Class 11 Maths · Chapter 9
Short answer:
Class 11 Maths Chapter 9 — Straight Lines NCERT Solutions: ye chapter Coordinate Geometry ka core hai — slope, do lines ke beech ka angle, line ki equation ke forms (point-slope, two-point, slope-intercept, intercept form), general equation, aur ek point se line ki perpendicular distance. Rationalised 2026-27 syllabus mein 3 exercises (9.1, 9.2, 9.3) + 1 Miscellaneous Exercise hain — total 47 + 23 = 70 questions. Neeche step-by-step working ke saath complete class 11 maths straight lines ncert solutions di gayi hain, latest rationalised syllabus ke hisaab se.
Chapter 9 "Straight Lines" class 11 maths ke Coordinate Geometry unit ka pehla chapter hai, aur is baat ka foundation hai ki har curve — chahe woh Chapter 10 ki Conic Sections ho ya calculus wale Chapter 12 ke Limits and Derivatives — algebra ki equation se represent hoti hai. Class 10 mein humne sirf distance formula aur section formula dekha tha; ab yahan se hum slope, do lines ke beech ka angle, aur line ki equation ke standard forms seekhte hain — jo aage Conic Sections aur higher coordinate geometry sab mein reuse hota hai.
Ye page complete class 11 maths ncert solutions deta hai Chapter 9 ke liye — Exercise 9.1, 9.2, 9.3 aur Miscellaneous Exercise ke saare 70 questions, step-by-step detailed working ke saath. Agar aap class 11 maths chapter 1 sets ncert solutions ya class 11 maths chapter 3 trigonometric functions ncert solutions dhoondh rahe hain, wo bhi is hi series mein available hain. Latest class 11 maths syllabus 2026-27 pdf ke hisaab se — Principle of Mathematical Induction aur Mathematical Reasoning wale purane chapters ab hata diye gaye hain, isliye chapter numbering shift hui hai aur Straight Lines ab Chapter 9 hai (purani 16-chapter edition mein ye Chapter 10 tha).
Agar aapko poora class 11 maths ncert book pdf download karna hai ya class 11 maths chapter 8 sequences and series important questions jaise dusre chapters chahiye, hamari series mein wo bhi cover hain.
Chapter 9 Summary — 5 Minute Revision
Straight Lines chapter mein sabse important concept hai slope (m) — line ka inclination θ (x-axis se anticlockwise angle) ka tan value. Do points se slope, do lines parallel/perpendicular hone ki condition, aur do lines ke beech acute angle — ye teeno formulae is chapter ki neev hain.
Uske baad line likhne ke 5 tarike aate hain: horizontal/vertical line (y=a, x=b), point-slope form, two-point form, slope-intercept form, aur intercept form — sab ek hi line ko alag-alag "diye gaye data" ke hisaab se likhne ke tarike hain. Inhe general form Ax+By+C=0 mein convert karke, ek point se line ki perpendicular distance aur do parallel lines ke beech ki distance nikalne ka formula milta hai.
Miscellaneous Exercise mein ye sab concepts mix hote hain — concurrent lines, reflection/image of a point, angle bisector jaisi HOTS-level problems. Ye chapter Chapter 10 Conic Sections (circle, parabola, ellipse, hyperbola) ki seedhi taiyari hai, kyunki wahan bhi line-point distance aur slope wahi tools use hote hain.
In-Text Questions — Solutions
[Ex 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.
Diagonal AC lekar quadrilateral ABCD ko do triangles ΔABC aur ΔACD mein baant dete hain, jahan A(–4,5), B(0,7), C(5,–5), D(–4,–2).
Area(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| = 29 sq units
Area(ΔACD) = ½|–4(–5–(–2))+5(–2–5)+(–4)(5–(–5))| = ½|12–35–40| = 31.5 sq units
Area of quadrilateral = 29 + 31.5 = 60.5 sq units
[Ex 9.1, 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.
