More practice questions

5. Q5. Find the distance between the parallel lines 3x − 4y + 7 = 0 and 6x − 8y − 1 = 0.

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   Divide the second by 2 so that A and B match: 3x − 4y − 1/2 = 0. Then d = |7 + 1/2|/5 = 3/2.

6. Q6. Find the foot of the perpendicular from the origin to 3x + 4y = 10, and the image of the origin in this line.

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   The line is 3x + 4y − 10 = 0, and at the origin 3x + 4y − 10 is −10. For the foot, h/3 = k/4 = 10/25 = 2/5, so the foot is (6/5, 8/5). Check: its distance from O is √(36 + 64)/5 = 2 = |−10|/5. The image is twice as far along: (12/5, 16/5).

7. Q7. Find c so that the line 3x − 4y + c = 0 is at distance 2 from (1, 1).

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   |3 − 4 + c|/5 = 2, so c − 1 = 10 or c − 1 = −10. So c = 11 or c = −9: two lines, one on each side of the point.

8. Q8. If p and q are the lengths of the perpendiculars from the origin to x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, with sin θ cos θ ≠ 0, prove that p² + 4q² = k² (NCERT Exercise 9.3, Q15).

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   For the first line, A² + B² = cos² θ + sin² θ = 1, so p = |k cos 2θ|. Multiplying the second by sin θ cos θ gives x sin θ + y cos θ = k sin θ cos θ, so

   q = |k sin θ cos θ| = |k sin 2θ|/2.

   So p² + 4q² is

   k²(cos² 2θ + sin² 2θ) = k².

9. Q9. Find the perpendicular distance from the origin to the line through the distinct points (cos θ, sin θ) and (cos φ, sin φ), so φ − θ is not an integer multiple of 2π (NCERT Miscellaneous Exercise, Q4).

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   By the two-point form, using sin φ cos θ − cos φ sin θ = sin(φ − θ), the line is (sin φ − sin θ)x − (cos φ − cos θ)y + sin(θ − φ) = 0. Here

   A² + B² = 2 − 2 cos(φ − θ)

   = 4 sin²((φ − θ)/2).

   So d = |sin(θ − φ)|/(2|sin((φ − θ)/2)|) = |cos((φ − θ)/2)|, using sin 2u = 2 sin u cos u. Both points are on the unit circle, so this is the distance of the chord from the centre.

10. Q10. For real non-zero a, b with a² ≥ b², prove that the product of the lengths of the perpendiculars from (√(a² − b²), 0) and (−√(a² − b²), 0) to the line (x/a) cos θ + (y/b) sin θ = 1 is b² (NCERT Miscellaneous Exercise, Q22).

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    Write the line as bx cos θ + ay sin θ − ab = 0 and let e = √(a² − b²). The product of the two distances is

    |be cos θ − ab|

    × |−be cos θ − ab|

    ÷ (b² cos² θ + a² sin² θ).

    The top is b²|a² − e² cos² θ| = b²(a² sin² θ + b² cos² θ), using e² = a² − b². It cancels with the bottom and leaves b².