Find the image when
the tollowing ase reflected
in line

Ask by Floyd Ball. in Ghana

Mar 30,2025

Question

Find the image when the tollowing ase reflected in line $$ 2 y+x=0 $$ $$ (-3,4) $$

UpStudy AI · Accepted Answer

We are given the point $$ P=(-3,4) $$ and the line 
$$
2y+x=0.
$$
To find the reflection of $$ P $$ in the line, we follow these steps:

1. **Express the line in standard form:**  
   The line is already written as  
   $$
   x+2y=0,
   $$
   which is in the form $$ ax+by+c=0 $$ with $$ a=1 $$, $$ b=2 $$, and $$ c=0 $$.

2. **Find the foot of the perpendicular $$ D $$ from $$ P $$ to the line:**  
   The formula for the foot $$ D=(x_D,y_D) $$ is  
   $$
   \left(x_0-\frac{ax_0+by_0+c}{a^2+b^2}\,a,\, y_0-\frac{ax_0+by_0+c}{a^2+b^2}\,b\right),
   $$
   where $$ P=(x_0,y_0) $$.

For $$ P=(-3,4) $$:
   - Compute the numerator:
     $$
     ax_0+by_0+c = 1\cdot(-3)+2\cdot 4+0 = -3+8 = 5.
     $$
   - Compute the denominator:
     $$
     a^2+b^2 = 1^2+2^2 = 1+4 = 5.
     $$
   - Thus, the scalar is:
     $$
     \frac{5}{5}=1.
     $$
   - Now, the coordinates of $$ D $$ are:
     $$
     x_D=-3-1\cdot 1 = -4,
     $$
     $$
     y_D=4-1\cdot 2 = 2.
     $$
   So, the foot of the perpendicular is $$ D=(-4,2) $$.

3. **Find the reflection point $$ Q $$:**  
   The reflection $$ Q $$ of $$ P $$ across $$ D $$ is given by
   $$
   Q=2D-P.
   $$
   Plugging in the coordinates:
   $$
   Q = 2(-4,2)-(-3,4) = (-8,4)-(-3,4).
   $$
   Compute coordinates component-wise:
   $$
   Q_x = -8 - (-3)= -8+3 = -5,
   $$
   $$
   Q_y = 4 - 4 = 0.
   $$
   Therefore, the reflection point is $$ Q=(-5,0) $$.

Thus, the image of $$ P=(-3,4) $$ when reflected in the line $$ 2y+x=0 $$ is  
$$
\boxed{(-5,0)}.
$$
The reflection of the point $$ (-3,4) $$ in the line $$ 2y+x=0 $$ is $$ (-5,0) $$.

Find the image when
the tollowing ase reflected
in line

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