NCERT Solutions for Class 11 Maths Chapter 9: Straight Lines – Free PDF Download

Chapter 9 of NCERT Class 11 Maths, Straight Lines, builds on the basics of coordinate geometry to study the different ways of writing the equation of a line — point-slope, two-point, slope-intercept, intercept and normal forms — along with the angle between two lines and the distance of a point from a line. These NCERT solutions, based on the current 2026-27 rationalised edition (Exercises 9.1 to 9.3 plus the Miscellaneous Exercise), give complete step-by-step answers to every question in the chapter. These Class 11 Maths Chapter 9 solutions are also useful as quick revision notes before exams.

Exercise 9.1

Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, –5) and (– 4, –2). Also, find its area.

Let the vertices be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2). Plotting these points and joining A→B→C→D→A gives a quadrilateral. To find its area, draw diagonal AC, which divides the quadrilateral into two triangles, ΔABC and ΔACD.

Area of ΔABC = ½|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)|
= ½|(–4)(7–(–5)) + 0(–5–5) + 5(5–7)|
= ½|–48 + 0 – 10| = ½(58) = 29 sq units

Area of ΔACD = ½|(–4)(–5–(–2)) + 5(–2–5) + (–4)(5–(–5))|
= ½|12 – 35 – 40| = ½(63) = 31.5 sq units

Area of quadrilateral ABCD = Area(ΔABC) + Area(ΔACD) = 29 + 31.5 = 60.5 sq units.

Q2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Since the base lies along the y-axis and its mid-point is the origin, the two base vertices are B(0, a) and C(0, –a) (distance BC = 2a as required).

The third vertex A lies on the x-axis (by symmetry) at a distance equal to the height of the equilateral triangle from BC. Height of an equilateral triangle of side 2a is (√3/2)(2a) = √3 a.

So the third vertex is A(√3 a, 0) or A(–√3 a, 0).

Vertices of the triangle: (0, a), (0, –a) and (√3 a, 0) [or (–√3 a, 0)].

Q3. Find the distance between P(x₁, y₁) and Q(x₂, y₂) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

1. If PQ is parallel to the y-axis, then x₁ = x₂. Using the distance formula, PQ = √[(x₂–x₁)² + (y₂–y₁)²] = √[0 + (y₂–y₁)²] = |y₂ – y₁|.
2. If PQ is parallel to the x-axis, then y₁ = y₂. So PQ = √[(x₂–x₁)² + 0] = |x₂ – x₁|.

Q4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Let the required point be (x, 0). Then, distance from (7, 6) = distance from (3, 4):

(x–7)² + (0–6)² = (x–3)² + (0–4)²
x² – 14x + 49 + 36 = x² – 6x + 9 + 16
–14x + 85 = –6x + 25
–8x = –60 ⟹ x = 15/2

The required point is (15/2, 0).

Q5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Mid-point of PB = ((0+8)/2, (–4+0)/2) = (4, –2).

Slope of line joining origin (0, 0) and (4, –2) = (–2 – 0)/(4 – 0) = –1/2.

Q6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Let A(4, 4), B(3, 5), C(–1, –1). Find the slopes of AB and BC:

Slope of AB = (5–4)/(3–4) = 1/(–1) = –1
Slope of BC = (–1–5)/(–1–3) = (–6)/(–4) = 3/2

Slope of AC = (–1–4)/(–1–4) = (–5)/(–5) = 1

Slope of AB × Slope of AC = (–1)(1) = –1.

Since the product of the slopes of AB and AC is –1, AB ⊥ AC. Hence the triangle has a right angle at A, so A, B, C are the vertices of a right-angled triangle.

Q7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

If a line makes an angle of 30° with the positive y-axis (measured anticlockwise), its inclination with the positive x-axis is 90° + 30° = 120°.

Slope, m = tan 120° = tan(180° – 60°) = –tan 60° = –√3.

Q8. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Let A(–2, –1), B(4, 0), C(3, 3), D(–3, 2).

Slope of AB = (0–(–1))/(4–(–2)) = 1/6
Slope of DC = (3–2)/(3–(–3)) = 1/6

Slope of AB = Slope of DC ⟹ AB ∥ DC.

Slope of BC = (3–0)/(3–4) = –3
Slope of AD = (2–(–1))/(–3–(–2)) = 3/(–1) = –3

Slope of BC = Slope of AD ⟹ BC ∥ AD.

Since both pairs of opposite sides are parallel, ABCD is a parallelogram.

Q9. Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

Slope of the line, m = (–2–(–1))/(4–3) = –1/1 = –1.

If θ is the angle the line makes with the x-axis, tan θ = m = –1, so θ = 135°.

The acute angle between the line and the x-axis is 45° (since tan 45° = 1, and the angle of inclination is 135°).

