Question

distance formula (-1 -3) (-3,x) distance 5 find the value of x

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Answer 1

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Answer 2

Given: Distance between the points (-1, -3) and (-3, x) is 5 units.

To find: The value of x.

Answer:

Distance formula:

$\tt \sqrt{\bigg(x_2\ -\ x_1\bigg)^2\ +\ \bigg(y_2\ -\ y_1\bigg)^2}$

From the given data, we have:

$\tt Distance\ =\ 5\ units\\\\x_1\ =\ -1\\\\x_2\ =\ -3\\\\y_1\ =\ -3\\\\y_2\ =\ x$

Using them in the formula,

$\tt 5\ =\ \sqrt{\bigg(-3\ +\ 1\bigg)^2\ +\ \bigg(x\ +\ 3\bigg)^2}\\\\\\5\ =\ \sqrt{\bigg(-2\bigg)^2\ +\ x^2\ +\ 6x\ +\ 9}\\\\\\5\ =\ \sqrt{\bigg(4\ +\ x^2\ +\ 6x\ +\ 9\bigg)$

Squaring both sides,

$\tt 25\ =\ 4\ +\ x^2\ +\ 6x\ +\ 9\\\\\\25\ =\ 13\ +\ x^2\ +\ 6x\\\\\\25\ -\ 13\ =\ x^2\ +\ 6x\\\\\\12\ =\ x^2\ +\ 6x\\\\\\x^2\ +\ 6x\ -\ 12$

Using the quadratic formula to obtain the value of x,

$\tt x\ =\ \bigg[\dfrac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}\bigg]$

From the equation we have,

• a = 1
• b = 6
• c = -12

Using them in the formula,

$\tt x\ =\ \dfrac{-6\ \pm\ \sqrt{(6)^2\ +\ a\ \times\ 1\ \times\ -12}}{2\ \times\ 1}\\\\\\x\ =\ \dfrac{-6\ \pm\ \sqrt{36\ +\ 48}}{2}\\\\\\x\ =\ \dfrac{-6\ \pm\ \sqrt{84}}{2}\\\\\\x\ =\ \dfrac{-6\ \pm\ 2\sqrt{21}}{2}\\\\\\x\ =\ \dfrac{2(-3\ \pm\ \sqrt{21})}{2}\\\\\\x\ =\ -3\ \pm\ \sqrt{21}\\\\\\\bf x\ =\ -3\ +\ \sqrt{21}\ \tt or\ \bf x\ =\ -3\ -\ \sqrt{21}$