1. Given a linear function $f(x)$. It is known that the distance between the points of intersection of the graphs $y=$ $x^{2}-1$ and $y=f(x)+1$ is $3 \sqrt{10}$, and the distance between the points of intersection of the graphs $y=x^{2}$ and $y=f(x)+3$ is $3 \sqrt{14}$. Find the distance between the points of intersection of the graphs of the functions $y=x^{2}$ and $y=f(x)$.
Answer: $3 \sqrt{2}$.

Solution. Let $f(x)=a x+b$. Then the abscissas of the points of intersection of the graphs in the first case are determined from the equation $x^{2}-1=a x+b+1$, and in the second case - from the equation $x^{2}=a x+b+3$.

Consider the first case in more detail. The equation has the form $x^{2}-a x-(b+2)=0$, from which $x_{1,2}=$ $\frac{a \pm \sqrt{a^{2}+4 b+8}}{2},\left|x_{2}-x_{1}\right|=\sqrt{a^{2}+4 b+8}$. Since the points of intersection of the graphs lie on a line with a slope of $a$, the distance between the points is $\sqrt{a^{2}+1}$ times greater than $\left|x_{2}-x_{1}\right|$. Therefore, the distance between the points is $\sqrt{\left(a^{2}+1\right)\left(a^{2}+4 b+8\right)}$. Similarly, we find that in the second case, the distance between the points is $\sqrt{\left(a^{2}+1\right)\left(a^{2}+4 b+12\right)}$. From the condition, we obtain the system of equations $\left\{\begin{array}{l}\left(a^{2}+1\right)\left(a^{2}+4 b+8\right)=90, \\ \left(a^{2}+1\right)\left(a^{2}+4 b+12\right)=126,\end{array} \quad\right.$ solving which, we find that $a^{2}=8, b=-\frac{3}{2}$.

Let's find the required distance. The abscissas of the points of intersection are determined by the equation $x^{2}-a x-b=0$, so $\left|x_{2}-x_{1}\right|=\sqrt{a^{2}+4 b}=\sqrt{2}$, and the distance between the points of intersection is

$\left|x_{2}-x_{1}\right| \sqrt{a^{2}+1}=\sqrt{2} \cdot 3=3 \sqrt{2}$.