Q5. Write the equations of the x-axis and the y-axis (NCERT Exercise 9.2, Q1).
Every point of the x-axis has y = 0, and every point of the y-axis has x = 0. So the x-axis is y = 0 and the y-axis is x = 0.
Q6. Find the equation of the line through (0, 0) with slope m (NCERT Exercise 9.2, Q3).
y − 0 = m(x − 0), that is y = mx.
Q7. Find the equation of the line through (−1, 1) and (2, −4) (NCERT Exercise 9.2, Q7).
The slope is (−4 − 1)/(2 + 1) = −5/3.
y − 1 = (−5/3)(x + 1)
3y − 3 = −5x − 5
So 5x + 3y + 2 = 0.
Q8. Find the equation of the line through (−3, 5) perpendicular to the line through (2, 5) and (−3, 6) (NCERT Exercise 9.2, Q9).
The given line has slope (6 − 5)/(−3 − 2) = −1/5, so the required slope is 5.
y − 5 = 5(x + 3)
So 5x − y + 20 = 0.
Q9. A line perpendicular to the segment joining (1, 0) and (2, 3) divides it internally in the ratio 1 : n, where n > 0. Find the equation of the line (NCERT Exercise 9.2, Q10).
The point of division is ((n + 2)/(n + 1), 3/(n + 1)). The segment has slope 3, so the line has slope −1/3:
y − 3/(n + 1)
= (−1/3)(x − (n + 2)/(n + 1))
Multiplying by 3(n + 1):
(1 + n)x + 3(1 + n)y = n + 11.
Q10. Find the equation of a line that cuts off equal intercepts on the axes and passes through (2, 3) (NCERT Exercise 9.2, Q11).
x/a + y/a = 1 gives x + y = a, and (2, 3) gives a = 5: the line is x + y = 5. If zero intercepts count as equal, the line y = 3x/2 through the origin also works.
Q11. Find the equation of the line through (0, 2) making an angle 2π/3 with the positive x-axis, and of the parallel line that crosses the y-axis 2 units below the origin (NCERT Exercise 9.2, Q13).
The slope is tan 120° = −√3. So the lines are y = −√3x + 2 and y = −√3x − 2, that is √3x + y − 2 = 0 and √3x + y + 2 = 0.
Q12. The length L (in cm) of a copper rod is a linear function of its Celsius temperature C. If L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C (NCERT Exercise 9.2, Q15).
The slope is (125.134 − 124.942)/
L = 124.942
+ (0.192/90)(C − 20).
Q13. P(a, b), with a, b ≠ 0, is the mid-point of a line segment between the axes. Show that the equation of the line is x/a + y/b = 2 (NCERT Exercise 9.2, Q17).
The ends are (2a, 0) and (0, 2b), since P is their mid-point. The intercept form gives x/(2a) + y/(2b) = 1; multiplying by 2 gives x/a + y/b = 2.
Q14. By using the equation of a line, prove that (3, 0), (−2, −2) and (8, 2) are collinear (NCERT Exercise 9.2, Q19).
The line through (3, 0) and (8, 2) has slope 2/5:
y = (2/5)(x − 3),
that is 2x − 5y − 6 = 0.
(−2, −2) satisfies it: −4 + 10 − 6 = 0. So the three points are collinear.
Q15. Reduce to slope-intercept form and find the slope and y-intercept: (i) x + 7y = 0; (iii) y = 0 (NCERT Exercise 9.3, Q1; part (ii) is in the general equation section above).
(i) y = −x/7: slope −1/7, y-intercept 0. (iii) y = 0·x + 0: slope 0, y-intercept 0.
Q16. Reduce to intercept form and find the intercepts: (ii) 4x − 3y = 6; (iii) 3y + 2 = 0 (NCERT Exercise 9.3, Q2; part (i) is above).
(ii) Dividing by 6: x/(3/2) + y/(−2) = 1, so the intercepts are 3/2 and −2. (iii) y = −2/3: the y-intercept is −2/3, and the line is parallel to the x-axis, so it has no x-intercept and no intercept form.
Q17. Find the equation of the line parallel to 3x − 4y + 2 = 0 through (−2, 3) (NCERT Exercise 9.3, Q6).
3(x + 2) − 4(y − 3) = 0, that is 3x − 4y + 18 = 0.
