Question

The distance between the points A (x, -1) and B (3, 2) is 5 units, find 'x'​

Answer 1

Answer:

Given that distance between A(x, -1) and B(3, 2) is 5 units.

we can solve this question that is find the value of x by using distance formula As given below.

So, using distance formula

$\sqrt{ {( x- 3)}^{2} + {( - 1 - 2)}^{2} } = 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: S.B.S. \\ {x}^{2} + 9 - 6x + 9 = 25 \\ \implies \: {x}^{2} - 6x - 7 = 0 \\ \implies {x}^{2} + x - 7x - 7 = 0\\ \implies x(x + 1) - 7(x + 1) = 0 \\ \implies(x + 1)(x - 7) = 0 \\ \implies \: x = - 1 \: or \: x = 7$

so possible values of x are - 1 and 7.

So, for x = -1, 7

given two points will be 5 unit distance from each other.

We can also verify the result by putting value of x and finding distance between the given points .

Answer 2

$\huge \red { \boxed{ \boxed{ \mathsf{ \mid \ulcorner Answer :\urcorner \mid }}}}$

Question :-

Distance between the points A(x,-1) and B(3,2) is 5 units, find 'x'

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Answer :-

We have distance formula

$\LARGE{\boxed{\boxed{\green{\sf{|AB| \: = \: \sqrt{{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} }}}}}}}$

Where,

x2 = 3

x1 = x

y2 = 2

y1 = -1

And |AB| = 5 units

___________________[Put Values]

$\large{\sf{5 \: = \: \sqrt{{(3 \: - \: x)}^{2} \: + \: {(2 \: -(-1))}^{2}}}}$

S.B.S

$\large{\sf{5^2 \: = \: (3 - x)^{2} \: + \: (2 + 1)^{2}}}$

$\large{\sf{25 \: = \: 9 \: + \: x^2 \: - \: 6x \: + 9}}$

$\large{\sf{25 \: = \: x^{2} \: - \: 6x \: + 18}}$

$\large{\sf{0 \: = \: x^{2} \: - \: 6x \: + 18 \: - \: 25}}$

$\large{\sf{0 \: = \: x^{2} \: - \: 6x \: - 7}}$

$\large{\sf{0 \: = \: x^{2} \: + \: x \: - 7x \: - \: 7}}$

$\large{\sf{0 \: = \: x(x \: + \: 1) \: -7(x \: + \: 1)}}$

$\large{\sf{0 \: = \: (x \: + \: 1)(x \: - \: 7)}}$

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$\large{\sf{0 \: = \: x \: + \: 1}}$

$\huge{\implies}{\boxed{\blue{\sf{x \: = \: -1}}}}$

OR

$\large{\sf{0 \: = \: x \: - \: 7}}$

$\huge{\implies}{\boxed{\blue{\sf{x \: = \: 7}}}}$