This MCQ module is based on: Introduction to Coordinate Geometry and Distance Formula
TOPIC 17 OF 35
Introduction to Coordinate Geometry and Distance Formula
🎓 Class 10
Mathematics
CBSE
Theory
Ch 7 — Coordinate Geometry
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Introduction to Coordinate Geometry and Distance Formula
Targeting Class 10 level in Coordinate Geometry, with Intermediate difficulty.
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7.1 Introduction
In Class IX you have studied how to locate the position of a point on a plane. You require two references — one perpendicular to the other — which are known as the x-coordinatei (abscissa) and the y-coordinatei (ordinate). An ordered pair \((x,y)\) fixes the point uniquely in the Cartesian plane.
In this chapter we build on that foundation. We shall learn to:
- Find the distance between two points whose coordinates are given.
- Find the coordinates of a point that divides a line segment in a given ratio — the section formula.
Coordinate geometry, invented in the 17th century by René Descartes, links algebra with geometry. Navigation, seismology, satellite imaging and architecture all rely on it.
Recall
If \(P(x,y)\) lies in the plane: \(x\) is the perpendicular distance from the \(y\)-axis, and \(y\) is the perpendicular distance from the \(x\)-axis. Quadrants: I(+,+), II(−,+), III(−,−), IV(+,−).
Fig. 7.1 — Cartesian plane with point P(4, 3)
7.2 Distance Formula
Consider the following situation. A town B is located 36 km east and 15 km north of a town A. How would you find the distancei from town A to town B without actually measuring it?
Place A at the origin. Then B has coordinates (36, 15). By the Pythagoras theorem applied to the right triangle formed,
\[AB=\sqrt{36^2+15^2}=\sqrt{1296+225}=\sqrt{1521}=39\text{ km}.\]Fig. 7.2 — Right triangle yields the distance by Pythagoras
Deriving the general formula
Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) be any two points in the plane. Draw PR and QS perpendicular to the x-axis, and PT perpendicular from P to QS (see Fig. 7.5).
Then \(PT=\text{OS}-\text{OR}=x_2-x_1\) and \(QT=\text{QS}-\text{TS}=y_2-y_1\). Triangle PTQ is right-angled at T. By Pythagoras:
\[PQ^2=PT^2+QT^2=(x_2-x_1)^2+(y_2-y_1)^2.\]Distance Formula
\[\boxed{\;PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\;}\]
Distance is always the positive square root.
Remarks. (1) The distance of \(P(x,y)\) from the origin O(0, 0) is \(OP=\sqrt{x^2+y^2}\). (2) The formula may be written as \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) since the squares remove sign.
Fig. 7.5 — General derivation of the distance formula
Activity — Discover the Distance Formula
Materials: graph paper, ruler, pencil
Predict: If a point P is 3 units right and 4 units up from Q, how far is P from Q?
- Plot Q(2, 1) and P(5, 5) on graph paper.
- Draw PQ, then drop a perpendicular from P to the horizontal through Q — call the foot T.
- Measure QT and PT. Compute \(\sqrt{QT^2+PT^2}\).
- Apply the distance formula directly and compare.
QT = 5 − 2 = 3, PT = 5 − 1 = 4, so PQ = √(9+16) = √25 = 5. The formula gives the same answer — the prediction is confirmed.
Example 1 — Quadrant check
Do the points (3, 2), (−2, −3) and (2, 3) form a triangle? If yes, of which type?
Solution. Let A(3, 2), B(−2, −3), C(2, 3). Using the distance formula:
\[AB=\sqrt{(3+2)^2+(2+3)^2}=\sqrt{50}=5\sqrt2,\;\;BC=\sqrt{16+36}=\sqrt{52},\;\;CA=\sqrt{1+1}=\sqrt2.\]Since the sum of any two side-lengths is greater than the third, the points form a triangle. Further, \(AB^2+CA^2=50+2=52=BC^2\); hence it is a right triangle, right-angled at A.
Example 2 — Collinearity
Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.
Solution. Label them A, B, C, D. Compute
\[AB=\sqrt{9+25}=\sqrt{34},\;\;BC=\sqrt{25+9}=\sqrt{34},\;\;CD=\sqrt{9+25}=\sqrt{34},\;\;DA=\sqrt{25+9}=\sqrt{34}.\]All four sides are equal. Diagonals: \(AC=\sqrt{4+64}=\sqrt{68}\), \(BD=\sqrt{64+4}=\sqrt{68}\) — equal diagonals with equal sides ⇒ a square.
