Chapter 7 — Coordinate Geometry

Concept Used

The distance between two points in a coordinate plane is obtained using the Distance Formula, which is derived from the Pythagoras Theorem.

If two points are A(x₁, y₁) and B(x₂, y₂), then

\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Solution Roadmap

• Identify coordinates of both points correctly.
• Substitute values carefully into the distance formula.
• Simplify step-by-step (avoid skipping).
• Handle negative signs properly (common mistake).
• Write final answer in simplest radical form.

Solution

Part (i)

Points A(2, 3) and B(4, 1)

$$\begin{aligned} AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &= \sqrt{(4 - 2)^2 + (1 - 3)^2} \\ &= \sqrt{(2)^2 + (-2)^2} \\ &= \sqrt{4 + 4} \\ &= \sqrt{8} \\ &= 2\sqrt{2} \end{aligned}$$

Final Answer: Distance = 2√2 units

Part (ii)

Points X(–5, 7) and Y(–1, 3)

$$\begin{aligned} XY &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &= \sqrt{(-1 - (-5))^2 + (3 - 7)^2} \\ &= \sqrt{(4)^2 + (-4)^2} \\ &= \sqrt{16 + 16} \\ &= \sqrt{32} \\ &= 4\sqrt{2} \end{aligned}$$

Final Answer: Distance = 4√2 units

Part (iii)

Points A(a, b) and B(–a, –b)

$$\begin{aligned} AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &= \sqrt{(-a - a)^2 + (-b - b)^2} \\ &= \sqrt{(-2a)^2 + (-2b)^2} \\ &= \sqrt{4a^2 + 4b^2} \\ &= \sqrt{4(a^2 + b^2)} \\ &= 2\sqrt{a^2 + b^2} \end{aligned}$$

Final Answer: Distance = 2√(a² + b²)

Exam Significance

• This is a direct formula-based question frequently asked in CBSE Board exams.
• Tests your accuracy in handling negative signs and squaring.
• Part (iii) is important for algebraic manipulation skills (common in competitive exams).
• Forms the base for advanced topics like:
  • Section Formula
  • Midpoint Formula
  • Coordinate Proofs

Common Mistakes to Avoid

• Writing multiplication instead of addition inside square root (very common error).
• Incorrect handling of negative signs.
• Skipping simplification steps.
• Not converting √8 into 2√2.

Concept Used

When one point is the origin (0,0), the distance formula simplifies significantly.

Distance from origin to a point (x, y) is:

\[ \text{Distance} = \sqrt{x^2 + y^2} \]

This is directly derived from the Pythagoras Theorem where x and y act as perpendicular sides.

Solution Roadmap

• Identify that one point is the origin (0,0).
• Use simplified distance formula √(x² + y²).
• Substitute values carefully.
• Perform square calculations step-by-step.
• Interpret the result in real-life context (town distance).

Solution

Let A(0, 0) and B(36, 15)

Distance between A and B (AB):

$$\begin{aligned} AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &= \sqrt{(36 - 0)^2 + (15 - 0)^2} \\ &= \sqrt{36^2 + 15^2} \\ &= \sqrt{1296 + 225} \\ &= \sqrt{1521} \\ &= 39 \end{aligned}$$

Final Answer: Distance = 39 units

Real-Life Interpretation

Since the coordinates represent positions of towns A and B, the calculated distance is:

Distance between town A and town B = 39 km

Exam Significance

• Very common 1–2 mark direct question in CBSE Board exams.
• Tests understanding of distance from origin shortcut.
• Important for competitive exams like NTSE, Olympiads.
• Forms conceptual base for:
  • Vector geometry
  • Coordinate proofs
  • Trigonometric applications

Common Mistakes to Avoid

• Forgetting that (0,0) simplifies the formula.
• Calculation errors in squares (36², 15²).
• Not interpreting answer in units (km).
• Skipping intermediate steps.

Concept Used

Three points are said to be collinear if they lie on the same straight line.

Using the distance method:

• If AB + BC = AC (or any such combination), then points are collinear.
• Otherwise, they are non-collinear.

Solution Roadmap

• Find distances AB, BC and CA using distance formula.
• Identify the largest distance.
• Check if sum of two smaller distances equals the largest.
• Conclude collinearity.

