311_Maths_Eng_Lesson13.pdf). Content covers sections 13.1–13.11.This NIOS Mathematics (311) lesson opens Module IV: Co-ordinate Geometry. It fixes how we locate a point in a plane using two perpendicular axes, then builds the tools you will reuse for straight lines, circles and conics: distance, section formula, area, collinearity, slope, intercepts, angle between lines, and translation of axes. Notes follow the textbook order from rectangular axes through shifting of origin.
Expected background (textbook): number system; plotting points; graphs of linear equations; solving linear systems. Treat every later coordinate problem as a combination of the formulae in this chapter.
To fix the position of a point in a plane we draw two mutually perpendicular lines intersecting at a fixed point O, the origin. These lines are the coordinate axes. The horizontal line X′OX is the x-axis (axis of x). The vertical line Y′OY is the y-axis (axis of y).
Let P be a point in the plane of axes X′OX and Y′OY. Drop perpendiculars from P to the axes. The directed distance along the x-axis is the abscissa (x-coordinate). The directed distance along the y-axis is the ordinate (y-coordinate). Together they form the ordered pair P(x, y).
Order matters: (3, 2) and (2, 3) are different points. In general (x, y) ≠ (y, x).
The axes divide the plane into four regions called quadrants (textbook spelling “quardrants”). Sign convention for P(x, y):
| Quadrant | x | y |
|---|---|---|
| I | x > 0 | y > 0 |
| II | x < 0 | y > 0 |
| III | x < 0 | y < 0 |
| IV | x > 0 | y < 0 |
Points on the axes have one coordinate zero (e.g. (a, 0) on x-axis; (0, b) on y-axis) and are not assigned to a single open quadrant in the usual sign table.
For P(x₁, y₁) and Q(x₂, y₂), form a right triangle by drawing a horizontal through P and a vertical through Q. Horizontal leg |x₂ − x₁|, vertical leg |y₂ − y₁|. By Pythagoras:
Example 13.1 (textbook): Distance A(14, 3) to B(10, 6) = √[(10−14)² + (6−3)²] = √(16+9) = √25 = 5. Distance M(−1, 2) to N(0, −6) = √[(0−(−1))² + (−6−2)²] = √[1 + 64] = √65.
Example 13.2: P(−1,−1), Q(2,3), R(−2,6). Compute PQ² = (2+1)²+(3+1)² = 9+16 = 25; QR² = (−2−2)²+(6−3)² = 16+9 = 25; RP² = (−1+2)²+(−1−6)² = 1+49 = 50. Since PQ² + QR² = RP², △PQR is right-angled at Q (converse of Pythagoras).
Example 13.3 (collinear by distance): A(1,2), B(4,5), C(−1,0). AB = 3√2, BC = 5√2, AC = 2√2. Then AB + AC = 5√2 = BC, so A, B, C are collinear.
Example 13.4 (equilateral): A(2a,4a), B(2a,6a), C(2a+√3 a, 5a) (as in text). All three sides equal 2a, and triangle inequalities hold, so the triangle is equilateral of side 2a.
Check Your Progress 13.1 ideas: Find distances such as (5,4) to (2,−3); (a,−a) to (b,b). Prove right-angled triangles by side squares; prove collinearity of (3,−6), (2,−4), (−4,8); recognise rectangle/square vertices by equal sides and right angles (or slopes).
When proving a rectangle, show opposite sides equal and diagonals equal (or adjacent sides perpendicular). For a square, add equal diagonals and adjacent sides equal with one right angle (or both pairs of adjacent sides equal and perpendicular).
Let R(x, y) divide the segment joining P(x₁, y₁) and Q(x₂, y₂) internally in the ratio m₁ : m₂ (PR : RQ = m₁ : m₂). By similar triangles:
Mid-point: m₁ = m₂ = 1 gives ((x₁+x₂)/2, (y₁+y₂)/2).
If R divides PQ externally in m₁ : m₂, the point lies on the line PQ but outside the segment. Formula:
Example 13.5 pattern: Find the point dividing the join of (x₁,y₁) and (x₂,y₂) in m₁:m₂ internally — substitute directly into the internal formula.