Base y-axis pe hai, midpoint origin pe, base ki length 2a — isliye base ke endpoints (0, a) aur (0, –a) hain. Symmetry ki wajah se third vertex x-axis pe hoga, aur equilateral triangle ki height = √3·a hoti hai.
Vertices: (0, a), (0, –a), (√3a, 0) — [ya symmetrically (–√3a, 0), triangle ki position ke hisaab se].
[Ex 9.1, 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.
(i) PQ, y-axis ke parallel ho toh x₁ = x₂. Distance formula mein x-term zero ho jata hai:
PQ = |y₂ – y₁|
(ii) PQ, x-axis ke parallel ho toh y₁ = y₂:
PQ = |x₂ – x₁|
[Ex 9.1, Q4] Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
x-axis par point (x, 0) lete hain. Distances barabar rakhte hain:
(x–7)² + 36 = (x–3)² + 16
–14x + 49 + 36 = –6x + 9 + 16 → –8x = –60 → x = 7.5
Point = (15/2, 0)
[Ex 9.1, 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).
Pehle PB ka midpoint nikalte hain: Mid-point of PB = ((0+8)/2, (–4+0)/2) = (4, –2)
Origin se is midpoint tak slope: m = (–2 – 0)/(4 – 0) = –1/2
[Ex 9.1, 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.
A(4,4), B(3,5), C(–1,–1) lete hain.
Slope AB = (5–4)/(3–4) = –1; Slope AC = (–1–4)/(–1–4) = 1
Slope AB × Slope AC = (–1)(1) = –1 ⇒ AB ⊥ AC
Isliye vertex A par 90° ka angle hai — triangle right-angled hai.
[Ex 9.1, Q7] Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Positive y-axis, x-axis se 90° pe hoti hai. y-axis se 30° aur anticlockwise ghumane par x-axis se inclination = 90° + 30° = 120°.
m = tan 120° = –√3
[Ex 9.1, Q8] Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.
A(–2,–1), B(4,0), C(3,3), D(–3,2)
Slope AB = (0–(–1))/(4–(–2)) = 1/6; Slope DC = (3–2)/(3–(–3)) = 1/6 ⇒ AB ∥ DC
Slope AD = (2–(–1))/(–3–(–2)) = –3; Slope BC = (3–0)/(3–4) = –3 ⇒ AD ∥ BC
Dono pairs of opposite sides parallel hain, isliye ABCD ek parallelogram hai.
[Ex 9.1, Q9] Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).
m = (–2–(–1))/(4–3) = –1
tan θ = –1 ⇒ θ = 135° (inclination x-axis ke positive direction se, 0° se 180° ke beech)
[Ex 9.1, 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.
Maan lo m₁ = m, m₂ = 2m.
tanθ = |(2m–m)/(1+2m²)| = |m/(1+2m²)| = 1/3
Case (+): 3m = 1+2m² ⇒ 2m²–3m+1=0 ⇒ (2m–1)(m–1)=0 ⇒ m = 1/2 ya m = 1
Case (–): 3m = –(1+2m²) ⇒ 2m²+3m+1=0 ⇒ (2m+1)(m+1)=0 ⇒ m = –1/2 ya m = –1
Slopes: m=1 ke liye (1, 2), ya m=–1 ke liye (–1, –2) [standard accepted pair]
[Ex 9.1, Q11] A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m(h – x₁).
Do points (x₁, y₁) aur (h, k) se guzarne wali line ki slope, definition se:
m = (k – y₁)/(h – x₁)
Dono side (h – x₁) se multiply karne par:
k – y₁ = m(h – x₁) — proved.
[Ex 9.2, Q1] Write the equations for the x- and y-axes.
x-axis ke har point ka y-coordinate zero hota hai, aur y-axis ke har point ka x-coordinate zero hota hai.
x-axis: y = 0 y-axis: x = 0
[Ex 9.2, Q2] Passing through the point (–4, 3) with slope 1/2.
Point-slope form use karte hain: y – y₀ = m(x – x₀)
y – 3 = ½(x + 4) → 2y – 6 = x + 4
x – 2y + 10 = 0
[Ex 9.2, Q3] Passing through (0, 0) with slope m.
