<p>A point on the straight line, \(3x + 5y = 15\) which is equidistant from the coordinate axes will lie only in</p>
<p>4th quadrant.</p>
<p>1st quadrant.</p>
<p>1st and 2nd quadrants.</p>
<p>1st, 2nd and 4th quadrants.</p>
Step-by-Step Solution
Key Concept: A point equidistant from both coordinate axes has |x| = |y|. Substitute this constraint into the line equation to find which quadrant(s) contain such points.
<p><strong>Step 1:</strong> A point equidistant from both coordinate axes satisfies |x| = |y|, which gives two cases: <strong>x = y</strong> or <strong>x = −y</strong>.</p><p><strong>Step 2:</strong> <strong>Case 1 (x = y):</strong> Substitute into 3x + 5y = 15:<br>3x + 5x = 15 ⟹ 8x = 15 ⟹ x = 15/8<br>Point: (15/8, 15/8) — lies in <strong>Quadrant I</strong> (both x > 0, y > 0)</p><p><strong>Step 3:</strong> <strong>Case 2 (x = −y):</strong> Substitute into 3x + 5y = 15:<br>3x + 5(−x) = 15 ⟹ 3x − 5x = 15 ⟹ −2x = 15 ⟹ x = −15/2<br>Point: (−15/2, 15/2) — lies in <strong>Quadrant II</strong> (x < 0, y > 0)</p><p><strong>Step 4:</strong> Verify that the line 3x + 5y = 15 passes through Quadrants I, II, and III only (y-intercept is 3, x-intercept is 5, both positive; negative x and positive y in Q2; negative x and negative y in Q3). The equidistant points lie only in <strong>Quadrants I and II</strong>.</p><p>∴ Answer: D (Quadrants I and II only)</p>
Correct Answer: D