## problem statement

Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?

$A\left(\frac{1}{4} ; \frac{1}{3} ; 1\right)$

a: $4 x-3 y+5 z-10=0$

$k=\frac{1}{2}$
## Solution

When transforming similarity with the center at the origin of the coordinate plane, the plane $a: A x + B y + C z + D = 0$ and the coefficient $k$ transitions to the plane $a^{\prime}: A x + B y + C z + k \cdot D = 0$. We find the image of the plane $a$:

$a^{\prime}: 4 x - 3 y + 5 z - 5 = 0$

Substitute the coordinates of point $A$ into the equation of $a^{\prime}$:

$4 \cdot \frac{1}{4} - 3 \cdot \frac{1}{3} + 5 \cdot 1 - 5 = 0$

$1 - 1 + 5 - 5 = 0$

$0 = 0$

Since $0 = 0$, point $A$ belongs to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry 12-12