Point-slope form use karte hain, origin (0,0) se guzarti hai:
y – 0 = m(x – 0)
y = mx
[Ex 9.2, Q4] Passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.
Pehle tan75° nikalte hain: tan75° = tan(45°+30°) = (1+1/√3)/(1–1/√3) = (√3+1)/(√3–1) = 2+√3 (rationalising karne par)
Point-slope form lagate hain: y – 2√3 = (2+√3)(x – 2)
y = (2+√3)x – 2(2+√3) + 2√3 = (2+√3)x – 4
(2+√3)x – y – 4 = 0
[Ex 9.2, Q5] Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
Diya gaya point = (–3, 0), slope = –2
y – 0 = –2(x – (–3)) → y = –2x – 6
2x + y + 6 = 0
[Ex 9.2, 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.
Diya gaya point = (0, 2), slope = tan30° = 1/√3
y – 2 = (1/√3)x → √3y – 2√3 = x
x – √3y + 2√3 = 0
[Ex 9.2, Q7] Passing through the points (–1, 1) and (2, –4).
Pehle dono points se slope nikalte hain:
m = (–4–1)/(2–(–1)) = –5/3
y – 1 = –5/3(x + 1) → 3y – 3 = –5x – 5
5x + 3y + 2 = 0
[Ex 9.2, Q8] The vertices of ΔPQR are P(2, 1), Q(–2, 3) and R(4, 5). Find equation of the median through the vertex R.
Median R se PQ ke midpoint tak jati hai. Midpoint of PQ = (0, 2)
m = (5–2)/(4–0) = 3/4
y – 2 = ¾(x – 0) → 4y – 8 = 3x
3x – 4y + 8 = 0
[Ex 9.2, Q9] Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Di gayi line ki slope = (6–5)/(–3–2) = –1/5. Perpendicular slope = 5.
y – 5 = 5(x + 3) → y – 5 = 5x + 15
5x – y + 20 = 0
[Ex 9.2, 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.
Section formula se, ratio 1:n mein divide karne wala point:
R = ((2+n)/(n+1), 3/(n+1))
Slope of segment (1,0)-(2,3) = 3, isliye perpendicular slope = –1/3
y – 3/(n+1) = –⅓[x – (n+2)/(n+1)]
3(n+1) se multiply karke simplify karne par:
(n+1)x + 3(n+1)y – (n+11) = 0
[Ex 9.2, Q11] Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Equal intercepts a: x/a + y/a = 1 ⇒ x + y = a
Point (2,3) satisfy karega: a = 2+3 = 5
x + y = 5
[Ex 9.2, Q12] Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
x/a + y/(9–a) = 1, point (2,2) se:
2(9–a) + 2a = a(9–a) → 18 = 9a – a² → a² – 9a + 18 = 0
(a–3)(a–6) = 0 ⇒ a = 3 ya a = 6
a=3, b=6: x/3+y/6=1 ⇒ 2x + y – 6 = 0
a=6, b=3: x/6+y/3=1 ⇒ x + 2y – 6 = 0
[Ex 9.2, 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 = tan120° = –√3
Line 1 (through (0,2)): y – 2 = –√3(x) ⇒ √3x + y – 2 = 0
Line 2 (parallel, through (0,–2)): y + 2 = –√3(x) ⇒ √3x + y + 2 = 0
[Ex 9.2, Q14] The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.
Slope of OP (origin to (–2,9)) = 9/(–2) = –9/2. Ye required line ke perpendicular hai, isliye required line ki slope = 2/9
y – 9 = 2/9(x + 2) → 9y – 81 = 2x + 4
2x – 9y + 85 = 0
[Ex 9.2, 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.
Do given values se slope nikalte hain: Slope = (125.134 – 124.942)/(110 – 20) = 0.192/90 ≈ 0.002133
Point-slope form lagate hain: L – 124.942 = 0.002133(C – 20)
L ≈ 0.002133C + 124.899
[Ex 9.2, 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?
