Question

3) Find the values of
for which the distance
between the points P(x, 4) and Q (9, 10) is 10 units​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:- the coordinates of points P(x,4) and Q(9,10) and distance between them is 10 units.

To find:- The value of x.

Solution :-

We know that distance between them is 10 units.

By using distance formula,

$\sf \: pq = \sqrt{( x_{2} - x_{2}) {}^{2} + ( y_{2} - y_{1}) {}^{2} }$

$\sf10 = \sqrt{(9 - x) {}^{2} + (10 - 4) {}^{2} }$

$\sf \: 10 = \sqrt{(9) {}^{2} + (x) {}^{2} - 2(9)(x) + (6) {}^{2} }$

$\sf \: 10 = \sqrt{81 + x {}^{2} - 18x + 36 }$

$\sf \: (10) {}^{2} = ( \sqrt{x {}^{2} - 18x + 117 }) {}^{2} ...( \because \: squaring \: both \: sides)$

$\sf \: 100 = x {}^{2} - 18x + 117$

$\sf \: x {}^{2} - 18 x + 17 = 0$

$\sf \: x {}^{2} - 17x - 1x + 17 = 0$

$\sf \: x(x - 17) - 1(x - 17) = 0$

$\sf \: (x - 17)(x - 1) = 0$

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Hence, value of x seems to be 17 or 1.