## Task Condition

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 $(7 ; 0 ;-1)$

$a: x-y-z-1=0$

$k=4$
## 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 $a:$

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

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

$7-0-(-1)-4=0$

$7-0+1-4=0$

$4=0$

Since $4 \neq 0$, point $A$ does not belong to the image of the plane $a$.

## Problem Kuznetsov Analytical Geometry 12-30