Example 13.6 pattern: Given a point R(3,−2) on the join of two points, find the ratio λ:1 by setting R’s coordinates equal to the section formula and solving for λ. If λ is positive, division is internal; if the form indicates external, use the external formula.
Example 13.7 (quadrilateral): Vertices A(1,4), B(−2,1), C(0,−1), D(…): show that diagonals bisect each other by proving mid-point of AC = mid-point of BD (parallelogram test).
Applications: centroid of a triangle divides each median in 2:1 (internal, m₁:m₂ = 2:1 from vertex to mid-point of opposite side). Mid-point formula is the special case used constantly in geometry proofs.
For vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃):
Example 13.8: Area of triangle with vertices A(3,4), B(6,−2), C(…): plug into the formula carefully with signs, then take absolute value and multiply by ½.
Example 13.9: Vertices (1, k), (4, −3), (−9, 7) and area 15. Compute ½|…| = 15 ⇒ |expression| = 30. In the textbook solution one branch yields k = −3 (and a second branch may appear from the ±). Always solve both cases unless the question restricts k.
Unit: square units. Order of vertices does not matter because of the absolute value, but keep a consistent cyclic order when expanding to avoid arithmetic slips.
Three points A, B, C are collinear if and only if the area of △ABC is zero:
x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) = 0 (without absolute value — the expression is zero).
Equivalent slope test (later section): slope of AB = slope of BC (when segments defined).
Use for finding k so that (1, 5), (k, 1), (4, 11) are collinear: plug into area = 0 and solve for k.
The inclination of a line is the angle θ that the line makes with the positive direction of the x-axis, 0 ≤ θ < 180° (or 0° ≤ θ < π). The slope (gradient) is:
If a line joins distinct points (x₁, y₁) and (x₂, y₂) with x₂ ≠ x₁:
If a line is equally inclined to both axes, θ = 45° or 135°, so m = ±1 (Example 13.15 idea).
Example patterns: Show line through A(5,6), B(2,3) is parallel to another join by equal slopes. Show A(2,−5), B(−2,5) is perpendicular to a given line by product −1. Using slopes, show A(4,4), B(3,5), C(−1,−1) form a right angle. Find y so that the line through A(3,y) and a second point is parallel (or perpendicular) to a given line.
Collinear by slope: Points A(6,−1), B(5,0), C(2,3) are collinear if slope AB = slope BC = slope CA (where defined).
Exam use: show three points form a right angle by product of slopes of two sides = −1; show a quadrilateral has perpendicular diagonals; find k so a join through (k,9) and (2,7) is parallel to the join of (2,−2) and (6,4).
If a line meets the x-axis at (a, 0) and the y-axis at (0, b), then a is the x-intercept and b is the y-intercept. For the line written ax + by + c = 0 (with a ≠ 0, b ≠ 0):
x-intercept = −c/a · y-intercept = −c/b.
Intercept form: x/a + y/b = 1 (when intercepts are a and b, nonzero).
If two lines have slopes m₁ and m₂ and are not perpendicular in a way that makes the formula fail (1 + m₁m₂ ≠ 0), the acute/obtuse angles φ between them satisfy:
If 1 + m₁m₂ = 0, the lines are perpendicular (φ = 90°). Textbook examples: slopes 3 and 1/2; angle between x-axis (m = 0) and a join; given tan of angle and one slope, find the other.
When the origin is translated to a new origin O′(h, k) without rotating the axes, a point with old coordinates (x, y) has new coordinates (x′, y′) related by:
Example 13.26 idea: Shift origin to (−3, 2); transform a point or equation by substitution. Example 13.27: Origin to (3, 4); line 3x + 2y − 5 = 0 becomes 3(x′+3) + 2(y′+4) − 5 = 0 → 3x′ + 2y′ + 12 = 0.
From the textbook terminal exercises and check-your-progress blocks, expect: (1) pure distance and collinearity; (2) section and mid-point in polygons; (3) area and parameter k; (4) slope, parallel/perpendicular, and angle between lines; (5) intercepts of ax+by+c=0; (6) translation of a point or a line equation after shifting origin to (h,k).
Worked habit: box the formula first, substitute with labelled x₁,y₁,x₂,y₂, simplify radicals only at the end, and write a one-line conclusion (“hence collinear”, “hence right-angled at Q”, “hence m₁m₂=−1”).
