If the line, 2x - y + 3 = 0 is at a distance
1/√5 and 2/√5 from the lines 4x - 2y + α = 0
and 6x - 3y + β = 0, respectively, then the sum of all possible values of α and β is :

Solution

Maths sir · 4 steps

First make the lines look identical on the left side, then compare only the constants.

1. Parallel lines. All three lines have slope 2, so they are parallel. For ax + by + c₁ = 0 and ax + by + c₂ = 0, the distance is (|c₁ - c₂|)/√(a² + b²). This works only when the x and y coefficients match exactly.
2. Find alpha. Multiply the given line by 2 to get 4x - 2y + 6 = 0. Then (|α - 6|)/√20 = (|α - 6|)/2√5 = 1/√5, so |α - 6| = 2. This gives α = 8 or α = 4, which add up to 12.
3. Find beta. Multiply the given line by 3 to get 6x - 3y + 9 = 0. Then (|β - 9|)/√45 = (|β - 9|)/3√5 = 2/√5, so |β - 9| = 6. This gives β = 15 or β = 3, which add up to 18.
4. So. The sum of all possible values is (8 + 4) + (15 + 3) = 30.

The common slip: 12. Comparing α and β with 3 before matching the coefficients gives α = 5, 1 and β = 9, -3, a total of 12 instead of 30.

The idea to keep. For the distance between parallel lines, scale one equation until the x and y coefficients match. Only then subtract the constants. The modulus gives two lines, one on each side.

Answer from the NTA final answer key. Solution by PrepWiser.