Unit 7: Coordinate Geometry — Class 11 Applied Maths 2026-27 | Boundless Maths
Unit 7 of 7 5 Marks CBSE 2026–27 Algebra Meets Geometry

Unit 7: Coordinate
Geometry

CBSE Class 11 Applied Mathematics · Unit 7 · Free MCQs, Solved Examples & Case Studies

Unit 7 carries 5 marks — the final unit of the syllabus. Complete free resources: 15 MCQs + 4 Assertion-Reason, 6 short answers, 6 long answers and 3 case studies. Covers Straight Lines, Circles and Parabola — fully aligned to CBSE 2026-27.

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Case Studies

Class 11 Applied Maths Unit 7: Coordinate Geometry — Complete Free Resources (CBSE 2026-27)

This page covers all topics in Unit 7 of CBSE Class 11 Applied Mathematics — the final unit of the syllabus, carrying 5 marks in the CBSE Class 11 annual exam. You'll find 15 MCQs and 4 Assertion-Reason questions with step-by-step answers, 12 solved examples, and 3 case studies. It covers Straight Lines (slope, equation forms, angle between lines, perpendicular distance, demand curve applications), Circles (locus definition, standard, central and general forms), and Parabola (focus, directrix, latus rectum, eccentricity). Free CBSE 2026-27 aligned practice on how to find the slope of a line, circle equation problems with solutions, and parabola focus-directrix questions explained step by step.

Straight Lines Circles Parabola 5 Marks
Unit 7 · 5 Marks

Topics & Key Formulas

Three topics — Straight Lines, Circles, and Parabola.

1. Straight Lines

Slope, equation forms, angle between lines, perpendicular distance.

  • Slope: m = (y₂−y₁)/(x₂−x₁) = tan θ
  • Point-slope: y−y₁ = m(x−x₁)
  • Intercept form: x/a + y/b = 1
  • Parallel: m₁=m₂  ·  Perpendicular: m₁×m₂=−1
  • Distance from point: d = |Ax₁+By₁+C|/√(A²+B²)
m₁×m₂ = −1 (perpendicular)  ·  d = |Ax₁+By₁+C|/√(A²+B²)

2. Circle

Locus definition, standard, central and general forms.

  • Central form: (x−h)²+(y−k)²=r²
  • Standard form (centre origin): x²+y²=r²
  • General form: x²+y²+2gx+2fy+c=0
  • Centre = (−g,−f)  ·  Radius = √(g²+f²−c)
Centre=(−g,−f)  ·  Radius=√(g²+f²−c)

3. Parabola

Focus-directrix definition, standard equation, key properties.

  • Standard equation: y² = 4ax (opens right, a>0)
  • Focus = (a, 0)  ·  Directrix: x = −a
  • Latus rectum length = 4a
  • Eccentricity e = 1 always (defining property)
y²=4ax  ·  Focus=(a,0)  ·  Latus Rectum=4a
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Interactive Practice