Solution

Let A(1, 5), B(2, 3) and C(–2, –11)

Step 1: Find AB

$$\begin{aligned} AB &= \sqrt{(2 - 1)^2 + (3 - 5)^2} \\ &= \sqrt{1^2 + (-2)^2} \\ &= \sqrt{1 + 4} \\ &= \sqrt{5} \end{aligned}$$

Step 2: Find BC

$$\begin{aligned} BC &= \sqrt{(-2 - 2)^2 + (-11 - 3)^2} \\ &= \sqrt{(-4)^2 + (-14)^2} \\ &= \sqrt{16 + 196} \\ &= \sqrt{212} \end{aligned}$$

Step 3: Find CA

$$\begin{aligned} CA &= \sqrt{(1 - (-2))^2 + (5 - (-11))^2} \\ &= \sqrt{(1 + 2)^2 + (5 + 11)^2} \\ &= \sqrt{3^2 + 16^2} \\ &= \sqrt{9 + 256} \\ &= \sqrt{265} \end{aligned}$$

Step 4: Check Collinearity

Largest distance = CA = \(\sqrt{265}\)

\[\sqrt{5} + \sqrt{212} \neq \sqrt{265}\]

Since the sum of two sides is NOT equal to the third side, the points do NOT lie on a straight line.

Final Conclusion: The given points are not collinear.

Exam Significance

• Very important conceptual question in CBSE Board exams.
• Tests understanding of distance formula + logic.
• Alternative method (slope method) is also frequently tested.
• Common in competitive exams like NTSE and Olympiads.

Common Mistakes to Avoid

• Incorrect subtraction of coordinates (especially negatives).
• Wrong calculation of CA (major error in many solutions).
• Forgetting to compare with the largest distance.
• Skipping verification step.

Concept Used

A triangle is said to be isosceles if any two of its sides are equal in length.

To verify this using coordinate geometry:

• Find all three side lengths using the distance formula.
• Compare the three distances.
• If any two distances are equal → triangle is isosceles.

Solution Roadmap

• Assign points A, B, C.
• Compute AB, BC and AC carefully.
• Handle negative signs correctly.
• Compare all three distances.
• Conclude the type of triangle.

Solution

Let A(5, –2), B(6, 4) and C(7, –2)

Step 1: Find AB

$$\begin{aligned} AB &= \sqrt{(6 - 5)^2 + (4 - (-2))^2} \\ &= \sqrt{(1)^2 + (6)^2} \\ &= \sqrt{1 + 36} \\ &= \sqrt{37} \end{aligned}$$

Step 2: Find BC

$$\begin{aligned} BC &= \sqrt{(7 - 6)^2 + (-2 - 4)^2} \\ &= \sqrt{(1)^2 + (-6)^2} \\ &= \sqrt{1 + 36} \\ &= \sqrt{37} \end{aligned}$$

Step 3: Find AC

$$\begin{aligned} AC &= \sqrt{(7 - 5)^2 + (-2 - (-2))^2} \\ &= \sqrt{(2)^2 + (0)^2} \\ &= \sqrt{4 + 0} \\ &= 2 \end{aligned}$$

Step 4: Compare the sides

AB = √37,    BC = √37,    AC = 2

Since AB = BC, two sides are equal.

Final Conclusion: The given points form an isosceles triangle.

Exam Significance

• Frequently asked 2–3 mark question in CBSE Board exams.
• Tests understanding of distance formula + triangle classification.
• Important for competitive exams (NTSE, Olympiads).
• Can be extended to identify:
  • Equilateral triangle
  • Right-angled triangle

Common Mistakes to Avoid

• Incorrect subtraction involving negative numbers.
• Not checking all three sides.
• Arithmetic mistakes in squaring.
• Skipping final comparison step.

Concept Used

A quadrilateral is a square if:

• All four sides are equal
• Both diagonals are equal

We verify this using the distance formula.

\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Solution Roadmap

• Find all four sides: AB, BC, CD, DA
• Check if all sides are equal
• Find diagonals AC and BD
• Check if diagonals are equal
• Conclude whether ABCD is a square

Solution

Given points:
A(3, 4), B(6, 7), C(9, 4), D(6, 1)

Step 1: Find AB

$$\begin{aligned} AB &= \sqrt{(6 - 3)^2 + (7 - 4)^2} \\ &= \sqrt{3^2 + 3^2} \\ &= \sqrt{9 + 9} \\ &= \sqrt{18} \end{aligned}$$