Q18. Find the equation of the line perpendicular to x − 7y + 5 = 0 with x-intercept 3 (NCERT Exercise 9.3, Q7).
The given slope is 1/7, so the required slope is −7. With x-intercept 3: y = −7(x − 3), that is 7x + y − 21 = 0.
Q19. Find the angles between the lines √3x + y = 1 and x + √3y = 1 (NCERT Exercise 9.3, Q8).
The slopes are −√3 and −1/√3. By tan θ = |(m1 − m2)/(1 + m1m2)| (NCERT Section 9.2.3),
tan θ = |(−1/√3 + √3)/(1 + 1)|
= (2/√3)/2 = 1/√3,
so the angles are 30° and 150°.
Q20. The line through (h, 3) and (4, 1) meets 7x − 9y − 19 = 0 at right angles. Find h (NCERT Exercise 9.3, Q9).
The given line has slope 7/9, so the other has slope −9/7:
(1 − 3)/(4 − h) = −9/7
14 = 9(4 − h), so h = 22/9.
Q21. Two lines through (2, 3) meet at an angle of 60°. If one has slope 2, find the equation of the other (NCERT Exercise 9.3, Q11).
By tan θ = |(m1 − m2)/(1 + m1m2)| (NCERT Section 9.2.3), with tan 60° = √3:
|(m − 2)/(1 + 2m)| = √3,
so m − 2 = ±√3(1 + 2m):
m = (2 + √3)/(1 − 2√3) or m = (2 − √3)/(1 + 2√3).
Putting each into y − 3 = m(x − 2) and clearing the denominators:
(√3 + 2)x + (2√3 − 1)y
= 8√3 + 1, or
(√3 − 2)x + (2√3 + 1)y
= 8√3 − 1.
Q22. Find the equation of the right bisector of the segment joining (3, 4) and (−1, 2) (NCERT Exercise 9.3, Q12).
The mid-point is (1, 3) and the segment has slope 1/2, so the bisector has slope −2:
y − 3 = −2(x − 1),
that is 2x + y − 5 = 0.
Q23. Find the equations of the lines whose intercepts on the axes have sum 1 and product −6 (NCERT Miscellaneous Exercise, Q2).
The intercepts are the roots of t2 − t − 6 = 0, that is 3 and −2. So the lines are x/3 + y/(−2) = 1 and x/(−2) + y/3 = 1, that is 2x − 3y − 6 = 0 and 3x − 2y + 6 = 0.
Q24. Find the equation of the line perpendicular to x/4 + y/6 = 1 through the point where it meets the y-axis (NCERT Miscellaneous Exercise, Q6).
The line meets the y-axis at (0, 6), and its slope is −6/4 = −3/2. The perpendicular has slope 2/3:
y = (2/3)x + 6,
that is 2x − 3y + 18 = 0.
Q25. Find the equations of the lines through (3, 2) which make an angle of 45° with the line x − 2y = 3 (NCERT Miscellaneous Exercise, Q10).
The given line has slope 1/2. By the angle formula, |(m − 1/2)/(1 + m/2)| = 1, so m = 3 or m = −1/3 (as in NCERT Example 2). The lines are
y − 2 = 3(x − 3), that is 3x − y − 7 = 0, and
y − 2 = (−1/3)(x − 3), that is x + 3y − 9 = 0.
Q26. Show that the line through the origin making an angle θ with the line y = mx + c is y/x = (m ± tan θ)/(1 ∓ m tan θ) (NCERT Miscellaneous Exercise, Q12).
A line through the origin is y = Mx, with slope M = y/x. It makes the angle θ with slope m when (M − m)/(1 + mM) = tan θ or −tan θ. Solving M − m = ±tan θ (1 + mM) for M:
M(1 ∓ m tan θ) = m ± tan θ,
so y/x = (m ± tan θ)/(1 ∓ m tan θ), for x ≠ 0, defined tan θ and non-zero denominator. Include the origin by writing the result as y = Mx. If 1 ∓ m tan θ = 0, that answer line is vertical, x = 0.
A form that also covers θ = 90° and vertical answers is
(cos θ ∓ m sin θ)y = (m cos θ ± sin θ)x.
For example, m = 1 and θ = 45° give both y = 0 and x = 0. At θ = 90°, the equation reduces to x + my = 0.