## 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 $a$?

$A(2; -5; 4)$

$a: 5x + 2y - z + 3 = 0$

$k = \frac{4}{3}$
## 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 $a$:

$a^{\prime}: 5 x+2 y-z+4=0$

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

$5 \cdot 2+2 \cdot(-5)-4+4=0$

$10-10-4+4=0$

$0=0$

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

## Problem Kuznetsov Analytical Geometry 12-9