## 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 $a$?

$A(-2; -1; 1)$

$a: x-2y+6z-10=0$

$k=\frac{3}{5}$
## Solution

When transforming similarity with the center at the origin of the plane

$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane

$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:

$a^{\prime}: x-2 y+6 z-6=0$

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

$-2-2 \cdot(-1)+6 \cdot 1-6=0$

$-2+2+6-6=0$

$0=0$

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

## Problem Kuznetsov Analytical Geometry 12-15