Question

Find the value of x, if the distance between the
points (x,-1) and (3, 2) is 5 units.​

Answer 1

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As per given

$= > = 5 \\ = > \sqrt{( x_{1} - x_{2}) {}^{2} + (y_{1} - y_{2}) {}^{2} } = 5 \\ = > \sqrt{(x - 3) {}^{2} + ( - 1 - 2) {}^{2} } = 5 \\ = > \sqrt{ {x}^{2} + 9 - 6x + ( - 3) {}^{2} } = 5 \\ = > \sqrt{x {}^{2} - 6x + 18 } = 5 \\ \: \: \: \: \: \: \: \: \: squaring \: both \: sides \\ = > {x}^{2} - 6x + 18 = 25 \\ = > {x}^{2} - 6x - 7 = 0 \\ = > {x}^{2} - 7x + 1x - 7 = 0 \\ = > x(x -7 ) + 1(x - 7) = 0 \\ = > (x + 1)(x - 7) \\ so \\ \\ x = 7 \: and \: - 1$

Answer 2

ANSWER:-

Given:

If the distance between the points (x,-1) & (3,2) is 5 units.

To find:

find the value of x.

Solution:

According to the distance formula,

⏺️x1= x & x2= 3

⏺️y1= -1 & y2= 2

$d = \sqrt{(x2 - x1) {}^{2} + {(y2 - y1)}^{2} } \\ \\ = > d = 5 = \sqrt{(x - 3) {}^{2} + ( - 1 - 2) {}^{2} } \\ \\Squaring \: both \: sides \\ \\ = > d = 25 = (x - 3) {}^{2} + ( - 3) {}^{2} \\ \\ = > d = 25 = (x - 3) {}^{2} + 9 \\ \\ = > d = 25 - 9 = (x - 3) {}^{2} \\ \\ = > d = 16 = (x - 3) {}^{2} \\ \\ = > {4}^{2} = (x - 3) {}^{2} \\ \\ = > ±4 = x - 3 \\ \\ = > x = 4 + 3 \\ \\ = > x = 7\\ \\ And, \\ x= -4+3 \\ \\=> x=-1$

Hope it helps ☺️