Step 2: Find BC

$$\begin{aligned} BC &= \sqrt{(9 - 6)^2 + (4 - 7)^2} \\ &= \sqrt{3^2 + (-3)^2} \\ &= \sqrt{9 + 9} \\ &= \sqrt{18} \end{aligned}$$

Step 3: Find CD

$$\begin{aligned} CD &= \sqrt{(6 - 9)^2 + (1 - 4)^2} \\ &= \sqrt{(-3)^2 + (-3)^2} \\ &= \sqrt{9 + 9} \\ &= \sqrt{18} \end{aligned}$$

Step 4: Find DA

$$\begin{aligned} DA &= \sqrt{(3 - 6)^2 + (4 - 1)^2} \\ &= \sqrt{(-3)^2 + 3^2} \\ &= \sqrt{9 + 9} \\ &= \sqrt{18} \end{aligned}$$

Therefore, AB = BC = CD = DA = √18

Step 5: Find diagonals AC

$$\begin{aligned} AC &= \sqrt{(9 - 3)^2 + (4 - 4)^2} \\ &= \sqrt{6^2 + 0^2} \\ &= \sqrt{36} \\ &= 6 \end{aligned}$$

Step 6: Find diagonal BD

$$\begin{aligned} BD &= \sqrt{(6 - 6)^2 + (7 - 1)^2} \\ &= \sqrt{0^2 + 6^2} \\ &= \sqrt{36} \\ &= 6 \end{aligned}$$

AC = BD = 6

Final Conclusion:

• All four sides are equal
• Both diagonals are equal

Therefore, ABCD is a square.

Champa is correct.

Exam Significance

• Very important 3–4 mark question in CBSE Board exams.
• Tests multi-step reasoning (sides + diagonals).
• Frequently used in coordinate geometry proofs.
• Important for competitive exams and Olympiads.

Common Mistakes to Avoid

• Wrong subtraction order in coordinates.
• Forgetting to check diagonals.
• Arithmetic mistakes in squaring.
• Incorrect final conclusion without full verification.

Concept Used

• All sides equal + diagonals equal → Square
• Opposite sides equal → Parallelogram
• Three collinear points → No quadrilateral

Solution Roadmap

• Find all side lengths using distance formula
• Check equality patterns
• Check diagonals if required
• Check collinearity if shape is doubtful

(i) A(–1, –2), B(1, 0), C(–1, 2), D(–3, 0)

$$\begin{aligned} AB &= \sqrt{(1+1)^2 + (0+2)^2} = \sqrt{4+4} = 2\sqrt{2} \\ BC &= \sqrt{(-1-1)^2 + (2-0)^2} = \sqrt{4+4} = 2\sqrt{2} \\ CD &= \sqrt{(-3+1)^2 + (0-2)^2} = \sqrt{4+4} = 2\sqrt{2} \\ DA &= \sqrt{(-1+3)^2 + (-2-0)^2} = \sqrt{4+4} = 2\sqrt{2} \end{aligned}$$

All sides are equal.

$$\begin{aligned} AC &= \sqrt{(-1+1)^2 + (2+2)^2} = \sqrt{0+16} = 4 \\ BD &= \sqrt{(1+3)^2 + (0-0)^2} = \sqrt{16} = 4 \end{aligned}$$

Conclusion: Square

(ii) A(–3,5), B(3,1), C(0,3), D(–1,–4)

$$\begin{aligned} AB &= \sqrt{(-6)^2 + 4^2} = \sqrt{52} \\ BC &= \sqrt{3^2 + (-2)^2} = \sqrt{13} \\ AC &= \sqrt{3^2 + (-2)^2} = \sqrt{13} \end{aligned}$$

Since AB ≠ BC + AC and points A, B, C lie on same line (slope method also confirms), three points are collinear.

Conclusion: No quadrilateral formed

(iii) A(4,5), B(7,6), C(4,3), D(1,2)

$$\begin{aligned} AB &= \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{10} \\ BC &= \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{18} \\ CD &= \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{10} \\ DA &= \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{18} \end{aligned}$$

AB = CD and BC = DA

Conclusion: Parallelogram

Exam Significance

• Very important 4–5 mark question in CBSE Board exams
• Tests classification logic of quadrilaterals
• Frequently asked in NTSE and Olympiads
• Requires multi-step reasoning (distance + logic)

Common Mistakes to Avoid

• Incorrect subtraction with negative numbers
• Skipping diagonal verification for square
• Not checking collinearity
• Wrong classification based only on sides

