Question

Find a
if the distance between the points P (11, -2) and Q (a, 1) is 5 units​

Answer 1

Answer:

$a = 15 \: and \: a = 7$

Step-by-step explanation:

$P=(11,-2), Q=(a,1) \\ x_{1} = 11 \: x_{2} = a \: y_{1} = ( - 2) \: y_{2} = 1 \\Distance \: between \: PQ = 5 \\ \sqrt{ {(x_{2} - x_{1}) }^{2} + {( y_{2} - y_{1})}^{2} } = 5 \\ \sqrt{ {(11 - a)}^{2} + {(1 + 2)}^{2} } = 5 \\ squaring \: on \: both \: the \: sides \\ {( \sqrt{ {(11 - a)}^{2} + {(3)}^{2} }) }^{2} = {(5)}^{2} \\ {(11 - a)}^{2} + 9 = 25 \\ 121 + {a}^{2} - 22a + 9 = 25 \\ {a}^{2} - 22a = 25 - 9 - 121 \\ {a}^{2} - 22a = ( - 105) \\ {a}^{2} - 22a + 105 = 0 \\ {a}^{2} - 15a - 7a + 105 = 0 \\ ( {a}^{2} - 15a) - (7a + 105) = 0 \\ a(a - 15) - 7(a - 15) = 0 \\ (a - 15)(a - 7) = 0 \\ (a - 15 = 0)(a - 7 = 0) \\ a = 15 \: and \: a = 7$