Question

Find the valur of a , if the distance between the points A(-3, -14) and B(a, -5) is 9 units​

Answer 1

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Q1) Find the value of a, if the distance between the point A(-3,-14) and B(a,-5) is 9units.

$\: \: \blue \bigstar{ \bf{ \: by \: using \: formula \: method}}$

$\: \\ \implies \displaystyle \sf{ab = \sqrt{ ({ - 5 - ( - 14)})^{2} + {(a - ( - 3)}^{2} ) } = 9}$

$\: \\ \implies \displaystyle \sf{ab = \sqrt{ ({ - 5 + 14})^{2} + {(a + 3)}^{2} } = 9}$

$\ \\ : \mapsto \displaystyle \sf{ab = ({ - 5 + 14})^{2} + {a + 3}^{2} = 81}$

$\: \mapsto \displaystyle \sf{ab = {(9)}^{2} + {a}^{2} + 9 + 6a = 81}$

$\: \mapsto \displaystyle \sf{81 = 81+ {a }^{2} + 6a + 9}$

$\: \mapsto \displaystyle \sf{81 = 90 + {a}^{2} + 6a}$

$\: \mapsto \displaystyle \sf{81 - 90 - {a}^{2} - 6a} = 0$

$\: \mapsto \displaystyle \sf{ - 1 - {a}^{2} - 6a = 0}$

$\: \: \mapsto \displaystyle \sf{ - {a}^{2} - 6a - 1 = 0}$

$\: \mapsto \displaystyle \sf{ {a}^{2} + 6a + 1 = 0}$

$\: \: \: \bigstar \bf{by \: using \: formula \:method}$

$\: \: \implies \displaystyle \sf{a = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\: \: \\ \implies \displaystyle \sf{a = \frac{ - 6 \pm \sqrt{ {(6)}^{2} - 4(1)(1) } }{2 \times 1} }$

$\: \: \implies \displaystyle \sf{a = \frac{ - 6 \pm \sqrt{36 - 4} }{2} }$

$\: \: \implies \displaystyle \sf{a = \frac{ - 6 \pm \sqrt{32} }{2} }$

$\: \: \implies \displaystyle \sf{a = \frac{ - 6 \pm \sqrt{16 \times 2} }{2} }$

$\: \: \implies \displaystyle \sf{a = \frac{ - 6 \pm4 \sqrt{2} }{2} }$

$\: \: \red \bigstar\underline{\boxed{ \displaystyle \sf{a = \frac{ - 6 + \sqrt{2} }{2} \: and \: a = \frac{ - 6 - \sqrt{2} }{2} \: is \: the \: answer }}}$