More practice questions

5. Q5. Find the slopes of the lines with inclinations 0°, 45°, 120° and 150°.

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   The slopes are 0, 1, −√3 and −1/√3, the values of tan 0°, tan 45°, tan 120° and tan 150°.

6. Q6. Is the line through (1, 2) and (3, 6) parallel to the line through (−1, 0) and (2, 6)?

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   Both slopes are 2:

   (6 − 2)/(3 − 1) = 2 and (6 − 0)/(2 + 1) = 2.

   The point (−1, 0) is not on the first line, because the slope from (1, 2) to (−1, 0) is

   (0 − 2)/(−1 − 1) = 1, not 2.

   So the lines are distinct and parallel.

7. Q7. The line through (2, a) and (5, 7) is parallel to the line through (−1, 3) and (1, 7). Find a.

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   The second slope is (7 − 3)/(1 + 1) = 2, so

   (7 − a)/(5 − 2) = 2

   7 − a = 6, so a = 1.

8. Q8. A(−2, 3), B(1, 1) and C(3, k) are such that AB is perpendicular to BC. Find k.

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   Slope of AB = (1 − 3)/(1 + 2) = −2/3, so the slope of BC must be 3/2:

   (k − 1)/(3 − 1) = 3/2

   k − 1 = 3, so k = 4.

9. Q9. Is the triangle with vertices A(1, 1), B(4, 3) and C(2, 6) right-angled? If so, where is the right angle?

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   Slope of AB = 2/3 and slope of BC = (6 − 3)/(2 − 4) = −3/2. Their product is −1, so AB is perpendicular to BC and the right angle is at B(4, 3).

10. Q10. Find the acute angle between the vertical line x = 2 and a line of slope √3.

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    The formula cannot be used, because x = 2 has no slope. The second line has inclination 60° (tan 60° = √3) and the vertical line 90°, so the angle between them is 30°, the difference 90° − 60°.

11. Q11. Show that the points (a, 0), (0, b) and (1, 1) are collinear if and only if 1/a + 1/b = 1 (a, b ≠ 0).

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    Slope from (a, 0) to (0, b): −b/a.

    Slope from (0, b) to (1, 1): 1 − b.

    They are equal exactly when

    −b = a − ab, that is ab = a + b.

    Dividing by ab gives 1/a + 1/b = 1.

12. Q12. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find m (NCERT Miscellaneous Exercise, Q18).

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    Solved for y, the equations show the slopes 3 and 1/2 as the coefficients of x. Equal inclinations mean

    |(3 − m)/(1 + 3m)|

    = |(1 − 2m)/(2 + m)|,

    after multiplying the top and bottom of (1/2 − m)/(1 + m/2) by 2. Cross-multiplying:

    Same sign:

    6 + m − m² = 1 + m − 6m²

    5m² + 5 = 0, no real root.

    Opposite signs:

    6 + m − m² = −1 − m + 6m²

    7m² − 2m − 7 = 0

    m = (1 ± 5√2)/7.