Straight Lines
Intercepts and Distance
Grade 11
Question:
<p>A line L has intercepts a and b on the coordinate axes. Keeping the origin fixed, the axes are rotated through a fixed angle. Now, the same line has intercepts p and q on the new axes. Then which relation holds?</p>
<p>(a) <p>a^2 + p^2 = b^2 + q^2</p></p>
<p>(b) <p>a^2 + b^2 = p^2 + q^2</p></p>
<p>(c) <p>\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} - \frac{1}{q^2}</p></p>
<p>(d) <p>\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}</p></p>
Step-by-Step Solution
Key Concept: The perpendicular distance from origin to a line is invariant under rotation of coordinate axes.
<p><strong>Solution:</strong> The line with intercepts a and b has equation: <p>\frac{x}{a} + \frac{y}{b} = 1</p></p><p>The perpendicular distance from origin to this line is: <p>d = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}</p></p><p>After rotation, the same line has intercepts p and q, so the perpendicular distance remains: <p>d = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}}</p></p><p>Since the perpendicular distance is invariant under rotation of axes:</p><p><p>\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}</p></p>
Correct Answer: D