Demand D, price P: slope = (1220–980)/(16–14) = 120
D – 980 = 120(P – 14)
P = 17 par: D = 980 + 120(3) = 980 + 360 = 1340 litres
[Ex 9.2, 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.
Line axes ko (p, 0) aur (0, q) par miley. Midpoint = (p/2, q/2) = (a, b) ⇒ p = 2a, q = 2b
x/(2a) + y/(2b) = 1
x/a + y/b = 2 — proved.
[Ex 9.2, Q18] Point R(h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.
Line axes A(p,0), B(0,q) ko meet kare. R, AB ko 1:2 mein divide kare:
R = (2p/3, q/3) = (h, k) ⇒ p = 3h/2, q = 3k
Equation: x/p + y/q = 1 ⇒ 2x/(3h) + y/(3k) = 1
2x/h + y/k = 3
[Ex 9.2, Q19] By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.
(3,0) aur (–2,–2) se line: slope = (–2–0)/(–2–3) = 2/5
y = 2/5(x–3) → 5y = 2x–6 → 2x – 5y – 6 = 0
Check (8,2): 2(8) – 5(2) – 6 = 16–10–6 = 0 ✓
Point (8,2) is bhi line par hai, isliye teeno points collinear hain.
[Ex 9.3, Q1] Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts: (i) x + 7y = 0, (ii) 6x + 3y – 5 = 0, (iii) y = 0.
Har equation ko slope-intercept form y = mx + c mein convert karte hain:
(i) y = –x/7 → slope = –1/7, y-intercept = 0
(ii) 3y = –6x+5 → y = –2x + 5/3 → slope = –2, y-intercept = 5/3
(iii) y = 0 → slope = 0, y-intercept = 0
[Ex 9.3, Q2] Reduce the following equations into intercept form and find their intercepts on the axes: (i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6, (iii) 3y + 2 = 0.
(i) 3x+2y=12 → x/4 + y/6 = 1 → x-intercept=4, y-intercept=6
(ii) x/(3/2) + y/(–2) = 1 → x-intercept=3/2, y-intercept=–2
(iii) y = –2/3 — ye x-axis ke parallel line hai, x-intercept exist nahi karta; y-intercept = –2/3
[Ex 9.3, Q3] Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
Pehle line ko general form mein laate hain: 12x + 72 = 5y – 10 → 12x – 5y + 82 = 0
Ab distance formula lagate hain: d = |12(–1) – 5(1) + 82| / √(144+25) = |65|/13 = 5 units
[Ex 9.3, Q4] Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.
Line ko general form mein likhte hain: 4x + 3y – 12 = 0. x-axis ka point (x, 0) lete hain:
|4x – 12|/5 = 4 ⇒ |4x – 12| = 20
Dono cases solve karne par: 4x–12=20 ⇒ x=8; 4x–12=–20 ⇒ x=–2
Points: (8, 0) and (–2, 0)
[Ex 9.3, 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.
(i) d = |–34–31|/√(225+64) = 65/17 = 65/17 units
(ii) lx+ly+p=0 aur lx+ly–r=0: d = |p–(–r)|/√(2l²) = |p+r|/(l√2)
[Ex 9.3, Q6] Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).
Parallel line: 3x – 4y + k = 0. Point (–2,3) se: 3(–2) – 4(3) + k = 0 ⇒ k = 18
3x – 4y + 18 = 0
[Ex 9.3, Q7] Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Di gayi line ki slope = 1/7 hai, isliye perpendicular slope = –7. Required line (3, 0) se guzarti hai:
y – 0 = –7(x – 3) → y = –7x + 21
7x + y – 21 = 0
[Ex 9.3, Q8] Find angles between the lines √3x + y = 1 and x + √3y = 1.
Dono lines ki slopes nikalte hain: m₁ = –√3, m₂ = –1/√3
Angle formula lagate hain: tanθ = |(m₂–m₁)/(1+m₁m₂)| = |(–1/√3+√3)/(1+1)| = |(2/√3)/2| = 1/√3
θ = 30° (obtuse angle = 150°)
[Ex 9.3, 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.
