between them $\sqrt{ {(a - p)}^{2} + {(b - q)}^{2} } \: units$

So, distance between A and B

$= \sqrt{ {(x + 2)}^{2} + {(7 - 3)}^{2} } \\ = \sqrt{ {(x + 2)}^{2} + {4}^{2} }$

According to question,

$\sqrt{ {x + 2)}^{2} + {4}^{2} } = 4 \sqrt{5} \\ \sqrt{ {x}^{2} + 4x + 4 + 16 } = 4 \sqrt{5} \\ \sqrt{ {x}^{2} + 4x + 20 } = 4 \sqrt{5} \\ {x}^{2} + 4x + 20 = {(4 \sqrt{5}) }^{2} \\ {x}^{2} + 4x + 20 = 80 \\ {x}^{2} + 4x - 60 = 0 \\ {x}^{2} + (10 - 6)x - 60 = 0 \\ {x}^{2} + 10x - 6x - 60 = 0 \\ x(x + 10) - 6(x + 10) = 0 \\ (x + 10)(x - 6) = 0 \\ x = - 10 \: or \: 6$

So, required value of x is (-10) or 6

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