Question

If the distance between the points (5, - 2) and (1, a) is 5, find the value(s) of a.

Answer 1

Answer:-

Given:

Distance between the points (5 , - 2) ; (1 , a) is 5 units.

We know that,

Distance between two points $\sf{(x_1 , y_1)}$ and $\sf{(x_2 , y_2)}$ = $\sf \large{\sqrt{{(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2}}}$

$\sf \implies \: \sqrt{ {(1 - 5)}^{2} + {(a + 2)}^{2} } = 5$

On squaring both sides we get,

$\sf \implies{( \sqrt{ {( - 4)}^{2} + {(a + 2)}^{2} } })^{2} = {5}^{2} \\ \\ \sf \implies \: 16 + {(a + 2)}^{2} = 25 \\ \\ \sf \implies \: {(a + 2)}^{2} = 25 - 16 \\ \\ \sf \implies \: {(a + 2)}^{2} = 9 \\ \\ \sf \implies \: a + 2 = \sqrt{9} \\ \\ \sf \implies \: a + 2 = \pm 3 \\ \\ \implies \sf \large{ a = 1 \: (or) \: - 5}$

Hence, the value of a is 1 or - 5.

Answer 2

Answer:

Step-by-step explanation:

using distance formula

Distance = 5$\sqrt[n]{(1-5)^{2} + (a+2)^2 } = 5$

$= 16 + a^2 + 4a +4 = 25\\a^2 + 4a - 5 = 0\\ (a+5)(a-1) = 0$

a is either -5 or 1

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