Q10. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Let the slope of one line be m, so the slope of the other is 2m.

tan θ = |(2m – m)/(1 + 2m·m)| = |m/(1 + 2m²)| = 1/3

Case 1: m/(1+2m²) = 1/3 ⟹ 3m = 1 + 2m² ⟹ 2m² – 3m + 1 = 0 ⟹ (2m–1)(m–1) = 0 ⟹ m = 1/2 or m = 1.
So the slopes are (1/2, 1) or (1, 2).

Case 2: m/(1+2m²) = –1/3 ⟹ 3m = –1–2m² ⟹ 2m² + 3m + 1 = 0 ⟹ (2m+1)(m+1) = 0 ⟹ m = –1/2 or m = –1.
So the slopes are (–1/2, –1) or (–1, –2).

Hence, the pairs of slopes are (1/2, 1), (1, 2), (–1/2, –1) or (–1, –2).

Q11. A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).

Since the line passes through (x₁, y₁) and (h, k), its slope, by definition, is:

m = (k – y₁)/(h – x₁)

Multiplying both sides by (h – x₁), we get:

m(h – x₁) = k – y₁, i.e., k – y₁ = m(h – x₁), which is the required result (this is simply the point-slope form of the equation of the line).

Exercise 9.2

In Exercises 1 to 8, find the equation of the line which satisfies the given conditions:

Q1. Write the equations for the x- and y-axes.

The x-axis is the set of all points where y = 0. So the equation of the x-axis is y = 0.

The y-axis is the set of all points where x = 0. So the equation of the y-axis is x = 0.

Q2. Passing through the point (– 4, 3) with slope 1/2.

Using point-slope form, y – y₁ = m(x – x₁):

y – 3 = (1/2)(x – (–4)) ⟹ 2y – 6 = x + 4 ⟹ x – 2y + 10 = 0.

Q3. Passing through (0, 0) with slope m.

y – 0 = m(x – 0) ⟹ y = mx.

Q4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.

Slope, m = tan 75° = tan(45° + 30°) = (tan45° + tan30°)/(1 – tan45°·tan30°) = (1 + 1/√3)/(1 – 1/√3) = (√3+1)/(√3–1) = 2 + √3.

Equation: y – 2√3 = (2 + √3)(x – 2)

y – 2√3 = (2 + √3)x – 2(2 + √3), i.e., (2+√3)x – y + 2√3 – 4 – 2√3 = 0 ⟹ (2+√3)x – y – 4 = 0.

Q5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

The point is (–3, 0). Using point-slope form:

y – 0 = –2(x – (–3)) ⟹ y = –2x – 6 ⟹ 2x + y + 6 = 0.

Q6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.

Slope, m = tan 30° = 1/√3, y-intercept c = 2. Using slope-intercept form y = mx + c:

y = x/√3 + 2 ⟹ √3y = x + 2√3 ⟹ x – √3y + 2√3 = 0.

Q7. Passing through the points (–1, 1) and (2, – 4).

Slope, m = (–4–1)/(2–(–1)) = –5/3.

Using two-point form: y – 1 = (–5/3)(x – (–1)) ⟹ 3y – 3 = –5x – 5 ⟹ 5x + 3y + 2 = 0.

Q8. The vertices of Δ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

The median through R passes through R and the mid-point of PQ.

Mid-point of PQ = ((2+(–2))/2, (1+3)/2) = (0, 2).

Slope of the median = (5–2)/(4–0) = 3/4.

Equation: y – 2 = (3/4)(x – 0) ⟹ 4y – 8 = 3x ⟹ 3x – 4y + 8 = 0.

Q9. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Slope of the line through (2, 5) and (–3, 6) = (6–5)/(–3–2) = 1/(–5) = –1/5.

Slope of the required (perpendicular) line = –1/(–1/5) = 5.

Equation: y – 5 = 5(x – (–3)) ⟹ y – 5 = 5x + 15 ⟹ 5x – y + 20 = 0.

Q10. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

The point R that divides the segment joining (1, 0) and (2, 3) in ratio 1:n is:

R = ((n·1 + 1·2)/(n+1), (n·0 + 1·3)/(n+1)) = ((n+2)/(n+1), 3/(n+1))

Slope of the given segment = (3–0)/(2–1) = 3, so the slope of the required perpendicular line = –1/3.

Equation: y – 3/(n+1) = (–1/3)[x – (n+2)/(n+1)]

Multiplying through by 3(n+1): 3(n+1)y – 9 = –(n+1)x + (n+2)

⟹ (n+1)x + 3(n+1)y = n + 11, i.e., x + 3y = (n + 11)/(n + 1).

Q11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

If intercepts on both axes are equal (say a), the equation is x/a + y/a = 1, i.e., x + y = a.

