Coordinate Geometry
Key Concepts
| # | Concept | Explanation |
|---|---|---|
| 1 | Distance Formula | Distance between A(x₁,y₁) & B(x₂,y₂) = √[(x₂–x₁)²+(y₂–y₁)²] |
| 2 | Section Formula | Point dividing AB in ratio m:n internally = [(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)] |
| 3 | Mid-Point | Mid-Point of AB = [(x₁+x₂)/2, (y₁+y₂)/2] |
| 4 | Centroid of Δ | Centroid G = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3] |
| 5 | Slope of Line | m = (y₂–y₁)/(x₂–x₁); horizontal→m=0, vertical→m=∞ |
| 6 | Equation of Line | y–y₁ = m(x–x₁) or ax+by+c=0 |
| 7 | Area of Triangle | ½ |
15 Practice MCQs
1. Distance between (3,4) and (–1,1) is
A) 3 B) 4 C) 5 D) 6 **Answer: C** Solution: √[(–1–3)²+(1–4)²]=√(16+9)=5 **Shortcut:** 3-4-5 Pythagorean triplet **Tag:** Distance Formula2. The mid-point of (7, –5) and (3, 9) is
A) (5,2) B) (5,–2) C) (2,5) D) (–2,5) **Answer: A** Solution: [(7+3)/2, (–5+9)/2] = (5,2) **Shortcut:** Average x and y separately **Tag:** Mid-Point3. A line makes 45° with the positive x-axis. Its slope is
A) 0 B) 1 C) –1 D) √3 **Answer: B** Solution: m = tan 45° = 1 **Shortcut:** tan θ gives slope directly **Tag:** Slope4. Area of triangle with vertices (0,0), (4,0), (0,3) is
A) 6 B) 12 C) 7 D) 5 **Answer: A** Solution: ½×base×height = ½×4×3 = 6 **Shortcut:** Right-angled ⇒ ½×product of legs **Tag:** Area5. The centroid of (2,1), (6,3), (4,9) is
A) (4,13/3) B) (12,13) C) (4,4) D) (4,13) **Answer: A** Solution: [(2+6+4)/3, (1+3+9)/3] = (12/3,13/3) **Shortcut:** Sum coordinates ÷3 **Tag:** Centroid6. Point dividing (1,2) and (7,8) internally in ratio 1:2 is
A) (3,4) B) (4,3) C) (5,6) D) (6,5) **Answer: A** Solution: x=(1×7+2×1)/3=9/3=3; y=(1×8+2×2)/3=12/3=4 **Shortcut:** Weighted average **Tag:** Section Formula7. Slope of line 3x–4y+12=0 is
A) 3/4 B) –3/4 C) 4/3 D) –4/3 **Answer: A** Solution: Rewrite y=(3/4)x+3 ⇒ m=3/4 **Shortcut:** For ax+by+c=0, m=–a/b **Tag:** Line Equation8. The value of k for which (2,k) lies on 5x–2y=10 is
A) 0 B) 5 C) –5 D) 2 **Answer: A** Solution: 5(2)–2k=10 ⇒ 10–2k=10 ⇒ k=0 **Shortcut:** Plug x, solve y **Tag:** Line Equation9. Distance of point (7,24) from origin is
A) 25 B) 24 C) 31 D) 30 **Answer: A** Solution: √(7²+24²)=√(49+576)=√625=25 **Shortcut:** 7-24-25 triplet **Tag:** Distance Formula10. If A(1,2), B(5,6), C(9,2), then ΔABC is
A) Equilateral B) Right C) Isosceles D) Scalene **Answer: C** Solution: AB=√32, BC=√32, AC=8 ⇒ AB=BC **Shortcut:** Compare distances **Tag:** Distance11. The line y=mx+c passes through (2,3) and (4,7). Find m.
A) 1 B) 2 C) 3 D) 4 **Answer: B** Solution: m=(7–3)/(4–2)=4/2=2 **Shortcut:** (y₂–y₁)/(x₂–x₁) **Tag:** Slope12. Area of quadrilateral with vertices (0,0), (3,0), (3,2), (0,2) is
A) 5 B) 6 C) 7 D) 8 **Answer: B** Solution: Rectangle 3×2 = 6 **Shortcut:** Count grid squares **Tag:** Area13. The reflection of (3,4) over x-axis is
A) (3,–4) B) (–3,4) C) (–3,–4) D) (4,3) **Answer: A** Solution: x same, y sign flipped **Shortcut:** x-axis refl ⇒ y→–y **Tag:** Reflection14. If slope of AB is ½ and A(2,–1), B(x,3), then x=
A) 6 B) 10 C) –6 D) 4 **Answer: B** Solution: (3–(–1))/(x–2)=½ ⇒ 4/(x–2)=½ ⇒ x–2=8 ⇒ x=10 **Shortcut:** Cross-multiply quickly **Tag:** Slope15. The point (–2,5) lies in which quadrant?
A) I B) II C) III D) IV **Answer: B** Solution: x negative, y positive **Shortcut:** II quadrant sign (–,+) **Tag:** QuadrantsSpeed Tricks
| Situation | Shortcut | Example |
|---|---|---|
| Right triangle vertices | Use ½×base×height instead of determinant | (0,0),(a,0),(0,b) ⇒ area=½ab |
| Collinearity check | Area = 0 or slopes AB = BC | Points (1,2),(3,4),(5,6) ⇒ slope always 1 |
| Centroid memory | “Add & divide by 3” | (1,1),(4,3),(7,5) ⇒ G=(12/3,9/3)=(4,3) |
| Slope from standard line | m = –(coeff of x)/(coeff of y) | 2x–5y+7=0 ⇒ m=2/5 |
| Distance on grid | Count Δx & Δy, look for Pythagorean triplets | (5,1)→(9,4): Δx=4, Δy=3 ⇒ dist=5 |
Quick Revision
| Point | Detail |
|---|---|
| 1 | Distance formula always gives positive value—take square root last. |
| 2 | Mid-point = average of coordinates. |
| 3 | Centroid divides median in 2:1; coordinates are simple mean. |
| 4 | Slope tan θ → θ=45° gives m=1; θ=0° gives m=0. |
| 5 | Horizontal line equation: y = k; vertical: x = k. |
| 6 | Area formula returns signed value—use absolute value for area. |
| 7 | Three points collinear ⇒ area of triangle = 0. |
| 8 | Reflection in x-axis: y → –y; in y-axis: x → –x. |
| 9 | Section formula works externally too—just use –m:n ratio. |
| 10 | Always sketch rough plot to visualise quadrant, slope or shape. |
BETA
