Question

The image of the origin with respect to the line $4x + 3y = 25$ is:

• A. $(4, 3)$
• B. $(3, 4)$
• C. $(6, 8)$
• D. $(4, 6)$
• E. $(8, 6)$

Hint:

The midpoint of the line segment connecting the point and its image must lie on the mirror line. For $(0, 0)$ and $(8, 6)$, the midpoint is $(4, 3)$. Plugging $(4, 3)$ into $4x + 3y = 25$ gives $4(4) + 3(3) = 16 + 9 = 25$, confirming the answer.

Answer 1

Answer: (E)

Concept: The image $(x_2, y_2)$ of a point $(x_1, y_1)$ with respect to a line $ax + by + c = 0$ is found using a formula similar to the foot of the perpendicular, but with a factor of 2: \[ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2\frac{(ax_1 + by_1 + c)}{a^2 + b^2} \]

Step 1: Identify the components.
Point $(x_1, y_1) = (0, 0)$. Line $4x + 3y - 25 = 0$ ($a=4, b=3, c=-25$).

Calculate the constant ratio:
\[ R = -2\frac{(4(0) + 3(0) - 25)}{4^2 + 3^2} = -2\frac{-25}{16 + 9} = \frac{50}{25} = 2 \]

Step 2: Solve for $x_2$ and $y_2$.
For $x_2$: \[ \frac{x_2 - 0}{4} = 2 \quad \Rightarrow \quad x_2 = 8 \] For $y_2$: \[ \frac{y_2 - 0}{3} = 2 \quad \Rightarrow \quad y_2 = 6 \]

Step 3: Conclusion.
The image of the origin $(0, 0)$ is $(8, 6)$.