Since the line passes through (2, 3): 2 + 3 = a ⟹ a = 5.

Required equation: x + y = 5.

Q12. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Let the intercepts be a and b, with a + b = 9, so b = 9 – a.

Equation: x/a + y/b = 1. Since it passes through (2, 2): 2/a + 2/(9–a) = 1.

2(9–a) + 2a = a(9–a) ⟹ 18 – 2a + 2a = 9a – a² ⟹ a² – 9a + 18 = 0 ⟹ (a–3)(a–6) = 0 ⟹ a = 3 or a = 6.

If a = 3, b = 6: x/3 + y/6 = 1 ⟹ 2x + y = 6.
If a = 6, b = 3: x/6 + y/3 = 1 ⟹ x + 2y = 6.

Q13. Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Slope, m = tan(2π/3) = tan 120° = –√3.

Equation of the required line (through (0, 2)): y – 2 = –√3(x – 0) ⟹ √3x + y – 2 = 0.

The parallel line has the same slope –√3 and y-intercept –2 (2 units below the origin):

y = –√3x – 2 ⟹ √3x + y + 2 = 0.

Q14. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Slope of the line joining origin and (–2, 9) = 9/(–2) = –9/2.

Since this segment is perpendicular to the required line, slope of required line = –1/(–9/2) = 2/9.

Equation: y – 9 = (2/9)(x – (–2)) ⟹ 9y – 81 = 2x + 4 ⟹ 2x – 9y + 85 = 0.

Q15. The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

Since L is a linear function of C, L = mC + c, where m = (125.134 – 124.942)/(110 – 20) = 0.192/90 = 0.00213 (approx.).

Using L = 124.942 at C = 20: 124.942 = 0.00213(20) + c ⟹ c = 124.942 – 0.0427 = 124.899 (approx.).

So, L = 124.899 + 0.00213 C (approximately).

Q16. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Let demand D be a linear function of price P: D = mP + c.

m = (1220 – 980)/(16 – 14) = 240/2 = 120.

Using D = 980 at P = 14: 980 = 120(14) + c ⟹ c = 980 – 1680 = –700.

So D = 120P – 700. At P = 17: D = 120(17) – 700 = 2040 – 700 = 1340 litres.

Q17. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b = 2.

Let the line meet the x-axis at (h, 0) and the y-axis at (0, k). Since P(a, b) is the mid-point:

a = h/2 ⟹ h = 2a, and b = k/2 ⟹ k = 2b.

Using the intercept form x/h + y/k = 1: x/(2a) + y/(2b) = 1 ⟹ x/a + y/b = 2, which is the required result.

Q18. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Let the line meet the x-axis at A(p, 0) and the y-axis at B(0, q). Since R(h, k) divides AB in the ratio 1:2 (from A to B):

h = (2·p + 1·0)/(1+2) = 2p/3 ⟹ p = 3h/2. k = (2·0 + 1·q)/3 = q/3 ⟹ q = 3k.

Using intercept form x/p + y/q = 1: x/(3h/2) + y/(3k) = 1 ⟹ 2x/h + y/k = 3 is the equation of the line (in terms of h, k).

Q19. By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.

Find the equation of the line through (3, 0) and (–2, –2):

Slope = (–2–0)/(–2–3) = –2/–5 = 2/5.

Equation: y – 0 = (2/5)(x – 3) ⟹ 5y = 2x – 6 ⟹ 2x – 5y – 6 = 0.

Check if (8, 2) satisfies this: 2(8) – 5(2) – 6 = 16 – 10 – 6 = 0. ✓

Since (8, 2) lies on the line through (3, 0) and (–2, –2), all three points are collinear.

Exercise 9.3

Q1. Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.

1. x + 7y = 0 ⟹ y = (–1/7)x + 0. Slope = –1/7, y-intercept = 0.
2. 6x + 3y – 5 = 0 ⟹ 3y = –6x + 5 ⟹ y = –2x + 5/3. Slope = –2, y-intercept = 5/3.
3. y = 0 ⟹ y = 0·x + 0. Slope = 0, y-intercept = 0.

Q2. Reduce the following equations into intercept form and find their intercepts on the axes.

1. 3x + 2y – 12 = 0 ⟹ 3x + 2y = 12 ⟹ x/4 + y/6 = 1. x-intercept = 4, y-intercept = 6.
2. 4x – 3y = 6 ⟹ x/(3/2) + y/(–2) = 1. x-intercept = 3/2, y-intercept = –2.
3. 3y + 2 = 0 ⟹ y = –2/3. This line is parallel to the x-axis, so it has no x-intercept; y-intercept = –2/3.

Q3. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Rewriting the line: 12x + 72 = 5y – 10 ⟹ 12x – 5y + 82 = 0.