Di gayi line ki slope = (7/9). Line through (h,3),(4,1) ki slope = (1–3)/(4–h) = –2/(4–h)
Perpendicular condition lagate hain:
[–2/(4–h)] × (7/9) = –1 ⇒ 14 = 9(4–h) ⇒ 14 = 36–9h
h = 22/9
[Ex 9.3, 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.
Parallel line same A, B ke saath: Ax + By + k = 0. Point (x₁,y₁) se guzarti hai:
Ax₁ + By₁ + k = 0 ⇒ k = –(Ax₁ + By₁)
Substitute karne par: Ax + By – (Ax₁+By₁) = 0
A(x – x₁) + B(y – y₁) = 0 — proved.
[Ex 9.3, 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.
tan60° = √3 = |(2–m)/(1+2m)|
Case (+): 2–m = √3(1+2m) ⇒ m(1+2√3) = 2–√3 ⇒ m = (2–√3)/(1+2√3) = (5√3–8)/11
Case (–): 2–m = –√3(1+2m) ⇒ m(1–2√3) = 2+√3 ⇒ m = –(8+5√3)/11
Equations: y – 3 = [(5√3–8)/11](x–2) or y – 3 = [–(8+5√3)/11](x–2)
[Ex 9.3, Q12] Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
Segment ka midpoint = (1, 3) hai. Slope of segment = (2–4)/(–1–3) = ½, isliye right bisector ki perpendicular slope = –2:
y – 3 = –2(x – 1) → y – 3 = –2x + 2
2x + y – 5 = 0
[Ex 9.3, Q13] Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Foot (x₁,y₁) = (–1,3) + t(3,–4) parametrize karte hain, jo line satisfy kare:
3(–1+3t) – 4(3–4t) – 16 = 0 ⇒ 25t – 31 = 0 ⇒ t = 31/25
x = –1 + 93/25 = 68/25; y = 3 – 124/25 = –49/25
Foot of perpendicular = (68/25, –49/25)
[Ex 9.3, 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.
(–1,2) line pe hai: 2 = –m + c ⇒ c = m + 2 ...(i)
Slope of O to (–1,2) = –2, ye given line ke perpendicular hai: m × (–2) = –1 ⇒ m = 1/2
(i) se: c = 1/2 + 2 = 5/2
m = 1/2, c = 5/2
[Ex 9.3, 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.
Dono lines ke liye origin se perpendicular distance nikalte hain:
Line 1: distance from origin p = |k cos2θ| ⇒ p² = k²cos²2θ
Line 2: q = k/√(sec²θ+cosec²θ)
q² = k²sin²θcos²θ/(sin²θ+cos²θ) = k²sin²θcos²θ
4q² = k²(2sinθcosθ)² = k²sin²2θ
Dono add karne par: p² + 4q² = k²cos²2θ + k²sin²2θ = k²(cos²2θ+sin²2θ) = k² — proved.
[Ex 9.3, 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.
Pehle slope BC nikalte hain: (2–(–1))/(1–4) = –1. Altitude, BC ke perpendicular hoti hai, isliye uski slope = 1
Altitude: y – 3 = 1(x–2) ⇒ x – y + 1 = 0
Ab line BC ki equation nikalte hain: y+1 = –1(x–4) ⇒ x+y–3=0
Point A se BC ki distance (altitude ki length): Length = |2+3–3|/√2 = √2 units
[Ex 9.3, 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.
Line ko intercept form se general form mein convert karte hain: x/a + y/b = 1 ⇒ bx + ay – ab = 0
Origin se distance formula lagate hain: p = |–ab|/√(a²+b²) ⇒ p² = a²b²/(a²+b²)
1/p² = (a²+b²)/(a²b²) = 1/a² + 1/b² — proved.
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Exercise Questions — Solutions (Q1–Q23)
[Misc, 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.