Practice MCQs — Unit 7: Coordinate Geometry

15 MCQs + 4 Assertion-Reason. Click Show Answer for the full explanation.

Straight Lines
Q1 Slope
A ramp is being built so that it makes an angle of 45° with the ground (the positive x-axis). What is the slope of this ramp?
(a) 0
(b) 1
(c) √3
(d) Undefined
Answer: (b) 1
Slope m = tan θ = tan 45° = 1.
Q2 Point-Slope
A surveyor needs the equation of a path that has a slope of 3 and passes through the marker point (1, −2). Which equation describes this path?
(a) y = 3x + 5
(b) y = 3x − 5
(c) y = −3x + 1
(d) y = 3x + 1
Answer: (b) y = 3x − 5
y−(−2) = 3(x−1) → y+2 = 3x−3 → y = 3x−5.
Q3 Perpendicular Distance
A delivery hub is located at the point (3, 4), and a straight highway is described by the line 3x − 4y + 5 = 0. What is the shortest (perpendicular) distance from the hub to the highway?
(a) 2/5
(b) 2
(c) 1
(d) 5
Answer: (a) 2/5
d = |3(3)−4(4)+5|/√(9+16) = |9−16+5|/5 = |−2|/5 = 2/5.
Q4 Parallel Lines
Two lines are parallel if their slopes m₁ and m₂ satisfy:
(a) m₁×m₂ = −1
(b) m₁ = m₂
(c) m₁+m₂ = 0
(d) m₁−m₂ = 1
Answer: (b) m₁ = m₂
Parallel lines have equal slopes. Perpendicular lines satisfy m₁×m₂=−1.
Q5 Intercept Form
The equation of a line with x-intercept 4 and y-intercept −3 is:
(a) x/4 + y/3 = 1
(b) x/4 − y/3 = 1
(c) 3x + 4y = 12
(d) x/3 + y/4 = 1
Answer: (b) x/4 − y/3 = 1
Intercept form x/a+y/b=1 with a=4, b=−3 gives x/4+y/(−3)=1 → x/4−y/3=1.
Circle
Q6 Central Form
Centre and radius of (x−3)²+(y+2)²=25 are:
(a) Centre (3,2), radius 25
(b) Centre (−3,2), radius 5
(c) Centre (3,−2), radius 5
(d) Centre (3,−2), radius 25
Answer: (c) Centre (3,−2), radius 5
h=3, k=−2, r²=25 → r=5.
Q7 Standard Form
The standard equation of a circle centred at origin with radius r is:
(a) x+y=r
(b) x²+y²=r
(c) x²+y²=r²
(d) (x−r)²+(y−r)²=0
Answer: (c) x²+y²=r²
From (x−0)²+(y−0)²=r².
Q8 General Form
In x²+y²+2gx+2fy+c=0, the radius is:
(a) √(g²+f²−c)
(b) √(g²+f²+c)
(c) g²+f²−c
(d) √(g+f−c)
Answer: (a) √(g²+f²−c)
Centre=(−g,−f). Condition g²+f²−c>0 ensures a real circle.
Q9 Diameter Endpoints
Circle with diameter endpoints A(2,0) and B(−2,0) has equation:
(a) x²+y²=2
(b) x²+y²=4
(c) x²+y²=16
(d) (x−2)(x+2)=0
Answer: (b) x²+y²=4
Centre=midpoint=(0,0). Radius=distance to A=2. Equation: x²+y²=4.
Q10 Locus Definition
A circle is the locus of a point whose distance from a fixed point is:
(a) Variable
(b) Zero
(c) Constant
(d) Equal to the x-coordinate
Answer: (c) Constant
The constant distance is called the radius; the fixed point is the centre.
Parabola
Q11 Standard Equation
The standard equation of a parabola opening right is:
(a) x²=4ay
(b) y²=4ax
(c) y²=−4ax
(d) x²=−4ay
Answer: (b) y²=4ax
y²=−4ax opens left; x²=4ay opens up; x²=−4ay opens down.
Q12 Focus
An engineer is designing a satellite dish whose cross-section follows the parabola y² = 12x. At which point should she place the receiver, given that it must sit exactly at the focus?
(a) (0,3)
(b) (3,0)
(c) (−3,0)
(d) (12,0)
Answer: (b) (3,0)
4a=12 → a=3. Focus=(a,0)=(3,0).
Q13 Eccentricity
The eccentricity of a parabola is always:
(a) 0
(b) Less than 1
(c) Greater than 1
(d) Equal to 1
Answer: (d) Equal to 1
e<1 → ellipse; e=1 → parabola; e>1 → hyperbola.
Q14 Latus Rectum
The length of the latus rectum of y²=8x is:
(a) 2
(b) 4
(c) 8
(d) 16
Answer: (c) 8
4a=8 → a=2. Latus rectum=4a=8.
Q15 Focus-Directrix
A parabola is the locus of a point equidistant from:
(a) Two fixed points
(b) A fixed point and a fixed circle
(c) A fixed point (focus) and a fixed line (directrix)
(d) A fixed line only
Answer: (c)
This focus-directrix property gives eccentricity e=1.
Assertion-Reason Questions (AR 1–4)
AR 1 Straight Line
Assertion (A): Lines with slopes m₁=2 and m₂=−½ are perpendicular.