Distance = |A x₁ + B y₁ + C| / √(A² + B²) = |12(–1) – 5(1) + 82| / √(144+25) = |–12–5+82|/13 = 65/13 = 5 units.

Q4. Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

The line x/3 + y/4 = 1 can be written as 4x + 3y – 12 = 0.

Let the point be (x, 0). Distance = |4x – 12|/5 = 4 ⟹ |4x – 12| = 20.

4x – 12 = 20 ⟹ x = 8, or 4x – 12 = –20 ⟹ x = –2.

The points are (8, 0) and (–2, 0).

Q5. Find the distance between parallel lines: (i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0, (ii) l(x+y) + p = 0 and l(x+y) – r = 0.

1. Distance = |C₁ – C₂|/√(A²+B²) = |–34 – 31|/√(15²+8²) = 65/√289 = 65/17 units.
2. Rewrite as lx + ly + p = 0 and lx + ly – r = 0. Distance = |p – (–r)|/√(l²+l²) = |p + r|/(l√2) units.

Q6. Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).

A line parallel to 3x – 4y + 2 = 0 has the form 3x – 4y + k = 0.

Since it passes through (–2, 3): 3(–2) – 4(3) + k = 0 ⟹ –6 – 12 + k = 0 ⟹ k = 18.

Required equation: 3x – 4y + 18 = 0.

Q7. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Slope of x – 7y + 5 = 0 is 1/7, so the slope of the perpendicular line is –7.

The line passes through (3, 0) (x-intercept 3): y – 0 = –7(x – 3) ⟹ y = –7x + 21.

Required equation: 7x + y – 21 = 0.

Q8. Find angles between the lines √3x + y = 1 and x + √3y = 1.

Slope of first line, m₁ = –√3. Slope of second line, m₂ = –1/√3.

tan θ = |(m₂ – m₁)/(1 + m₁m₂)| = |(–1/√3 – (–√3))/(1 + (–√3)(–1/√3))| = |(–1/√3 + √3)/(1 + 1)| = |(2/√3)/2| = 1/√3.

θ = 30°. The angles between the lines are 30° and 150° (its supplement).

Q9. The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

Slope of the line through (h, 3) and (4, 1) = (1–3)/(4–h) = –2/(4–h).

Slope of 7x – 9y – 19 = 0 is 7/9.

Since the lines are perpendicular: [–2/(4–h)] × (7/9) = –1 ⟹ –14 = –9(4–h) ⟹ –14 = –36 + 9h ⟹ 9h = 22 ⟹ h = 22/9.

Q10. Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x – x₁) + B (y – y₁) = 0.

Any line parallel to Ax + By + C = 0 has the form Ax + By + k = 0, for some constant k.

Since this line passes through (x₁, y₁): Ax₁ + By₁ + k = 0 ⟹ k = –(Ax₁ + By₁).

Substituting back: Ax + By – (Ax₁ + By₁) = 0 ⟹ A(x – x₁) + B(y – y₁) = 0, which is the required result.

Q11. Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

Let the slope of the other line be m. Using tan 60° = √3:

√3 = |(m – 2)/(1 + 2m)|

Case 1: (m–2)/(1+2m) = √3 ⟹ m – 2 = √3 + 2√3m ⟹ m(1–2√3) = √3+2 ⟹ m = (√3+2)/(1–2√3) = –(8+5√3)/11 (after rationalising).

Case 2: (m–2)/(1+2m) = –√3 ⟹ m – 2 = –√3–2√3m ⟹ m(1+2√3) = 2–√3 ⟹ m = (2–√3)/(1+2√3) = (5√3–8)/11 (after rationalising).

So the equation of the other line, through (2, 3), is y – 3 = m(x – 2) with m = –(8+5√3)/11 or m = (5√3–8)/11.

Q12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Mid-point of the segment = ((3–1)/2, (4+2)/2) = (1, 3).

Slope of the segment = (2–4)/(–1–3) = (–2)/(–4) = 1/2, so the slope of the perpendicular bisector = –2.

Equation: y – 3 = –2(x – 1) ⟹ y – 3 = –2x + 2 ⟹ 2x + y – 5 = 0.

Q13. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Using the foot-of-perpendicular formula, for line Ax+By+C=0 and point (x₁,y₁), with t = –(Ax₁+By₁+C)/(A²+B²):

t = –(3(–1) – 4(3) – 16)/(9+16) = –(–3–12–16)/25 = 31/25.

Foot = (x₁ + At, y₁ + Bt) = (–1 + 3(31/25), 3 – 4(31/25)) = (–1 + 93/25, 3 – 124/25) = (68/25, –49/25).

Foot of perpendicular = (68/25, –49/25).

