Question:
A point on the straight line \( 3x + 5y = 15 \) which is equidistant from the coordinate axes will lie in
A point on the straight line \( 3x + 5y = 15 \) which is equidistant from the coordinate axes will lie in
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A point equidistant from the coordinate axes has coordinates \( (a, a) \) or \( (a, -a) \). Substitute these forms into the equation of the given line and solve for \( a \). Determine the quadrant in which the resulting points lie based on the signs of their coordinates.
Updated On: May 12, 2025
- either \( 1^{st} \) quadrant or \( 2^{nd} \) quadrant
- \( 4^{th} \) quadrant only
- \( 3^{rd} \) quadrant only
- either in the \( 3^{rd} \) or in the \( 4^{th} \) quadrant
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The Correct Option is A
Solution and Explanation
Let the point on the line \( 3x + 5y = 15 \) be \( (a, b) \).
Since the point is equidistant from the coordinate axes, we have \( |a| = |b| \), which means \( b = a \) or \( b = -a \).
Case 1: \( b = a \) Substitute \( y = x \) into the equation of the line: \( 3x + 5x = 15 \) \( 8x = 15 \) \( x = \frac{15}{8} \) So, \( a = \frac{15}{8} \) and \( b = \frac{15}{8} \).
The point is \( (\frac{15}{8}, \frac{15}{8}) \), which lies in the \( 1^{st} \) quadrant.
Case 2: \( b = -a \) Substitute \( y = -x \) into the equation of the line: \( 3x + 5(-x) = 15 \) \( 3x - 5x = 15 \) \( -2x = 15 \) \( x = -\frac{15}{2} \) So, \( a = -\frac{15}{2} \) and \( b = -(-\frac{15}{2}) = \frac{15}{2} \).
The point is \( (-\frac{15}{2}, \frac{15}{2}) \), which lies in the \( 2^{nd} \) quadrant.
Therefore, the point lies in either the \( 1^{st} \) quadrant or the \( 2^{nd} \) quadrant.
Since the point is equidistant from the coordinate axes, we have \( |a| = |b| \), which means \( b = a \) or \( b = -a \).
Case 1: \( b = a \) Substitute \( y = x \) into the equation of the line: \( 3x + 5x = 15 \) \( 8x = 15 \) \( x = \frac{15}{8} \) So, \( a = \frac{15}{8} \) and \( b = \frac{15}{8} \).
The point is \( (\frac{15}{8}, \frac{15}{8}) \), which lies in the \( 1^{st} \) quadrant.
Case 2: \( b = -a \) Substitute \( y = -x \) into the equation of the line: \( 3x + 5(-x) = 15 \) \( 3x - 5x = 15 \) \( -2x = 15 \) \( x = -\frac{15}{2} \) So, \( a = -\frac{15}{2} \) and \( b = -(-\frac{15}{2}) = \frac{15}{2} \).
The point is \( (-\frac{15}{2}, \frac{15}{2}) \), which lies in the \( 2^{nd} \) quadrant.
Therefore, the point lies in either the \( 1^{st} \) quadrant or the \( 2^{nd} \) quadrant.
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