Question 50 of 70intermediate🔧 ApplyShort Answer4 marks

What are the points on the yy-axis whose distance from the line x3+y4=1\frac{x}{3} + \frac{y}{4} = 1 is 4 units.

Correct Answer

The points on the y-axis are (0, 32/3) and (0, -8/3).

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 3 | (Page 22)
For More Understanding

Explanation

This question tests the concept of perpendicular distance from a point to a line. Points on the y-axis have x-coordinate equal to 0, so we represent them as (0, k). The perpendicular distance formula from a point (x₁, y₁) to a line ax + by + c = 0 is d = |ax₁ + by₁ + c|/a2+b2\sqrt{a² + b²}. After converting the given line to standard form and applying this formula, we get two possible values for k, giving us two points on the y-axis that are 4 units away from the given line.

Solution Steps

  1. Step 1: Convert the line equation to standard form. Given line: x/3 + y/4 = 1. Multiplying by 12: 4x + 3y = 12, or 4x + 3y - 12 = 0.

  2. Step 2: Let the point on y-axis be (0, k) since all points on y-axis have x-coordinate = 0.

  3. Step 3: Apply the perpendicular distance formula. Distance from (0, k) to line 4x + 3y - 12 = 0 is d = |4(0) + 3k - 12|/42+32\sqrt{4² + 3²} = |3k - 12|/5.

  4. Step 4: Set distance equal to 4 units. |3k - 12|/5 = 4, which gives |3k - 12| = 20.

  5. Step 5: Solve for k. Case 1: 3k - 12 = 20 → 3k = 32 → k = 32/3. Case 2: 3k - 12 = -20 → 3k = -8 → k = -8/3.

  6. Step 6: The required points are (0, 32/3) and (0, -8/3).

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