Q14. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

Slope of the line joining origin and (–1, 2) is 2/(–1) = –2. This segment is perpendicular to y = mx + c, so:

m × (–2) = –1 ⟹ m = 1/2.

Since (–1, 2) lies on y = mx + c: 2 = (1/2)(–1) + c ⟹ c = 2 + 1/2 = 5/2.

So m = 1/2 and c = 5/2.

Q15. If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos2θ and x secθ + y cosecθ = k, respectively, prove that p² + 4q² = k².

Distance of origin from x cosθ – y sinθ – k cos2θ = 0 is p = |–k cos2θ|/√(cos²θ+sin²θ) = |k cos2θ| (since cos²θ+sin²θ=1). So p² = k²cos²2θ.

The second line, x secθ + y cosecθ – k = 0, can be written as x sinθ + y cosθ – k sinθ cosθ = 0 (multiplying through by sinθcosθ). So q = |–k sinθ cosθ|/√(sin²θ+cos²θ) = |k sinθ cosθ| = |k(sin2θ)/2|. So q² = k²sin²2θ/4, i.e., 4q² = k²sin²2θ.

Adding: p² + 4q² = k²cos²2θ + k²sin²2θ = k²(cos²2θ + sin²2θ) = k², as required.

Q16. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Slope of BC = (2–(–1))/(1–4) = 3/(–3) = –1, so the slope of the altitude from A (perpendicular to BC) is 1.

Equation of altitude through A(2, 3): y – 3 = 1(x – 2) ⟹ x – y + 1 = 0.

Equation of BC: y – (–1) = –1(x – 4) ⟹ y + 1 = –x + 4 ⟹ x + y – 3 = 0.

Length of altitude = distance from A(2, 3) to line BC = |2 + 3 – 3|/√(1²+1²) = 2/√2 = √2 units.

Q17. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p² = 1/a² + 1/b².

The line with intercepts a and b is x/a + y/b = 1, i.e., bx + ay – ab = 0.

Distance from origin, p = |–ab|/√(a²+b²) = ab/√(a²+b²).

So p² = a²b²/(a²+b²) ⟹ 1/p² = (a²+b²)/(a²b²) = a²/(a²b²) + b²/(a²b²) = 1/b² + 1/a², as required.

Miscellaneous Exercise

Q1. Find the values of k for which the line (k–3) x – (4 – k²) y + k² –7k + 6 = 0 is (a) Parallel to the x-axis, (b) Parallel to the y-axis, (c) Passing through the origin.

1. A line is parallel to the x-axis when the coefficient of x is 0 (and coefficient of y is not 0): k – 3 = 0 ⟹ k = 3.
2. A line is parallel to the y-axis when the coefficient of y is 0 (and coefficient of x is not 0): 4 – k² = 0 ⟹ k = 2 or k = –2.
3. A line passes through the origin when the constant term is 0: k² – 7k + 6 = 0 ⟹ (k–1)(k–6) = 0 ⟹ k = 1 or k = 6.

Q2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

Let the intercepts be a and b. Given a + b = 1 and ab = –6. So a and b are roots of t² – t – 6 = 0 ⟹ (t–3)(t+2) = 0 ⟹ t = 3 or t = –2.

So (a, b) = (3, –2) or (–2, 3).

For a=3, b=–2: x/3 + y/(–2) = 1 ⟹ 2x – 3y – 6 = 0.
For a=–2, b=3: x/(–2) + y/3 = 1 ⟹ 3x – 2y + 6 = 0.

Q3. What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.

The line can be written as 4x + 3y – 12 = 0. Let the point be (0, y).

Distance = |3y – 12|/5 = 4 ⟹ |3y – 12| = 20.

3y – 12 = 20 ⟹ y = 32/3, or 3y – 12 = –20 ⟹ y = –8/3.

Points are (0, 32/3) and (0, –8/3).

Q4. Find perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosφ, sinφ).

Equation of the line joining (cosθ, sinθ) and (cosφ, sinφ), using the two-point form and simplifying using sum-to-product formulas, gives:

x cos((θ+φ)/2) + y sin((θ+φ)/2) = cos((θ–φ)/2)

This is already in normal form, so the perpendicular distance from the origin is simply the constant on the right-hand side:

Distance = cos((θ–φ)/2).

Q5. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

From 3x + y = 0, y = –3x. Substituting in the first equation: x – 7(–3x) + 5 = 0 ⟹ x + 21x + 5 = 0 ⟹ 22x = –5 ⟹ x = –5/22.

A line parallel to the y-axis through this point of intersection is simply x = –5/22, i.e., 22x + 5 = 0.

Q6. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis.

The line x/4 + y/6 = 1 meets the y-axis at (0, 6).

Rewriting the line as 3x + 2y – 12 = 0, its slope = –3/2, so the perpendicular line has slope 2/3.