Teeno conditions ko coefficient/constant term zero karke check karte hain:
(a) Parallel to x-axis: coefficient of x = 0 ⇒ k–3=0 ⇒ k = 3
(b) Parallel to y-axis: coefficient of y = 0 ⇒ 4–k²=0 ⇒ k = 2 or k = –2
(c) Through origin: constant term = 0 ⇒ k²–7k+6=0 ⇒ (k–1)(k–6)=0 ⇒ k = 1 or k = 6
[Misc, Q2] Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.
a+b=1, ab=–6 ⇒ a,b, equation t²–t–6=0 ke roots hain: (t–3)(t+2)=0 ⇒ t=3,–2
Case (a=3,b=–2): x/3 – y/2 = 1 ⇒ 2x – 3y – 6 = 0
Case (a=–2,b=3): –x/2 + y/3 = 1 ⇒ 3x – 2y + 6 = 0
[Misc, Q3] What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.
Line: 4x+3y–12=0. Point (0,y):
|3y–12|/5 = 4 ⇒ 3y–12=±20
y = 32/3 ya y = –8/3
Points: (0, 32/3) and (0, –8/3)
[Misc, Q4] Find perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosφ, sinφ).
Do-point form se line nikal ke sum-to-product identities lagane par, line ki equation normal form mein simplify ho jati hai:
x cos((θ+φ)/2) + y sin((θ+φ)/2) = cos((θ–φ)/2)
Ye already normal form (a cosα + b sinα = p) mein hai, isliye origin se distance seedhe p ke barabar hai:
Distance = cos((θ–φ)/2)
[Misc, 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.
3x+y=0 ⇒ y=–3x. Substitute: x–7(–3x)+5=0 ⇒ 22x=–5 ⇒ x=–5/22
y-axis ke parallel line ki form x = constant hoti hai:
x = –5/22
[Misc, 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.
Given line y-axis ko x=0 par milti hai: y=6, point (0,6)
Given line: 3x+2y=12, slope=–3/2. Perpendicular slope = 2/3
y – 6 = (2/3)x ⇒ 3y – 18 = 2x
2x – 3y + 18 = 0
[Misc, Q7] Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.
Teeno lines y=x, y=–x, x=k ke intersection points (vertices) nikalte hain: (0,0), (k,k), (k,–k)
Area formula lagate hain: Area = ½|0(k–(–k)) + k(–k–0) + k(0–k)| = ½|–2k²|
Area = k² sq units
[Misc, 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.
Line 1 aur Line 3 solve karte hain: 3x+y=2, 2x–y=3. Add: 5x=5 ⇒ x=1, y=–1. Point (1,–1)
Line 2 mein substitute: p(1)+2(–1)–3=0 ⇒ p–5=0
p = 5
[Misc, 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.
Line1 aur Line2 ka intersection point: x = (c₂–c₁)/(m₁–m₂), y = (m₁c₂–c₁m₂)/(m₁–m₂)
Ye point line 3 (y=m₃x+c₃) bhi satisfy karega. Substitute karke simplify karne par:
m₁c₂ – m₁c₃ + m₂c₃ – m₂c₁ + m₃c₁ – m₃c₂ = 0
Regroup karne par:
m₁(c₂–c₃) + m₂(c₃–c₁) + m₃(c₁–c₂) = 0 — proved.
[Misc, 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 given line = 1/2. tan45°=1=|(m–½)/(1+m/2)|
Case(+): m–½=1+m/2 ⇒ m/2 = 3/2 ⇒ m=3
Case(–): m–½=–(1+m/2) ⇒ (3/2)m=–1/2 ⇒ m=–1/3
(m=3): y–2=3(x–3) ⇒ 3x – y – 7 = 0
(m=–1/3): 3y–6=–(x–3) ⇒ x + 3y – 9 = 0
[Misc, 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.
Dono equations solve karne par: 4x+7y=3 and 4x–6y=–2. Subtract: 13y=5 ⇒ y=5/13, x=1/13
Equal intercepts: x+y=a. Point (1/13,5/13) se: a = 6/13
13x + 13y – 6 = 0
[Misc, 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).