Reason (R): Two lines are perpendicular if and only if the product of their slopes equals −1.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
m₁×m₂=2×(−½)=−1 → perpendicular ✓. R states the exact criterion and directly explains A.
AR 2 Circle
Assertion (A): x²+y²−4x+6y+13=0 represents a point circle (radius=0).

Reason (R): For x²+y²+2gx+2fy+c=0, if g²+f²−c=0, the circle degenerates to a single point.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
g=−2, f=3, c=13. Radius=√(4+9−13)=√0=0 ✓. R states the exact condition that explains A.
AR 3 Parabola
Assertion (A): The parabola y²=−8x opens to the left.

Reason (R): For y²=4ax, if a>0 it opens right; if a<0 it opens left.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (a)
4a=−8 → a=−2 (negative) → opens left ✓. R correctly explains the direction rule.
AR 4 Straight Line
Assertion (A): The slope of a vertical line is zero.

Reason (R): Slope m=tan θ; for a vertical line θ=90° and tan 90° is undefined.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R does NOT explain A
(c) A is true, R is false
(d) A is false, R is true
Answer: (d)
A is false — a vertical line's slope is undefined, not zero (it's a horizontal line that has slope zero). R is true: tan 90° is undefined, which is exactly why a vertical line's slope cannot be defined.

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Step-by-Step Solutions

Short Answer Questions

Click Show Solution to reveal complete working.

Q1 Straight Line
Find the slope of the line passing through (2, 3) and (−1, −6).
m = (y₂−y₁)/(x₂−x₁) = (−6−3)/(−1−2) = −9/−3
✓ Slope m = 3
Q2 Straight Line
Find the equation of a line parallel to 2x−3y+5=0 and passing through (1, 4).
Slope of given line: 2x−3y+5=0 → y=(2/3)x+5/3, so m=2/3
Parallel line has the same slope m=2/3
y−4 = (2/3)(x−1) → 3(y−4)=2(x−1) → 3y−12=2x−2 → 2x−3y+10=0
✓ Required equation: 2x − 3y + 10 = 0
Q3 Circle
Find the equation of a circle with centre (−3, 4) and radius 6.
Central form: (x−h)²+(y−k)²=r², h=−3, k=4, r=6
(x−(−3))²+(y−4)² = 6²
✓ (x+3)² + (y−4)² = 36
Q4 Circle
Find the centre and radius of the circle x²+y²−6x+8y−11=0.
2g=−6 → g=−3; 2f=8 → f=4; c=−11
Centre = (−g,−f) = (3,−4)
Radius = √(g²+f²−c) = √(9+16+11) = √36 = 6
✓ Centre = (3, −4), Radius = 6
Q5 Parabola
For the parabola y²=20x, find the focus, directrix, and length of the latus rectum.
4a=20 → a=5
Focus = (a,0) = (5,0)
Directrix: x = −a → x = −5
Latus rectum = 4a = 20
✓ Focus: (5,0)  |  Directrix: x=−5  |  Latus Rectum: 20 units
Q6 Straight Line · Economics
A demand equation is p = −2q + 80, where p is price (₹) and q is quantity demanded. Find the slope and interpret it.
p = −2q + 80 is in slope-intercept form (p = mq + c)
Slope = −2 (negative — demand curve slopes downward)
Interpretation: for every 1 unit increase in quantity demanded, price decreases by ₹2
p-intercept = 80 (price when q=0)  ·  q-intercept = 40 (quantity when p=0)
✓ Slope = −2, confirming the law of demand: as quantity increases, price falls

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5-Mark Questions

Long Answer Questions

Complete solutions for higher-order, multi-step questions.