Equation: y – 6 = (2/3)(x – 0) ⟹ 3y – 18 = 2x ⟹ 2x – 3y + 18 = 0.

Q7. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Line y = x and x = k intersect at (k, k). Line x + y = 0 and x = k intersect at (k, –k). Line y = x and x + y = 0 intersect at (0, 0).

So the vertices of the triangle are (0, 0), (k, k) and (k, –k).

Area = ½|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)| = ½|0(k–(–k)) + k(–k–0) + k(0–k)| = ½|–k² – k²| = ½(2k²) = k² sq units.

Q8. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

First find the intersection of 3x + y – 2 = 0 and 2x – y – 3 = 0: Adding, 5x – 5 = 0 ⟹ x = 1, and y = 2 – 3(1) = –1. So the point is (1, –1).

Since all three lines are concurrent, (1, –1) must satisfy px + 2y – 3 = 0: p(1) + 2(–1) – 3 = 0 ⟹ p – 2 – 3 = 0 ⟹ p = 5.

Q9. If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0.

Solving the first two lines, m₁x + c₁ = m₂x + c₂ gives the point of intersection x = (c₂–c₁)/(m₁–m₂), y = (m₁c₂–m₂c₁)/(m₁–m₂).

Since the three lines are concurrent, this point also lies on the third line, y = m₃x + c₃:

(m₁c₂–m₂c₁)/(m₁–m₂) = m₃(c₂–c₁)/(m₁–m₂) + c₃

m₁c₂ – m₂c₁ = m₃(c₂–c₁) + c₃(m₁–m₂)

m₁c₂ – m₂c₁ – m₃c₂ + m₃c₁ – c₃m₁ + c₃m₂ = 0

Rearranging by grouping m₁, m₂, m₃: m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0, as required.

Q10. Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.

Slope of x – 2y = 3 is 1/2. Let the slope of the required line be m. Using tan45° = 1:

1 = |(m – 1/2)/(1 + m/2)|

Case 1: m – 1/2 = 1 + m/2 ⟹ m/2 = 3/2 ⟹ m = 3. Equation: y – 2 = 3(x – 3) ⟹ 3x – y – 7 = 0.

Case 2: m – 1/2 = –1 – m/2 ⟹ (3/2)m = –1/2 ⟹ m = –1/3. Equation: y – 2 = (–1/3)(x – 3) ⟹ 3y – 6 = –x + 3 ⟹ x + 3y – 9 = 0.

Q11. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Solving the two equations: from 2x – 3y + 1 = 0, multiply by 2: 4x – 6y + 2 = 0. Subtracting from 4x + 7y – 3 = 0: 13y – 5 = 0 ⟹ y = 5/13.

4x = 3 – 7(5/13) = (39–35)/13 = 4/13 ⟹ x = 1/13.

For equal intercepts, the line is x + y = a. Since it passes through (1/13, 5/13): a = 1/13 + 5/13 = 6/13.

Required equation: x + y = 6/13 ⟹ 13x + 13y – 6 = 0.

Q12. Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = (m ± tanθ)/(1 ∓ m tanθ).

Let the slope of the required line (through origin) be m′, so its equation is y = m′x, i.e., y/x = m′.

Since it makes an angle θ with y = mx + c (slope m):

tanθ = |(m′ – m)/(1 + mm′)|

Taking the two sign possibilities and solving for m′: m′ – m = ±tanθ(1 + mm′) ⟹ m′ ∓ m′m tanθ = m ± tanθ ⟹ m′(1 ∓ m tanθ) = m ± tanθ

⟹ m′ = (m ± tanθ)/(1 ∓ m tanθ), i.e., y/x = (m ± tanθ)/(1 ∓ m tanθ), as required.

Q13. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Let the required point divide the segment joining (–1, 1) and (5, 7) in ratio k : 1. Then the point is:

((5k–1)/(k+1), (7k+1)/(k+1))

Since this point lies on x + y = 4: (5k–1)/(k+1) + (7k+1)/(k+1) = 4 ⟹ (12k)/(k+1) = 4 ⟹ 12k = 4k + 4 ⟹ 8k = 4 ⟹ k = 1/2.

So the required ratio is 1 : 2.

Q14. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

First find the point where the line 2x – y = 0 meets 4x + 7y + 5 = 0. Since y = 2x: 4x + 7(2x) + 5 = 0 ⟹ 18x = –5 ⟹ x = –5/18, y = –5/9.

So the point of intersection is (–5/18, –5/9).

Required distance = distance between (1, 2) and (–5/18, –5/9):

= √[(1+5/18)² + (2+5/9)²] = √[(23/18)² + (23/9)²] = √[(23²/18²) + (46²/18²)] = (23/18)√(1+4) = 23√5/18 units.