Required line ki slope m' ho. tanθ = |(m'–m)/(1+mm')|
Positive case: tanθ(1+mm') = m'–m ⇒ m'(1–m tanθ) = m+tanθ ⇒ m' = (m+tanθ)/(1–m tanθ)
Negative case: m' = (m–tanθ)/(1+m tanθ)
Line through origin: y = m'x, isliye
y/x = (m ± tanθ)/(1 ∓ m tanθ) — proved.
[Misc, Q13] In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?
Ratio k:1 lete hain. Dividing point = ((5k–1)/(k+1), (7k+1)/(k+1))
x+y=4 mein substitute:
(5k–1+7k+1)/(k+1) = 4 ⇒ 12k = 4k+4 ⇒ k = 1/2
Ratio = 1 : 2
[Misc, Q14] Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.
Check (1,2): 2(1)–2=0, isliye point line 2x–y=0 par hi hai.
2x–y=0 aur 4x+7y+5=0 ka intersection: y=2x se, 4x+14x+5=0 ⇒ x=–5/18, y=–5/9
Distance between (1,2) and (–5/18,–5/9):
Δx=23/18, Δy=23/9 ⇒ d = 23√((1/18)²+(1/9)²) = 23√(5/324)
d = 23√5/18 units
[Misc, 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.
Parametric form: x=–1+r cosθ, y=2+r sinθ. r=3 par x+y=4 satisfy karna hai:
(–1+3cosθ)+(2+3sinθ)=4 ⇒ cosθ+sinθ=1
√2 sin(θ+45°)=1 ⇒ θ+45°=45° or 135° ⇒ θ=0° or 90°
Line x-axis ke parallel (θ=0°) ya y-axis ke parallel (θ=90°) khinchni chahiye.
[Misc, 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.
Legs axes ke parallel hone chahiye — right angle vertex do possible positions par ho sakta hai: (1,1) ya (–4,3).
Case 1 (vertex (1,1)): legs — x = 1 and y = 1
Case 2 (vertex (–4,3)): legs — x = –4 and y = 3
[Misc, 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.
Image (h,k) ho. Midpoint line pe hoga: (3+h)+3(8+k)=14 ⇒ h+3k=–13 ...(i)
Segment ⊥ line: slope of segment=3 (since line's slope=–1/3): (k–8)/(h–3)=3 ⇒ k=3h–1 ...(ii)
(i),(ii) solve: h+3(3h–1)=–13 ⇒ 10h=–10 ⇒ h=–1, k=–4
Image = (–1, –4)
[Misc, 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.
m₁=3, m₂=1/2. Equal inclination: |(3–m)/(1+3m)| = |(1/2–m)/(1+m/2)|
Cross-multiply karke dono cases solve karne par (positive case mein no real solution milta hai):
7m² – 2m – 7 = 0 ⇒ m = (2 ± √200)/14
m = (1 ± 5√2)/7
[Misc, 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.
Distances: |x+y–5|/√2 aur |3x–2y+7|/√13. Ek fixed sign-combination maan kar (jaise dono expressions positive):
(x+y–5)/√2 + (3x–2y+7)/√13 = 10
Ye x aur y mein linear equation hai (koi x², y², xy term nahi) — har possible sign-combination ke liye bhi yehi hota hai.
Isliye P ek straight line par move karta hai — proved.
[Misc, Q20] Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
9x+6y–7=0 ko 3 se divide: 3x+2y–7/3=0. Doosri line: 3x+2y+6=0
Equidistant line ka constant, in dono ka average hoga:
k = (–7/3 + 6)/2 = 11/6
3x+2y+11/6=0, 6 se multiply karne par:
18x + 12y + 11 = 0
[Misc, 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.
(1,2) ka x-axis ke against image = (1,–2). Reflected ray, image point se (5,3) tak jaane wali line ke barabar hoti hai.
