Question

Distance between two points (8,-4) and (0, a) is 10. All the values are in the same unit of length. Find the positive value of a.​

Answer 1

Answer:

$\boxed{a = 2}$

Explanation:

Let P(8,-4) and Q(0,a) be two points distance PQ is 10.

According to distance formula

$\boxed{PQ = \sqrt{ {(x_{2} - x_{1})}^{2} + ( {y_{2} - y_{1})}^{2} } } \\ = > PQ = \sqrt{ {(0 - 8)}^{2} + {(a + 4)}^{2} } \\ = > PQ = \sqrt{64 + ( {a }^{2} + 16 + 8a) } \\ = > PQ = \sqrt{80 + {a}^{2} + 8a}$

And we have value of PQ so

On putting the value of PQ

$= > 10 = \sqrt{80 + {a}^{2} + 8a} \\ = > {a}^{2} + 8a + 80 = 100 \\ = > {a}^{2} + 8a = 100 - 80 \\ = > {a}^{2} + 8a = 20 \\ = > {a}^{2} + 10a - 2a - 20 = 0 \\ = > (a - 2)(a + 10) = 0 \\ \bold{when} \\ = > a - 2 = 0 \\ = > \boxed{a = 2} \\ \bold{when} \\ = > a + 10 = 0 \\ = > \boxed{a = - 10}$

Answer 2

$\large \underline{ \underline{ \sf \: Answer : \: \: \: }}$

$\star$ a = 2

$\large \underline{ \underline{ \sf \: Explaination : \: \: \: }}$

Given ,

$\star \sf \: \: Distance = 10 \\</p><p> \star \: \: \sf X_{1} = 8 \\ \sf \star \: \: X_{2} = 0 \\ \star \: \: \sf Y_{1} = - 4 \\ \star \: \: \sf Y_{2} = a$

We Know that ,

$\large \fbox{ \fbox{ \sf \: \: Distance = \sf \sqrt{ {( X_{2} - X_{1} )}^{2} + {(Y_{2} - Y_{1} )}^{2} \: } \: \: \: \: }}$

$\sf \to 10 = \sqrt{ {(0 - 8)}^{2} + { \bigg(a - ( - 4) \bigg)}^{2} } \\ \\ \sf \to 10 = \sqrt{(64) + {(a + 4)}^{2} } \\ \\ \sf \to {(10 )}^{2} = { \bigg(\sqrt{(64) + {(a + 4)}^{2} } \: \bigg)}^{2} \\ \\ \sf \to 100 = 64 + {(a + 4)}^{2} \\ \\ \sf \to 36 = {a}^{2} + {(2)}^{2} + 2 \times a \times 4 \\ \\ \sf \to 36 = {a}^{2} + 16 + 8a \\ \\ \sf \to {a}^{2} + 8a - 20 = 0 \\ \\ \sf \to {a}^{2} - 2a + 10a - 20 = 0 \\ \\ \sf \to a(a - 2) + 10(a - 2) = 0 \\ \\ \sf \to (a + 10)(a - 2) = 0 \\ \\ \sf \to a = - 10 \: \: or \: \: a = 2$

Thus , the required positive value of a is 2