Q15. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Using the parametric (distance) form of a line: x = –1 + r cosθ, y = 2 + r sinθ, where r is the distance from (–1, 2).

Substituting into x + y = 4 with r = 3: (–1 + 3cosθ) + (2 + 3sinθ) = 4 ⟹ 3(cosθ + sinθ) = 3 ⟹ cosθ + sinθ = 1.

√2 sin(θ + 45°) = 1 ⟹ sin(θ+45°) = 1/√2 ⟹ θ + 45° = 45° or 135° ⟹ θ = 0° or θ = 90°.

So the line must be drawn either parallel to the x-axis (θ = 0°) or parallel to the y-axis (θ = 90°).

Q16. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (–4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

If the legs are parallel to the axes, the third vertex (the right angle) must be at (1, 1) or (–4, 3) — i.e., where a horizontal line through one end meets a vertical line through the other.

Taking the third vertex as (1, 1): the legs are the vertical line x = 1 (through (1,3) and (1,1)) and the horizontal line y = 1 (through (–4,1) and (1,1)).

[Equivalently, taking the third vertex as (–4, 3): the legs would be x = –4 and y = 3.]

Q17. Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

For line ax + by + c = 0 and point (x₁, y₁), the image (x′, y′) is given by using t = –2(ax₁+by₁+c)/(a²+b²), with x′ = x₁ + at, y′ = y₁ + bt.

Here a=1, b=3, c=–7, (x₁,y₁)=(3,8): a x₁+b y₁+c = 3+24–7 = 20.

t = –2(20)/(1+9) = –40/10 = –4.

x′ = 3 + 1(–4) = –1, y′ = 8 + 3(–4) = –4.

The image of the point is (–1, –4).

Q18. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

Slopes: m₁ = 3, m₂ = 1/2. Since y = mx + 4 makes equal angles with both lines:

(m–3)/(1+3m) = –(m–1/2)/(1+m/2)

Cross-multiplying and simplifying: (m–3)(1+m/2) = –(m–1/2)(1+3m)

0.5m² – 0.5m – 3 = –3m² + 0.5m + 0.5

3.5m² – m – 3.5 = 0 ⟹ 7m² – 2m – 7 = 0

m = [2 ± √(4+196)]/14 = [2 ± √200]/14 = [2 ± 10√2]/14 = (1 ± 5√2)/7.

Q19. If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

The perpendicular distance from P(x, y) to x + y – 5 = 0 is |x+y–5|/√2, and to 3x – 2y + 7 = 0 is |3x–2y+7|/√13.

Given that the sum of these distances is always 10, and taking the point P to lie on the side of each line where both expressions have a fixed (say positive) sign:

(x+y–5)/√2 + (3x–2y+7)/√13 = 10

This is a linear equation in x and y (all coefficients are constants), of the form Ax + By + C = 0. Hence, P always moves on a straight line.

Q20. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Rewrite the second line with the same coefficients as the first: multiplying by 3, 9x + 6y + 18 = 0.

The line equidistant from 9x + 6y – 7 = 0 and 9x + 6y + 18 = 0 has constant term equal to the average of –7 and 18:

k = (–7+18)/2 = 11/2. So the equidistant line is 9x + 6y + 11/2 = 0, i.e., 18x + 12y + 11 = 0.

Q21. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

By the law of reflection, the image of (1, 2) in the x-axis is (1, –2), and A, this image point, and (5, 3) are collinear (the reflected ray appears to come from the image point).

Find where the line joining (1, –2) and (5, 3) meets the x-axis (y = 0):

Slope = (3–(–2))/(5–1) = 5/4. Equation: y – (–2) = (5/4)(x–1) ⟹ y + 2 = (5/4)(x–1).

At y = 0: 2 = (5/4)(x–1) ⟹ x – 1 = 8/5 ⟹ x = 13/5.

Coordinates of A = (13/5, 0).

Q22. Prove that the product of the lengths of the perpendiculars drawn from the points (√(a²–b²), 0) and (–√(a²–b²), 0) to the line (x/a)cosθ + (y/b)sinθ = 1 is b².

Rewrite the line as: (cosθ/a)x + (sinθ/b)y – 1 = 0.

Perpendicular distance from (√(a²–b²), 0): d₁ = |(cosθ/a)√(a²–b²) – 1| / √(cos²θ/a² + sin²θ/b²)

Perpendicular distance from (–√(a²–b²), 0): d₂ = |–(cosθ/a)√(a²–b²) – 1| / √(cos²θ/a² + sin²θ/b²)

d₁ · d₂ = |1 – (cos²θ/a²)(a²–b²)| / (cos²θ/a² + sin²θ/b²) [using (1–u)(1+u) type expansion with u = (cosθ/a)√(a²–b²)]

Numerator = |1 – cos²θ + (b²/a²)cos²θ| = |sin²θ + (b²/a²)cos²θ| = (b²/a²)cos²θ + sin²θ (all positive terms).