Concept Used

• Any point on x-axis has coordinates (x, 0)
• Equidistant condition → distances are equal
• Use distance formula and solve algebraically

Solution Roadmap

• Assume point on x-axis as (x, 0)
• Apply distance formula from both given points
• Equate the two distances
• Solve the resulting equation step-by-step
• Write final coordinate

Solution

Let the required point on x-axis be P(x, 0)

Since P(x,0) is equidistant from A(2, –5) and B(–2, 9),

$$ \begin{aligned} \text{Distance PA} &= \text{Distance PB} \end{aligned} $$ $$ \begin{aligned} \sqrt{(x-2)^2 + (0+5)^2} &= \sqrt{(x+2)^2 + (0-9)^2} \end{aligned} $$

Squaring both sides:

$$ \begin{aligned} (x-2)^2 + 5^2 &= (x+2)^2 + (-9)^2 \end{aligned} $$ $$ \begin{aligned} (x-2)^2 + 25 &= (x+2)^2 + 81 \end{aligned} $$

Expanding both sides:

$$ \begin{aligned} (x^2 - 4x + 4) + 25 &= (x^2 + 4x + 4) + 81 \end{aligned} $$ $$ \begin{aligned} x^2 - 4x + 29 &= x^2 + 4x + 85 \end{aligned} $$

Bring like terms together:

$$ \begin{aligned} x^2 - 4x + 29 - x^2 - 4x - 85 &= 0 \\ -8x - 56 &= 0 \end{aligned} $$ $$ \begin{aligned} -8x &= 56 \\ x &= -7 \end{aligned} $$

Therefore, required point is:

(x, 0) = (–7, 0)

Final Answer

The required point on the x-axis is (–7, 0).

Exam Significance

• Very important 2–3 mark question in CBSE Board exams
• Tests equation formation using distance formula
• Frequently appears in NTSE and Olympiads
• Foundation for locus and coordinate geometry proofs

Common Mistakes to Avoid

• Forgetting that y = 0 for x-axis
• Sign errors in (0 + 5) and (0 − 9)
• Incorrect expansion of (x ± 2)²
• Skipping algebra steps leading to wrong answer

Concept Used

• Distance between two points is given by distance formula
• Equation involving unknown coordinate leads to quadratic form
• Taking square root introduces two possible cases (±)

Solution Roadmap

• Apply distance formula between P and Q
• Equate it to given distance (10 units)
• Square both sides to remove root
• Solve resulting equation carefully
• Consider both + and – cases

Solution

Given points: P(2, –3) and Q(10, y)

Using distance formula:

$$\begin{aligned} PQ &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \end{aligned}$$ $$\begin{aligned} 10 &= \sqrt{(10 - 2)^2 + (y - (-3))^2} \end{aligned}$$ $$\begin{aligned} 10 &= \sqrt{8^2 + (y + 3)^2} \end{aligned}$$

Squaring both sides:

$$\begin{aligned} 10^2 &= 8^2 + (y + 3)^2 \\ 100 &= 64 + (y + 3)^2 \end{aligned}$$ $$\begin{aligned} 100 - 64 &= (y + 3)^2 \\ 36 &= (y + 3)^2 \end{aligned}$$

Taking square root:

$$\begin{aligned} y + 3 &= \pm 6 \end{aligned}$$

Case 1:

$$\begin{aligned} y + 3 &= 6 \\ y &= 3 \end{aligned}$$

Case 2:

$$\begin{aligned} y + 3 &= -6 \\ y &= -9 \end{aligned}$$

Final Answer

Required values of y are: 3 and –9

Geometric Insight

The point Q lies on a vertical line x = 10. There are two possible points on this line at equal distance (10 units) from point P, one above and one below.

Exam Significance

• Common 2–3 mark question in CBSE Board exams
• Tests equation solving + distance formula
• Important for NTSE and Olympiads
• Builds base for locus problems

Common Mistakes to Avoid

• Sign error in (y + 3)
• Forgetting ± while taking square root
• Arithmetic mistakes in subtraction (100 − 64)
• Missing one of the two solutions

Concept Used

• Equidistant condition → QP = QR
• Use distance formula
• Square both sides to eliminate root
• Evaluate distances for each value of x

Solution Roadmap

• Apply distance formula between QP and QR
• Equate and simplify
• Solve for x
• Substitute each value of x to find QR and PR

Solution

Given: Q(0, 1), P(5, –3), R(x, 6)

