Question

The value of a, if the points (3,4) and (7,1) are eqidistant from the point (a,0)​

Answer 1

Step-by-step explanation:

given

point

A=(3,4)

O=(7,1)

now according to formula

AO=

$ao = \sqrt{(3 - 7) {}^{2} + (4 - 1) {}^{2} } \\ = \sqrt{( - 4) {}^{2} + (3) {}^{2} } \\ = \sqrt{16 + 9} \\ = \sqrt{25} \\ = 5units$

I hope it will be helpful to you dear

Answer 2

Answer:

$a = \frac{25}{8}$

Step-by-step explanation:

Let, p=(3,4) and r=(7,1) and q=(a,0)

$d(pq) = \sqrt{ {(3 - a)}^{2} + {(4 - 0)}^{2} } \\ d(pq) = \sqrt{ {(3 - a)}^{2} + 16 } \\ \\ d(qr) = \sqrt{ {(a - 7)}^{2} + {(0 - 1)}^{2} } \\ d(qr) = \sqrt{ {(a - 7)}^{2} + 1 } \\ \\ by \: given \: condition \\ d(pq) = d(qr) \\ \sqrt{ {(3 - a)}^{2} + 16} = \sqrt{ {(a - 7)}^{2} + 1 } \\ squaring \: both \: side \\ {(3 - a)}^{2} + 16 = {(a - 7)}^{2} + 1 \\ {a}^{2} - 6a + 9 + 16 = {a}^{2} - 14a + 49 + 1 \\ {a }^{2} - {a}^{2} - 6a + 14a = 50 - 25 \\ 8a = 25 \\ a = \frac{25}{8}$

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