Q5. If the lines y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, show that m1(c2 − c3) + m2(c3 − c1) + m3(c1 − c2) = 0 (NCERT Miscellaneous Exercise, Q9).
For three distinct concurrent lines, m1 ≠ m2. The first two lines meet where m1x + c1 = m2x + c2:
x = (c2 − c1)/(m1 − m2),
y = (m1c2 − m2c1)/(m1 − m2).
Putting this point into the third equation and multiplying by m1 − m2:
m1c2 − m2c1
= m3(c2 − c1) + c3(m1 − m2).
Collecting the terms of each m on one side gives the result.
Q6. Show that the area of the triangle formed by the lines y = m1x + c1, y = m2x + c2 and x = 0, with m1 ≠ m2 and c1 ≠ c2, is (c1 − c2)2/(2|m1 − m2|) (NCERT Miscellaneous Example 14).
Two vertices are on the y-axis, at (0, c1) and (0, c2), so the base is |c1 − c2|. The third vertex is where the two lines meet, at x = (c2 − c1)/(m1 − m2), so the height is |c1 − c2|/|m1 − m2|. Half the base times the height gives the result.
Q7. A line is such that its segment between the lines 5x − y + 4 = 0 and 3x + 4y − 4 = 0 is bisected at (1, 5). Find its equation (NCERT Miscellaneous Example 15).
Let the line meet the first at (α1, 5α1 + 4) and the second at (α2, (4 − 3α2)/4). The mid-point is (1, 5), so
α1 + α2 = 2,
20α1 − 3α2 = 20.
So α1 = 26/23 and the first point is (26/23, 222/23). The line through it and (1, 5) is 107x − 3y − 92 = 0.
Q8. Find the distance of the line 4x + 7y + 5 = 0 from (1, 2) measured along the line 2x − y = 0 (NCERT Miscellaneous Exercise, Q14).
(1, 2) lies on 2x − y = 0. Putting y = 2x into 4x + 7y + 5 = 0 gives 18x = −5, so the lines meet at (−5/18, −5/9). The distance is
√((23/18)2 + (46/18)2)
= 23√5/18.
Q9. In which direction must a straight line be drawn through (−1, 2) so that its intersection with x + y = 4 is 3 units from (−1, 2) (NCERT Miscellaneous Exercise, Q15)?
A point 3 units away in the direction θ is (−1 + 3 cos θ, 2 + 3 sin θ). It is on x + y = 4 when cos θ + sin θ = 1, so θ = 0° or 90° (by the sum formula, √2 sin(θ + 45°) = 1). The line is parallel to the x-axis or to the y-axis.
Q10. The hypotenuse of a right triangle has its ends at (1, 3) and (−4, 1). Find equations of the legs if they are parallel to the axes (NCERT Miscellaneous Exercise, Q16).
The right-angle vertex is (1, 1) or (−4, 3). So the legs are x = 1 and y = 1, or x = −4 and y = 3.
Q11. A ray of light through (1, 2) reflects on the x-axis at A and the reflected ray passes through (5, 3). Find A (NCERT Miscellaneous Exercise, Q21).
Reflecting the ray from A to (5, 3) in the x-axis gives the ray from A to (5, −3), so the incident ray, extended beyond A, passes through (5, −3). So A is where the line from (1, 2) to (5, −3) meets the x-axis. Its slope is −5/4, and y = 0 gives x − 1 = 8/5. So A = (13/5, 0).
Q12. A(1, 2) and B(3, 5) are two vertices of a triangle whose centroid is (2, 2). Find the third vertex.
(1 + 3 + x)/3 = 2 and (2 + 5 + y)/3 = 2, so the third vertex is (2, −1).
Q13. Find the circumcentre of the triangle with vertices (1, 1), (5, 1) and (1, 7), and check that it is equally far from the three vertices.
The right angle is at (1, 1), so the circumcentre is the mid-point of the hypotenuse from (5, 1) to (1, 7): (3, 4). Each vertex is √13 from it, since 4 + 9 = 13.