Since Q is equidistant from P and R:

$$\begin{aligned} QP = QR \end{aligned}$$

Step 1: Find QP

$$\begin{aligned} QP &= \sqrt{(5 - 0)^2 + (-3 - 1)^2} \\ &= \sqrt{5^2 + (-4)^2} \\ &= \sqrt{25 + 16} \\ &= \sqrt{41} \end{aligned}$$

Step 2: Find QR

$$\begin{aligned} QR &= \sqrt{(x - 0)^2 + (6 - 1)^2} \\ &= \sqrt{x^2 + 5^2} \\ &= \sqrt{x^2 + 25} \end{aligned}$$

Step 3: Equate distances

$$\begin{aligned} \sqrt{41} &= \sqrt{x^2 + 25} \end{aligned}$$

Squaring both sides:

$$\begin{aligned} 41 &= x^2 + 25 \\ x^2 &= 16 \\ x &= \pm 4 \end{aligned}$$

Step 4: Find distances

For x = 4:

$$\begin{aligned} QR &= \sqrt{4^2 + 25} = \sqrt{16 + 25} = \sqrt{41} \\ PR &= \sqrt{(5 - 4)^2 + (-3 - 6)^2} \\ &= \sqrt{1^2 + (-9)^2} \\ &= \sqrt{1 + 81} \\ &= \sqrt{82} \end{aligned}$$

For x = –4:

$$\begin{aligned} QR &= \sqrt{(-4)^2 + 25} = \sqrt{16 + 25} = \sqrt{41} \\ PR &= \sqrt{(5 - (-4))^2 + (-3 - 6)^2} \\ &= \sqrt{9^2 + (-9)^2} \\ &= \sqrt{81 + 81} \\ &= 9\sqrt{2} \end{aligned}$$

Final Answer

• Values of x: 4 and –4
• Distance QR = √41 (for both cases)
• PR = √82 (when x = 4)
• PR = 9√2 (when x = –4)

Exam Significance

• Important 3–4 mark question in CBSE exams
• Tests equation formation + multi-case analysis
• Common in NTSE and Olympiads
• Builds base for locus and coordinate proofs

Common Mistakes to Avoid

• Forgetting to square both sides correctly
• Sign mistakes in (-3 − 1), (6 − 1)
• Not evaluating both values of x
• Incorrect PR calculation (very common error)

Concept Used

• Equidistant condition → distances are equal
• Distance formula is used to form equation
• Result represents a straight line (locus of points)

Geometric Insight

The required relation represents the perpendicular bisector of the line segment joining (3,6) and (–3,4).

Solution Roadmap

• Assume general point (x, y)
• Apply distance formula from both given points
• Equate both distances
• Square and simplify step-by-step
• Obtain linear relation in x and y

Solution

Let point P(x, y) be equidistant from A(3, 6) and B(–3, 4)

Since P(x, y) is equidistant from A and B:

$$\begin{aligned} PA = PB \end{aligned}$$ $$\begin{aligned} \sqrt{(x-3)^2 + (y-6)^2} = \sqrt{(x+3)^2 + (y-4)^2} \end{aligned}$$

Squaring both sides:

$$\begin{aligned} (x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2 \end{aligned}$$

Expanding both sides:

$$\begin{aligned} (x^2 - 6x + 9) + (y^2 - 12y + 36) \\ = (x^2 + 6x + 9) + (y^2 - 8y + 16) \end{aligned}$$ $$\begin{aligned} x^2 + y^2 - 6x - 12y + 45 \\ = x^2 + y^2 + 6x - 8y + 25 \end{aligned}$$

Cancel common terms:

$$\begin{aligned} -6x - 12y + 45 = 6x - 8y + 25 \end{aligned}$$

Bring like terms together:

$$\begin{aligned} -6x - 6x -12y + 8y + 45 - 25 = 0 \\ -12x - 4y + 20 = 0 \end{aligned}$$ $$\begin{aligned} 3x + y - 5 = 0 \end{aligned}$$

Final Answer

Required relation: 3x + y – 5 = 0

Exam Significance

• Very important 3–4 mark question in CBSE exams
• Tests equation formation and simplification
• Direct application of locus concept
• Frequently asked in NTSE and Olympiads

Common Mistakes to Avoid

• Sign errors in expansion (very common)
• Incorrect simplification after cancelling terms
• Missing final linear form
• Not identifying geometric meaning (perpendicular bisector)