Question

find a point on x axis which is equidistant from the points (7, 6 )and (3,4)
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Answer 1

Answer:

(15/2,0)

Step-by-step explanation:

let the point be (x,0)

then

(x-7)²+6²=(x-3)²+4²

8x=36+49-9-16

8x=60

x=15/2

Answer 2

Given:-

• A point on x axis is equidistant from the points (7, 6 )and (3,4)

To find:-

• Find the point on x axis..?

Solutions:-

• Let (a, 0) be the point on the x axis that is equidistant from the point (7, 6) and (3, 4).

According to the questions,

=> √(7 - a)² + (6 - 0)² = √(3 - a)² + (4 - 0)²

=> √49 + a² - 14a + 36 = √9 + a² - 6a + 16

=> √a² - 14a + 85 = √a² - 6a + 25

On squaring both sides, we obtain.

=> a² - 14a + 85 = a² - 6a + 25

=> a² - a² - 14a + 6a = 25 - 85

=> -8a = -60

=> a = -60/-8

=> a = 15/2

Hence, the required point on the x - axis is (15/2 , 0)