Example 3 — Using the distance formula to find an unknown
Find a point on the y-axis which is equidistant from A(6, 5) and B(−4, 3).
Solution. Any point on the y-axis is P(0, y). Then \(PA=PB\Rightarrow PA^2=PB^2\):
\[(0-6)^2+(y-5)^2=(0+4)^2+(y-3)^2\] \[36+y^2-10y+25=16+y^2-6y+9\Rightarrow -4y=-36\Rightarrow y=9.\]Required point: (0, 9).
Competency-Based Questions
A city planner places three landmarks on a map at L(2, 3), M(8, 11) and N(14, 3). All distances are in km.
Q1. Classify triangle LMN by its sides and justify numerically.
L3 ApplyLM = √(36+64) = 10, MN = √(36+64) = 10, LN = √(144+0) = 12. Since LM = MN, triangle LMN is isosceles.
Q2. Is angle at M a right angle? Justify.
L4 AnalyseFor a right angle at M we would need LM² + MN² = LN². Here 100 + 100 = 200 ≠ 144. So angle M is not a right angle.
Q3. A water-pump must be installed equidistant from all three landmarks on the line x = 8. Find its y-coordinate.
L5 EvaluateLet P(8, k). PL² = PN² gives 36 + (k−3)² = 36 + (k−3)², already equal. PL² = PM²: 36 + (k−3)² = 0 + (k−11)² ⇒ 36 + k² − 6k + 9 = k² − 22k + 121 ⇒ 16k = 76 ⇒ k = 4.75. P(8, 4.75).
Q4. Propose a new landmark K such that LMNK is a parallelogram with K opposite M. Justify by checking opposite-side lengths.
L6 CreateIn parallelogram LMNK, diagonals bisect. Midpoint of LN = midpoint of MK ⇒ ((2+14)/2, (3+3)/2) = ((8+x)/2, (11+y)/2) ⇒ K(8, −5). Check: LK = √(36+64) = 10 = MN ✓, MK = √(0+256) = 16… wait recompute: MK = √(0+256)=16 while LN=12 — diagonals need not be equal in a parallelogram. Opposite sides LM=10=NK and MN=10=LK ✓. So K(8, −5) works.
Assertion & Reason
Assertion (A): The point (0, −5) is equidistant from (3, 0) and (−3, 0).
Reason (R): A point on the y-axis is equidistant from two points symmetric about the y-axis.
Reason (R): A point on the y-axis is equidistant from two points symmetric about the y-axis.
Answer: A. Distance from (0, −5) to each of (3, 0) and (−3, 0) is √(9+25) = √34. The y-axis is the perpendicular bisector of a segment whose ends are symmetric about it, so every y-axis point is equidistant — R is the explanation.
Assertion (A): The points (1, 2), (4, 6) and (7, 10) are collinear.
Reason (R): Three points are collinear if the sum of the distances between two pairs equals the third pair's distance.
Reason (R): Three points are collinear if the sum of the distances between two pairs equals the third pair's distance.
Answer: D. AB = 5, BC = 5, AC = √(36+64) = 10. Here AB + BC = 10 = AC, so the three points A, B, C are indeed collinear — wait then A is true too. Re-examine: points (1,2),(4,6),(7,10) — AB=√(9+16)=5, BC=5, AC=√(36+64)=10. Since AB+BC=AC, they ARE collinear. So Answer = A (both true, R explains A).
Frequently Asked Questions
What is coordinate geometry?
Coordinate geometry is the branch of mathematics that uses a pair of numbers (coordinates) to describe the position of a point in a plane, enabling us to study geometric figures through algebra.
How is the distance formula derived?
By applying the Pythagoras theorem to the right triangle whose hypotenuse joins the two given points, with horizontal leg |x2 minus x1| and vertical leg |y2 minus y1|.
What is the distance between the origin and a point (a, b)?
It is the square root of (a squared plus b squared), obtained by applying the distance formula with the origin (0, 0) as one of the points.
Can the distance formula give a negative result?
No. Distance is always non-negative because squares of real numbers are non-negative and the square root returns the positive value.
How do you verify if three points are collinear using the distance formula?
Compute the three pairwise distances. If the sum of the two smaller distances exactly equals the largest distance, the three points lie on a straight line.
What are the applications of coordinate geometry?
Coordinate geometry is used in computer graphics, GPS navigation, physics, engineering design, robotics and wherever positions and shapes must be represented numerically.
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