This MCQ module is based on: Distance Between Two Points in the Plane
TOPIC 2 OF 25
Distance Between Two Points in the Plane
🎓 Class 9
Mathematics
CBSE
Theory
Ch 1 — Orienting Yourself: The Use of Coordinates
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Distance Between Two Points in the Plane
Targeting Class 9 level in Coordinate Geometry, with Intermediate difficulty.
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1.4 Distance Between Two Points on an Axis
Suppose two points lie on the same axis. Their separation is just the absolute difference of the non-zero coordinates.
Distance on an Axis
For points \((x_1, y)\) and \((x_2, y)\) on a line parallel to the x-axis, distance \(= |x_2 - x_1|\).For points \((x, y_1)\) and \((x, y_2)\) on a line parallel to the y-axis, distance \(= |y_2 - y_1|\).
Example: Distance between \(A(2, 3)\) and \(B(7, 3)\) is \(|7-2| = 5\) units. Distance between \(P(4, -1)\) and \(Q(4, 6)\) is \(|6-(-1)| = 7\) units.
1.5 Distance Between Two Points in the Plane
What happens when the two points have both coordinates different? We borrow an idea more than 2,500 years old: the Baudhāyana–Pythagoras Theorem?.
Historical Note
The Baudhāyana Sulba-sūtra (c. 800 BCE), an ancient Indian geometry manual for altar construction, states: "The diagonal of a rectangle produces by itself both the areas which its two sides produce separately." This is precisely \(c^2 = a^2 + b^2\) — written more than 200 years before Pythagoras.
Deriving the Distance Formula
Let \(A(x_1, y_1)\) and \(B(x_2, y_2)\) be any two points. Drop a perpendicular from \(A\) and another from \(B\); they meet at \(C(x_2, y_1)\). Triangle \(ACB\) is right-angled at \(C\) (Fig 1.7).
Fig 1.7: Distance derivation using a right triangle
By Baudhāyana–Pythagoras: \(AB^2 = AC^2 + BC^2\)
\(AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\)
Taking the positive square root (because distance cannot be negative):
\(\boxed{\;AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\;}\)
Distance Formula
The distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) in the plane is
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
The order of subtraction does not matter because of the squaring: \((x_2-x_1)^2 = (x_1-x_2)^2\).
Worked Examples
Example 1: Find the distance between \(L(2, 5)\) and \(M(8, 5)\).
Both have \(y = 5\), so \(LM = |8-2| = 6\) units. (Using the formula: \(\sqrt{36 + 0} = 6\).)
Example 2: Find the distance between \(D(7, 1)\) and \(M(5, 5)\).
\(DM = \sqrt{(5-7)^2 + (5-1)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\) units \(\approx 4.47\).
Example 3: Find the distance from origin \(O(0,0)\) to \(A(3, 4)\).
\(OA = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) units. (The classic 3-4-5 triangle.)
Example 4: Show that \(P(1, 5)\), \(Q(2, 3)\), \(R(-2, -11)\) are not collinear (they don't lie on a single line).
\(PQ = \sqrt{1 + 4} = \sqrt{5}\). \(QR = \sqrt{16 + 196} = \sqrt{212}\). \(PR = \sqrt{9 + 256} = \sqrt{265}\). For collinearity we'd need \(PQ + QR = PR\), but \(\sqrt{5}+\sqrt{212} \approx 2.24+14.56 = 16.8\) while \(\sqrt{265}\approx 16.28\). The three are not collinear.
Section 1.5 Exercises
Q1. Find the distance between \(A(2, 3)\) and \(B(4, 1)\).
\(\sqrt{(4-2)^2+(1-3)^2}=\sqrt{4+4}=2\sqrt{2}\approx 2.83\) units.
Q2. Find a point on the y-axis equidistant from \((-5, -2)\) and \((3, 2)\).
Let the point be \((0, y)\). Setting squares of distances equal: \(25 + (y+2)^2 = 9 + (y-2)^2\). Expanding: \(25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4\), so \(8y = -16\), \(y = -2\). Point: \((0, -2)\).
Q3. Determine whether \(A(1, 5)\), \(B(2, 3)\), \(C(-2, -11)\) form a right triangle.
\(AB^2 = 5\), \(BC^2 = 212\), \(AC^2 = 265\). Since \(5+212 = 217 \neq 265\), they don't satisfy Pythagoras → not a right triangle.
Activity: Verifying Pythagoras with String
L3 Apply
Materials: Graph paper, ruler, string, scissors
Predict: If you draw a right triangle with legs 6 cm and 8 cm, can you guess the hypotenuse without measuring?
- On graph paper, mark \(O(0,0)\), \(A(6,0)\), \(B(0,8)\). Cut a string equal to \(OA\) and another equal to \(OB\).
- Compute \(AB\) using the distance formula: \(\sqrt{36+64} = \sqrt{100} = 10\) cm.