Write every answer with clear substitution of (x₁,y₁), (x₂,y₂). State which formula (distance, section, area, slope, translation). For geometry proofs (right triangle, rectangle, square, collinear), compute sides or slopes and quote the criterion. Keep signs careful in external section and area absolute value. Next lessons (straight lines, circles) reuse slope and distance constantly — this chapter is the foundation of Module IV.
Link forward: Lesson 14 (Straight Lines) will write equations using slope and intercepts you now know; distance from a point to a line builds on this geometry. Keep the distance and section formulae on your formula sheet until they become automatic.
Most exam-important points from this chapter:
Always write (x, y) in that order. (3,2) and (2,3) are different points.
Use √[(Δx)²+(Δy)²] in all quadrants; check right triangles with PQ²+QR²=RP².
Internal: plus in numerator and denominator. External: minus; m₁ ≠ m₂.
Area formula uses |…|; collinearity when the expression inside is zero.
m = tan θ = rise/run; parallel m₁=m₂; perpendicular m₁m₂=−1; angle via tan φ formula.
Substitute x = x′+h, y = y′+k into the equation; do not rotate axes in this lesson.
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
5 question(s) · Sources: Apr 2024, Oct 2024, Sample QP 2024
PYQ1. The coordinates of the mid-point of A(4, −1) and B(7, 2) are:
Model Answer
1) (11/2, 1/2)
Explanation
1) Mid-point M = ((x₁+x₂)/2, (y₁+y₂)/2) = ((4+7)/2, (−1+2)/2) = (11/2, 1/2).
PYQ2. The slope of the line segment joining A(2, 3) and B(6, −7) is:
Model Answer
1) −5/2
Explanation
1) m = (y₂−y₁)/(x₂−x₁) = (−7−3)/(6−2) = (−10)/4 = −5/2.
PYQ3. The coordinates of the centroid of the triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) are:
Model Answer
1) ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
Explanation
1) The centroid is the average of the three vertices’ coordinates (section formula with equal weights).
PYQ4. The points A(−1, −1), B(2, 3) and C(−2, 6) are the vertices of:
Model Answer
1) an isosceles right triangle
Explanation
1) AB²=(3)²+(4)²=25, BC²=(−4)²+(3)²=25, AC²=(−1)²+(7)²=
2) 5
0. AB=BC and AB²+BC²=AC² ⇒ isosceles right-angled at B.
PYQ5. Find the distance between the points A(−2, 3) and B(4, −5).
Model Answer
1) 10 units
Explanation
1) PQ = √[(4−(−2))²+(−5−3)²] = √(6²+(−8)²) = √(36+64) = √100 = 10.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Find the distance between A(14, 3) and B(10, 6).
1) Use PQ = √[(x₂−x₁)² + (y₂−y₁)²] with A(14,3), B(10,6).
2) Δx = 10 − 14 = −4, Δy = 6 − 3 = 3.
3) PQ = √[(−4)² + 3²] = √(16+9) = √25 = 5.
Answer: 5
By the distance formula in the Cartesian plane, the length of segment AB is √[(10−14)²+(6−3)²]=√25=5.
Horizontal gap 4, vertical gap 3 → 3-4-5 triangle, so length is 5.
Distance works in all quadrants because squares remove sign.
Square first, then add, then square-root.
Find the mid-point of the segment joining (−1, 2) and (5, −4).
1) M_x = (−1+5)/2 = 4/2 = 2.
2) M_y = (2+(−4))/2 = (−2)/2 = −1.
3) Hence M = (2, −1).
Answer: (2, −1)
The mid-point formula averages corresponding coordinates: ((−1+5)/2,(2−4)/2)=(2,−1).
Average the x’s: 2. Average the y’s: −1.
Mid-point is the internal section formula with m₁=m₂=1.
Watch the sign of y when adding 2+(−4).
Find the point that divides the join of P(2, −3) and Q(6, 5) internally in the ratio 1 : 3.
1) Here m₁=1, m₂=3, (x₁,y₁)=(2,−3), (x₂,y₂)=(6,5).
2) x = (1·6 + 3·2)/(1+3) = (6+6)/4 = 12/4 = 3.