Q1 Straight Line · 5M
A new footpath needs to cut directly across an existing road described by 4x−3y=10, meeting it at a perpendicular angle. The footpath must pass through the point (3, −2). Find the equation of the footpath, and also find its perpendicular distance from the origin.
Slope of given line: m₁ = 4/3
Slope of perpendicular: m₂ = −1/m₁ = −3/4
y−(−2) = (−3/4)(x−3) → 4(y+2) = −3(x−3) → 4y+8 = −3x+9 → 3x+4y−1=0
Distance from origin: d = |3(0)+4(0)−1|/√(9+16) = 1/5
✓ Required line: 3x+4y−1=0  |  Distance from origin = 1/5 units
Q2 Straight Line · 5M
Two roads 3x+y=12 and x+3y=12 intersect. Find the intersection point and the angle between them.
From (1): y=12−3x. Sub into (2): x+3(12−3x)=12 → −8x=−24 → x=3, y=3
Intersection point: (3, 3)
m₁=−3, m₂=−1/3
tan θ = |(m₁−m₂)/(1+m₁m₂)| = |(−3+1/3)/2| = 4/3
✓ Intersection: (3,3)  |  θ = tan⁻¹(4/3) ≈ 53.13°
Q3 Circle · 5M
A landscape designer is laying out a circular garden whose diameter runs between two corner posts, A(−4,3) and B(4,−3). She wants to know the full boundary equation of the garden, and also wants to check whether a proposed fountain location at P(0,5) falls inside, on, or outside this boundary.
Centre = midpoint of AB = (0,0)
Radius = √((−4)²+3²) = √(16+9) = 5
Equation: x²+y²=25
For P(0,5): 0²+5²=25 = r² → P lies exactly ON the circle
✓ Equation: x²+y²=25  |  P(0,5) lies ON the boundary
Q4 Parabola · 5M
An engineer is designing a parabolic reflector with its vertex at the origin and its axis running along the x-axis. The curve of the reflector must pass through the point (4, 6). Find the equation of this parabola, along with its focus and directrix.
Equation form: y²=4ax
Passes through (4,6): 6²=4a(4) → 36=16a → a=9/4
Equation: y²=4(9/4)x → y²=9x
Focus = (9/4, 0)  ·  Directrix: x=−9/4
✓ Equation: y²=9x  |  Focus: (9/4,0)  |  Directrix: x=−9/4
Q5 Circle · 5M
A water sprinkler is installed at the point (2, −3) and is set so that its spray reaches exactly to the origin, where a flower bed begins. Find the equation of the circular area covered by the sprinkler, and express it in general form too.
Radius = distance from (2,−3) to origin = √(4+9) = √13
Central form: (x−2)²+(y+3)²=13
Expanding: x²−4x+4+y²+6y+9=13
✓ General form: x² + y² − 4x + 6y = 0
Q6 Straight Line · Economics · 5M
A demand curve is a straight line. At price ₹50, quantity demanded is 200; at price ₹30, quantity demanded is 400. Find the demand equation and estimate quantity at price ₹40.
Points: (q₁,p₁)=(200,50), (q₂,p₂)=(400,30)
Slope = (30−50)/(400−200) = −20/200 = −1/10
p−50 = (−1/10)(q−200) → p = −q/10 + 70
Demand equation: p = −q/10 + 70 (or q = −10p + 700)
At p=40: q = −10(40)+700 = 300
✓ Demand equation: p = −q/10 + 70  |  At ₹40, quantity demanded = 300 units

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4-Mark Questions

Case Studies

Board-pattern case questions across Lines, Circles and Parabola. Click Show Answer for each part.

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Case Study 1: Demand Curve Analysis (Straight Line)

A market research firm found that when price p=₹100, quantity demanded q=50 units; when p=₹60, q=90 units. The demand curve is assumed to be a straight line.
(i)

What is the slope of the demand curve?

(ii)

What is the demand equation?

(iii)

At what price will quantity demanded be zero?

(iv)

How many units will be demanded at price ₹80?

(i)m = (60−100)/(90−50) = −40/40 = −1. Negative slope reflects the law of demand.
(ii)p−100 = −1(q−50) → p = −q + 150.
(iii)When q=0: p=−(0)+150 = ₹150 — the price at which demand drops to zero.
(iv)80=−q+150 → q = 70 units.
⛲

Case Study 2: Circular Fountain Design (Circle)

An architect designs a circular fountain centred at (5, −3) on the plan, passing through (2, 1). A lamp post is placed at (5, 2).
(i)

What is the radius of the fountain?

(ii)

What is the equation of the fountain's boundary?