Slope = (3–(–2))/(5–1) = 5/4
y+2 = 5/4(x–1). At y=0: 2 = 5/4(x–1) ⇒ x = 13/5
A = (13/5, 0)
[Misc, 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 cosθ/a + y sinθ/b = 1 is b².
Let c = √(a²–b²), D = √(cos²θ/a² + sin²θ/b²)
d₁d₂ = |(c cosθ/a – 1)(–c cosθ/a – 1)| / D² = |1 – c²cos²θ/a²| / D²
Numerator ko simplify karne par a²sin²θ+b²cos²θ milta hai jo denominator ke saath cancel ho jata hai:
d₁ × d₂ = b² — proved.
[Misc, 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.
Junction point (dono lines solve karke): x=–1/17, y=22/17
Least time = shortest (perpendicular) path. 6x–7y+8=0 ki slope=6/7, perpendicular slope=–7/6
y – 22/17 = –7/6(x + 1/17)
Simplify karne par:
119x + 102y – 125 = 0
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Slope from inclination θ | m = tanθ, θ ≠ 90° |
| Slope from two points | m = (y₂–y₁)/(x₂–x₁), x₁≠x₂ |
| Parallel lines | m₁ = m₂ |
| Perpendicular lines | m₁ × m₂ = –1 |
| Angle between two lines | tanθ = |(m₂–m₁)/(1+m₁m₂)|, 1+m₁m₂≠0 |
| Collinearity of A, B, C | Slope AB = Slope BC |
| Horizontal line (distance a) | y = a or y = –a |
| Vertical line (distance b) | x = b or x = –b |
| Point-slope form | y – y₀ = m(x – x₀) |
| Two-point form | y – y₁ = [(y₂–y₁)/(x₂–x₁)](x – x₁) |
| Slope-intercept form | y = mx + c |
| Slope + x-intercept d | y = m(x – d) |
| Intercept form | x/a + y/b = 1 |
| General equation of a line | Ax + By + C = 0 (A, B not both zero) |
| Distance of point (x₁,y₁) from Ax+By+C=0 | d = |Ax₁+By₁+C| / √(A²+B²) |
| Distance between parallel lines Ax+By+C₁=0, Ax+By+C₂=0 | d = |C₁–C₂| / √(A²+B²) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Slope formula ulta likh dena — (x₂–x₁)/(y₂–y₁) likh dete hain jabki sahi formula m = (y₂–y₁)/(x₂–x₁) hai.
- Angle-between-lines formula mein modulus (| |) lagana bhool jaate hain, jisse acute aur obtuse angle mix ho jate hain.
- Point-slope form y–y₀=m(x–x₀) mein fixed point (x₀,y₀) aur variable point (x,y) ko galti se swap kar dete hain.
- Intercept form x/a+y/b=1 mein a ya b ka negative sign bhool jaate hain (jaise a=–3 ko +3 maan lena).
- Perpendicular condition m₁m₂=–1 ko parallel condition m₁=m₂ se confuse kar dete hain, khaaskar 'perpendicular' aur 'parallel' wale word-problems mein.
- Distance-from-line formula lagane se pehle line ko general form Ax+By+C=0 mein poora convert nahi karte — constant term ko doosri side hi chhod dete hain, jisse C ki value galat aa jati hai.
Board-Style Important Questions
- 1 mark: Find the slope of the line joining the points (2, –1) and (5, 3).
- 1 mark: Write the equation of the line parallel to the x-axis at a distance of 5 units below the origin.
- 2 marks: Find the acute angle between the lines whose slopes are 2 and 1/3.
- 4 marks: Find the equation of the line passing through (2, –3) and perpendicular to the line joining the points (1, 2) and (3, 6).
- 4 marks: Find the distance between the parallel lines 3x + 4y – 7 = 0 and 3x + 4y + 8 = 0.
- 6 marks: Find the coordinates of the foot of perpendicular from the point (2, 3) to the line x – 3y + 4 = 0, and hence find the distance of the point from the line.
Aksar Poochhe Jaane Wale Sawaal
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Class 11 Maths — Saare Chapters
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