Denominator = cos²θ/a² + sin²θ/b² = (b²cos²θ + a²sin²θ)/(a²b²).

d₁d₂ = [(b²cos²θ + a²sin²θ)/a²] ÷ [(b²cos²θ + a²sin²θ)/(a²b²)] = a²b²/a² = b², as required.

Q23. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

First, find the point of intersection (the junction) of 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0:

Multiply the first by 4 and the second by 3: 8x – 12y + 16 = 0 and 9x + 12y – 15 = 0. Adding: 17x + 1 = 0 ⟹ x = –1/17.

From 2x – 3y + 4 = 0: 3y = 2(–1/17) + 4 = 66/17 ⟹ y = 22/17. Junction point: (–1/17, 22/17).

To reach the path 6x – 7y + 8 = 0 in the least time, the person must travel along the perpendicular to that path. Slope of 6x – 7y + 8 = 0 is 6/7, so the perpendicular path has slope –7/6.

Equation of the path through (–1/17, 22/17) with slope –7/6:

7x + 6y = 7(–1/17) + 6(22/17) = (–7+132)/17 = 125/17

Multiplying both sides by 17: 119x + 102y = 125.

Class 11 Maths Chapter 9 – Notes and Extra Questions

• Slope formula: The slope of a line through (x₁, y₁) and (x₂, y₂) is m = (y₂ – y₁)/(x₂ – x₁), x₁ ≠ x₂. If a line makes an angle α with the positive x-axis, m = tan α.
• Parallel and perpendicular lines: Two lines are parallel if and only if their slopes are equal (m₁ = m₂). Two lines are perpendicular if and only if the product of their slopes is –1 (m₁m₂ = –1).
• Angle between two lines: If two lines have slopes m₁ and m₂, the acute angle θ between them satisfies tan θ = |(m₂ – m₁)/(1 + m₁m₂)|.
• Point-slope form: y – y₀ = m(x – x₀), the equation of a line with slope m passing through a fixed point (x₀, y₀).
• Two-point form: y – y₁ = [(y₂–y₁)/(x₂–x₁)](x – x₁), the equation of a line through two given points (x₁,y₁) and (x₂,y₂).
• Slope-intercept form: y = mx + c, where m is the slope and c is the y-intercept.
• Intercept form: x/a + y/b = 1, where a and b are the x-intercept and y-intercept respectively.
• General linear equation: Ax + By + C = 0 (with A, B not both zero) represents every straight line in the plane.
• Distance of a point from a line: The perpendicular distance of (x₁, y₁) from Ax + By + C = 0 is d = |Ax₁ + By₁ + C| / √(A² + B²).
• Distance between two parallel lines: For y = mx + c₁ and y = mx + c₂, distance = |c₁ – c₂|/√(1+m²); for Ax+By+C₁=0 and Ax+By+C₂=0, distance = |C₁–C₂|/√(A²+B²).

Frequently Asked Questions

How many exercises are there in Class 11 Maths Chapter 9, Straight Lines?

As per the current (2026-27 rationalised) NCERT textbook, Chapter 9 Straight Lines has three exercises — Exercise 9.1 (11 questions), Exercise 9.2 (19 questions) and Exercise 9.3 (17 questions) — plus a Miscellaneous Exercise with 23 questions at the end of the chapter, covering all the various forms of the equation of a line and the distance formulas.

What are the important formulas in Chapter 9 Straight Lines?

The key formulas are the slope formula, the conditions for parallel and perpendicular lines, the angle between two lines, the five standard forms of the equation of a line (point-slope, two-point, slope-intercept, intercept, and general/normal form), the distance of a point from a line, and the distance between two parallel lines. These are listed in the Notes section above.

What is the difference between the slope-intercept form and the intercept form of a line?

The slope-intercept form, y = mx + c, expresses a line using its slope m and the point where it cuts the y-axis (0, c). The intercept form, x/a + y/b = 1, instead expresses a line using the two points where it cuts the x-axis (a, 0) and the y-axis (0, b). Both forms describe the same family of straight lines, just using different known information.

Are these Straight Lines NCERT solutions useful for competitive exams like JEE?

Yes. Coordinate geometry of straight lines is a foundational topic for JEE Main, JEE Advanced and other engineering entrance exams, and it also builds directly into later topics like conic sections and 3D geometry. Practising every question of Exercises 9.1 to 9.3 and the Miscellaneous Exercise thoroughly — especially the perpendicular distance and angle-between-lines problems — builds the problem-solving speed and accuracy needed for competitive exams.