Straight Lines
Image of a Point in a Line
Grade 11
Question:
<p>Let <i>L</i> denote the line in the <i>xy</i>-plane with <i>x</i> and <i>y</i> intercepts as 3 and 1 respectively. Then, the image of the point (−1, −4) in this line is</p>
<p>(a) $\left(\frac{11}{5}, \frac{28}{5}\right)$</p>
<p>(b) $\left(\frac{29}{5}, \frac{8}{5}\right)$</p>
<p>(c) $\left(\frac{8}{5}, \frac{29}{5}\right)$</p>
<p>(d) $\left(\frac{29}{5}, \frac{11}{5}\right)$</p>
Step-by-Step Solution
Key Concept: To find the image of a point with respect to a line, use the reflection formula: move twice the perpendicular distance from the point to the line along the normal direction.
<p><strong>Step 1:</strong> Find the equation of line <i>L</i>. With <i>x</i>-intercept 3 and <i>y</i>-intercept 1:</p><p>$$\frac{x}{3} + \frac{y}{1} = 1$$</p><p>$$x + 3y = 3 \text{ ... (i)}$$</p><p><strong>Step 2:</strong> To find the image of point (−1, −4) with respect to line (i), use the reflection formula. The perpendicular distance from (−1, −4) to the line is:</p><p>$$d = \frac{|(-1) + 3(-4) - 3|}{\sqrt{1^2 + 3^2}} = \frac{|-1 - 12 - 3|}{\sqrt{10}} = \frac{16}{\sqrt{10}}$$</p><p><strong>Step 3:</strong> The image point $(x_1, y_1)$ is obtained by moving twice this distance along the perpendicular direction to the line.</p><p>$$x_1 + 1 = \frac{-2(1)}{1 + 9} \times 16 = \frac{-32}{10}$$</p><p>$$y_1 + 4 = \frac{-2(3)}{1 + 9} \times 16 = \frac{-96}{10}$$</p><p><strong>Step 4:</strong> Solving:</p><p>$$x_1 = 1 - 3.2 = \frac{11}{5}$$</p><p>$$y_1 = 4 - 9.6 = \frac{28}{5}$$</p><p>∴ Answer is A.</p>
Correct Answer: A