(iii)

Does the lamp post at (5, 2) lie inside, on, or outside the fountain?

(iv)

What is the general form of the fountain's equation?

(i)r = √((5−2)²+(−3−1)²) = √(9+16) = √25 = 5 units.
(ii)h=5, k=−3, r=5: (x−5)² + (y+3)² = 25.
(iii)(5−5)²+(2+3)² = 0+25 = 25 = r² → the point lies exactly ON the boundary.
(iv)Expanding (x−5)²+(y+3)²=25: x²+y²−10x+6y+34−25=0 → x² + y² − 10x + 6y + 9 = 0.
🌉

Case Study 3: Parabolic Arch Design (Parabola)

A civil engineer designs a parabolic arch for a bridge. Satellite dishes and headlights also use parabolic shapes. Answer based on the parabola y² = 16x.
(i)

What is the value of 'a' for this parabola?

(ii)

Where is the focus located?

(iii)

What is the equation of the directrix?

(iv)

What are the latus rectum length and eccentricity of this parabola?

(i)4a=16 → a=4.
(ii)Focus = (a,0) = (4, 0). In satellite dishes, the receiver sits at the focus to collect reflected signals.
(iii)Directrix: x = −a = x = −4.
(iv)Latus rectum = 4a = 16. Eccentricity = e = 1 (always, for any parabola).

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Score Full Marks

Exam Tips — Unit 7: Coordinate Geometry

What separates full 5-mark answers from partial answers in this unit.

✅ Tip 1 — Straight Lines

Always state which equation form you're using before substituting. Write "Using point-slope form: y−y₁=m(x−x₁)" explicitly. This earns the formula mark even with a small arithmetic slip later.

✅ Tip 2 — Circles

Convert general form to central form carefully. Always double-check the signs: centre = (−g, −f), not (g, f). This sign error is the single most common mistake in circle questions.

✅ Tip 3 — Parabola

Identify the orientation first. Check the sign of 'a' and which variable is squared before writing focus/directrix — this determines whether the parabola opens left, right, up, or down.

✅ Tip 4 — Point on Circle/Line

To check if a point lies inside, on, or outside a circle, substitute it into the equation: if LHS = r², it's ON the circle; if LHS < r², inside; if LHS > r², outside.

❌ Common Mistakes to Avoid in Unit 7

  • Sign errors when finding centre from general form: centre is (−g,−f), not (g,f)
  • Confusing perpendicular condition (m₁×m₂=−1) with parallel condition (m₁=m₂)
  • Forgetting the directrix sign: for y²=4ax, directrix is x=−a (negative)
  • Mixing up which axis the parabola opens along based on which variable is squared
  • Not simplifying the perpendicular distance formula fraction fully
  • Assuming eccentricity varies for parabolas — it is always exactly 1
Common Questions

Frequently Asked Questions

Questions students ask most about Class 11 Applied Maths Unit 7.

Unit 7 (Coordinate Geometry) covers Straight Lines (slope, equation forms, angle between lines, perpendicular distance), Circles (locus definition, standard, central and general forms), and Parabola (focus, directrix, latus rectum, eccentricity). It carries 5 marks — the lowest-weightage unit, but still important for full marks.
The slope of a line through two points (x₁,y₁) and (x₂,y₂) is m = (y₂−y₁)/(x₂−x₁). Alternatively, slope equals tan θ, where θ is the angle the line makes with the positive x-axis.
Unit 7 Coordinate Geometry carries 5 marks in the CBSE Class 11 Applied Maths annual exam.
The general equation is x²+y²+2gx+2fy+c=0. The centre is at (−g, −f) and the radius equals √(g²+f²−c). The condition g²+f²−c > 0 must hold for a real circle to exist.
The eccentricity of every parabola is always exactly 1. This is the defining property — eccentricity less than 1 gives an ellipse/circle, equal to 1 gives a parabola, and greater than 1 gives a hyperbola.
Two lines with slopes m₁ and m₂ are parallel if m₁ = m₂. They are perpendicular if m₁ × m₂ = −1. These two conditions are the most frequently tested concepts in this unit.
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