3) y = (1·5 + 3·(−3))/(1+3) = (5−9)/4 = (−4)/4 = −1.
4) Required point is (3, −1).
Answer: (3, −1)
Using the internal section formula with ratio 1:3 gives ((1·6+3·2)/4, (1·5+3·(−3))/4)=(3,−1).
Weighted average: closer to P because the ratio toward Q is smaller (1 vs 3).
Internal uses + in numerator and denominator; external uses −.
Keep the order m₁:m₂ matching P:Q as stated in the formula.
Show that A(1,2), B(4,5), C(−1,0) are collinear using the area formula.
1) Compute Δ = 1(5−0)+4(0−2)+(−1)(2−5) = 1·5 + 4·(−2) + (−1)·(−3).
2) Δ = 5 − 8 + 3 = 0.
3) Area = ½|0| = 0 ⇒ points are collinear.
Answer: Area = 0 ⇒ A, B, C collinear
The determinant/area expression vanishes, so the three points determine a degenerate triangle and are collinear.
Plug into the area formula; get zero, so they lie on one straight line.
Collinearity ⇔ area zero ⇔ slopes AB = AC (alternative test).
Include the absolute value only for positive area; for collinearity, zero is enough.
Find the slope of the line joining (2, 3) and (6, −1). Hence find the slope of a line perpendicular to it.
1) m = (−1 − 3)/(6 − 2) = (−4)/4 = −1.
2) If m₁ = −1 and the perpendicular has slope m₂, then m₁m₂ = −1 ⇒ (−1)m₂ = −1 ⇒ m₂ = 1.
Answer: m = −1; perpendicular slope = 1
Slope of the given line is −1; any perpendicular line has slope 1 because the product of slopes is −1.
Rise −4 over run 4 gives slope −1; flip and change sign → perpendicular slope 1.
Vertical/horizontal special cases: undefined slope ⟂ slope 0.
Use m₁m₂=−1 only when both slopes are defined.
If the origin is shifted to (2, −1) without rotating axes, find the new coordinates of the point (5, 3).
1) Here (h,k)=(2,−1) and old coordinates (x,y)=(5,3).
2) x′ = x − h = 5 − 2 = 3.
3) y′ = y − k = 3 − (−1) = 4.
4) New coordinates are (3, 4).
Answer: (3, 4)
Under translation of origin to (h,k), x′=x−h and y′=y−k, giving (3,4).
Subtract the new origin from the old coordinates: (5−2, 3−(−1))=(3,4).
Translation does not change distances or slopes of lines.
Remember y′ = y − k, so minus a negative becomes plus.
Two towns are marked on a map at A(−2, 3) and B(4, −5). Find the map distance between them.
1) Δx = 4 − (−2) = 6, Δy = −5 − 3 = −8.
2) Distance = √(6² + (−8)²) = √(36+64) = √100 = 10.
Answer: 10 units
By the Cartesian distance formula, AB = √[(4+2)²+(−5−3)²] = 10.
Gaps 6 and 8 form a 6-8-10 triangle.
Squares remove signs of Δx and Δy.
Simplify the square root fully.
Find the centroid of the triangle with vertices (1, 2), (3, 4) and (5, 0).
1) x̄ = (1+3+5)/3 = 9/3 = 3.
2) ȳ = (2+4+0)/3 = 6/3 = 2.
3) Centroid G = (3, 2).
Answer: (3, 2)
The centroid is the average of the three vertices’ coordinates.
Average x is 3; average y is 2.
Centroid divides each median in 2:1 ratio.
Do not use mid-point formula alone.
Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 6).
1) Δ = 0(0−6)+4(6−0)+0(0−0) = 0 + 24 + 0 = 24.
2) Area = ½|24| = 12.
Answer: 12 square units
Using the coordinate area formula gives 12.
Base 4, height 6 ⇒ area ½×4×6=12 (same result).
Right triangle on the axes.
Include the factor ½.
A ramp joins floor point (1, 2) to platform (5, 10). Find the slope of the ramp.
1) m = (10 − 2)/(5 − 1) = 8/4 = 2.
Answer: 2
Slope is rise over run: 8/4 = 2.
For every 1 unit right, height rises by 2.
Positive slope means the line rises to the right.
Order of points does not change the slope.