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Straight Lines and Pair of Straight Lines question

2019 · 8 Apr · Shift 1 · Q36
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Straight Lines and Pair of Straight Lines question

2019 · 8 Apr · Shift 1 · Q36

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only in :
  1. A
    1st and 2nd qudratants
  2. B
    4th qudratant
  3. C
    1st and 2nd and 4th qudratants
  4. D
    1st qudratant
View written solutionFree

Correct answer: A

  1. Let the point on the line be (x,y)(x,y)(x,y).

    The line is: 3x+5y=153x+5y=153x+5y=15

  2. A point equidistant from the coordinate axes means:

    • distance from the xxx-axis =∣y∣=|y|=∣y∣
    • distance from the yyy-axis =∣x∣=|x|=∣x∣

    So, ∣x∣=∣y∣|x|=|y|∣x∣=∣y∣ which gives: y=xory=−xy=x \quad \text{or} \quad y=-xy=xory=−x

  3. Check both cases with the given line.

    Case 1: y=xy=xy=x

    Substitute in 3x+5y=153x+5y=153x+5y=15: 3x+5x=153x+5x=153x+5x=15 8x=158x=158x=15 x=158x=\frac{15}{8}x=815​ Hence, y=158y=\frac{15}{8}y=815​ So the point is: (158,158)\left(\frac{15}{8},\frac{15}{8}\right)(815​,815​) This lies in the 1st quadrant.

    Case 2: y=−xy=-xy=−x

    Substitute in 3x+5y=153x+5y=153x+5y=15: 3x+5(−x)=153x+5(-x)=153x+5(−x)=15 3x−5x=153x-5x=153x−5x=15 −2x=15-2x=15−2x=15 x=−152x=-\frac{15}{2}x=−215​ Hence, y=152y=\frac{15}{2}y=215​ So the point is: (−152,152)\left(-\frac{15}{2},\frac{15}{2}\right)(−215​,215​) This lies in the 2nd quadrant.

  4. Therefore, such points lie only in the: 1st and 2nd quadrants\text{1st and 2nd quadrants}1st and 2nd quadrants

  5. Comparing with options:

    • A: 1st and 2nd quadrants ✅
    • B: 4th quadrant ❌
    • C: 1st and 2nd and 4th quadrants ❌
    • D: 1st quadrant ❌

Hence, the correct option is A.

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