Find a point on the x-axis,which is equidista | vedclass

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(A) Let $(a, 0)$ be the point on the $x$-axis that is equidistant from the points $P(7, 6)$ and $Q(3, 4)$.According to the distance formula,the distan

Vedclass · Accepted Answer

$(15/2, 0)$

Vedclass · Answer

(A) Let $(a, 0)$ be the point on the $x$-axis that is equidistant from the points $P(7, 6)$ and $Q(3, 4)$.According to the distance formula,the distance from $(a, 0)$ to $(7, 6)$ is equal to the distance from $(a, 0)$ to $(3, 4)$.$\sqrt{(7-a)^2 + (6-0)^2} = \sqrt{(3-a)^2 + (4-0)^2}$Squaring both sides:$(7-a)^2 + 36 = (3-a)^2 + 16$$(49 - 14a + a^2) + 36 = (9 - 6a + a^2) + 16$$85 - 14a = 25 - 6a$$85 - 25 = 14a - 6a$$60 = 8a$$a = \frac{60}{8} = \frac{15}{2}$Thus,the required point on the $x$-axis is $(\frac{15}{2}, 0)$.

Find a point on the $x$-axis,which is equidistant from the points $(7, 6)$ and $(3, 4)$.

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