- Cut a third string of length 10 cm and place it from \(A\) to \(B\). Does it fit exactly?
- Repeat for points \(P(2,3)\) and \(Q(5,7)\). Predict, then verify.
For \(P(2,3)\) and \(Q(5,7)\): \(\sqrt{9 + 16} = 5\) cm. The string method always matches the formula because the formula is the Pythagorean theorem in coordinate language.
Competency-Based Questions
Scenario: A city park is laid out on a coordinate grid (1 unit = 100 m). Three monuments stand at \(M_1(2, 3)\), \(M_2(8, 3)\), and \(M_3(5, 7)\). A walking path connects them in a triangle.
Q1. What is the total length of the walking path \(M_1 M_2 M_3 M_1\)?
L3 Apply\(M_1M_2 = 6\), \(M_2M_3 = \sqrt{9+16} = 5\), \(M_3M_1 = \sqrt{9+16} = 5\). Total \(= 16\) units = 1600 m.
Q2. Analyse what kind of triangle the three monuments form.
L4 AnalyseTwo sides are 5 units long → it is isosceles. Checking Pythagoras: \(5^2 + 5^2 = 50 \neq 6^2 = 36\) → not right-angled. So an isosceles, acute triangle.
Q3. A planner claims, "If we move \(M_3\) to \((5, 6)\), the path length stays the same." Evaluate.
L5 EvaluateNew \(M_2M_3 = \sqrt{9+9} = \sqrt{18} \approx 4.24\) and new \(M_3M_1 = \sqrt{9+9} \approx 4.24\). Total \(\approx 6 + 4.24 + 4.24 = 14.48\) — not 16. The planner is wrong; lowering \(M_3\) by 1 unit shortens both connecting paths.
Q4. Design coordinates for a fourth monument \(M_4\) so that \(M_1 M_2 M_3 M_4\) becomes a parallelogram. Show your reasoning.
L6 CreateFor a parallelogram \(M_1 M_2 M_3 M_4\), opposite sides must be equal & parallel. Vector \(M_1 M_2 = (6, 0)\). So \(M_4 = M_3 + (-6, 0) = (-1, 7)\) — many other valid arrangements exist depending on vertex order.
Assertion–Reason Questions
A: The distance from \((-3, 4)\) to the origin is 5 units.
R: Distance from origin = \(\sqrt{x^2 + y^2}\).
R: Distance from origin = \(\sqrt{x^2 + y^2}\).
Answer: (a) — \(\sqrt{9+16}=5\). R explains A.
A: Distance is always positive.
R: The square root in the distance formula always returns a non-negative real number.
R: The square root in the distance formula always returns a non-negative real number.
Answer: (a) — Both correct, and R is the algebraic reason for A. (Also distance can be 0 only if the points coincide.)
A: The distance between \((1, 2)\) and \((4, 6)\) equals the distance between \((4, 6)\) and \((1, 2)\).
R: \((x_2 - x_1)^2 = (x_1 - x_2)^2\) since squaring removes the sign.
R: \((x_2 - x_1)^2 = (x_1 - x_2)^2\) since squaring removes the sign.
Answer: (a) — Both 5 units. R correctly explains why order doesn't matter.
Frequently Asked Questions
What is the distance formula for two points in Class 9 Maths?
The distance between P(x1, y1) and Q(x2, y2) is d = sqrt((x2-x1)^2 + (y2-y1)^2). It is derived using the Baudhayana-Pythagoras theorem on a right triangle formed by the horizontal and vertical gaps.
How is the distance formula derived?
Drop a perpendicular from each point so they meet at C(x2, y1), forming a right triangle. The legs are |x2-x1| and |y2-y1|. By Pythagoras, AB^2 = (x2-x1)^2 + (y2-y1)^2, and taking the positive square root gives the distance formula.
Why does the order of subtraction not matter in the distance formula?
Because (x2-x1)^2 = (x1-x2)^2 (squaring removes the sign). So whether you compute the gap from P to Q or Q to P, the distance is the same non-negative value.
What is the distance from the origin to a point (x, y)?
The distance from the origin O(0,0) to any point A(x, y) is OA = sqrt(x^2 + y^2). For example, the distance from the origin to (3, 4) is sqrt(9+16) = 5 units.
How do you check if three points are collinear using the distance formula?
Compute the three pairwise distances. If the sum of the two smaller distances equals the largest distance (PQ + QR = PR), the three points are collinear. Otherwise they form a triangle.
What is the link between the distance formula and the Baudhayana Sulba-sutra?
The Baudhayana Sulba-sutra (c. 800 BCE), an ancient Indian altar-construction manual, states that the square on the diagonal of a rectangle equals the sum of the squares on its two sides, which is exactly c^2 = a^2 + b^2. The distance formula is this rule expressed in coordinates.
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