3. Find the point $T(x, y)$ on the x-axis that is equidistant from points $A(1,4)$ and $B(7,2)$, and accurately calculate the length of the altitude of triangle $A T B$ to side $A B$.


3. Since point $T(x, y)$ is equally distant from points $\mathrm{A}$ and $\mathrm{B}$, we write the equation $d(A, T)=d(B, T)$. The distance between two points is calculated using the formula $d(A, T)=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$. Point T lies on the x-axis, so $T(x, 0)$. We equate $\sqrt{(x-1)^{2}+(0-4)^{2}}=\sqrt{(x-7)^{2}+(0-2)^{2}}$. We square and simplify the equation to get $12 x=36$. From this, we get $x=3$. Therefore, point $\mathrm{T}$ has coordinates $T(3,0)$. Since $\triangle A B T$ is an isosceles triangle, the height bisects the base, and the height is equal to the distance between points $\mathrm{S}$ and $\mathrm{T}$, where $\mathrm{S}$ is the midpoint of side $\mathrm{AB}$. The coordinates of the midpoint are $S\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. Thus, the point $S\left(\frac{1+7}{2}, \frac{4+2}{2}\right)$ is (4,3). The distance $d(S, T)=\sqrt{10}$.

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Squaring the equation $x^{2}-2 x+1+16=x^{2}-14 x+49+4 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$................................

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