Class 10 Maths Coordinate Geometry – NCERT Formulas & Solutions

Class 10 Maths — Chapter 7: Coordinate Geometry

Prepared by FUZY MATH ACADEMY — Classes 5 to 12. Full chapter guide, 25 solved problems (NCERT → higher), and 15 FAQs. Try our LMS and 24/7 AI Math Solver at www.fuzymathacademy.com.

Key formulas (keep these handy)

• Distance between points A(x₁, y₁), B(x₂, y₂): d = √[(x₂ − x₁)² + (y₂ − y₁)²]
• Midpoint: ( (x₁+x₂)/2, (y₁+y₂)/2 )
• Slope (gradient) of AB: m = (y₂−y₁)/(x₂−x₁)
• Slope-intercept form: y = mx + c
• Two-point form: y − y₁ = (y₂ − y₁)/(x₂ − x₁) × (x − x₁)
• Section formula (internal): ( (m x₂ + n x₁)/(m+n), (m y₂ + n y₁)/(m+n) )
• Area of triangle:

  Area = ½ | x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) |

---

25 Solved Questions (NCERT → Higher)

1. Q1 (easy): Find the distance between (−1, 4) and (3, −2).
   d = √[(3 − (−1))² + (−2 − 4)²] = √[4² + (−6)²] = √(16 + 36) = √52 = 2√13.
2. Q2 (easy): Find the midpoint of A(2,3) and B(6,7).
   Midpoint = ((2+6)/2, (3+7)/2) = (4, 5).
3. Q3 (NCERT): Find the coordinates of point P which divides AB internally in the ratio 2:3 where A(1,2), B(6,7).
   Using section formula: x = (2×6 + 3×1)/5 = 3; y = (2×7 + 3×2)/5 = 4. So P(3,4).
4. Q4 (slope): Find slope of line through (−2,5) and (3, −1).
   m = (−1 − 5)/(3 − (−2)) = (−6)/5 = −6/5.
5. Q5 (equation of line): Find equation of line with slope 2 passing through (1,−3).
   y − (−3) = 2(x − 1) → y + 3 = 2x − 2 → y = 2x − 5.
6. Q6 (two-point form): Find equation of line through A(0,2) and B(3, −1).
   Slope m = (−1 − 2)/(3 − 0) = −3/3 = −1.
   y − 2 = −1(x − 0) → y − 2 = −x → y = −x + 2.
7. Q7 (perpendicular slopes): Find slope of line perpendicular to line joining (1,1) and (4,5).
   Slope = (5−1)/(4−1) = 4/3. Perpendicular slope = −3/4.
8. Q8 (distance to origin): Find distance of (−3,4) from origin.
   d = √((−3)² + 4²) = √(9 + 16) = √25 = 5.
9. Q9 (area of triangle): Find area of triangle with vertices (0,0), (4,0), (0,3).
   Area = ½ × 4 × 3 = 6.
10. Q10 (area via determinant): Area of triangle with vertices A(1,2), B(4,6), C(7,2).
    Area = ½ | 1(6−2) + 4(2−2) + 7(2−6) | = ½ |4 + 0 + 7(−4)| = ½ |4 −28| = ½ × 24 = 12.
11. Q11 (point on line test): Is (2,3) on line 3x − 2y = 0?
    Plugging: 3×2 − 2×3 = 6 − 6 = 0 → Yes, (2,3) lies on the line.
12. Q12 (section formula external): Point dividing AB externally in ratio 2:1 where A(2,3), B(5,1). Find coordinates.
    External section formula: x = (m x₂ − n x₁)/(m − n); m:n = 2:1.
    x = (2×5 − 1×2)/1 = 8; y = (2×1 − 1×3)/1 = −1. Point (8, −1).
13. Q13 (parallel & perpendicular lines): Are lines 2x + 3y + 1 = 0 and 4x + 6y − 5 = 0 parallel, coincident, or intersecting?
    Compare a₁/a₂ = 2/4 = ½, b₁/b₂ = 3/6 = ½, c₁/c₂ = 1/(−5) = −1/5.
    a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so lines are parallel.
14. Q14 (foot of perpendicular): Find equation of line through (2,−1) perpendicular to y = (½)x + 3.
    Slope of given line = ½ → perpendicular slope = −2.
    y + 1 = −2(x − 2) → y = −2x + 3.
15. Q15 (distance from point to line): Distance from (3,4) to 3x − 4y + 12 = 0.
    
    Distance = |3×3 − 4×4 + 12| / √(9+16) = |9 − 16 + 12| / 5 = |5|/5 = 1.
16. Q16 (area doubles): Vertices (0,0), (x,0), (0,3) area is 6, find x.
    
    Area = ½ × |x| × 3 = (3|x|)/2. Set equal to 6 → |x| = 4. x = ±4.
17. Q17 (collinearity): Are points (1,2), (2,3), (3,4) collinear?
    
    Slope 1-2: (3−2)/(2−1) = 1; Slope 2-3: (4−3)/(3−2)=1. Points are collinear.
18. Q18 (coordinates from conditions): Midpoint of AB is (3,5) and A(1,2), find B.
    
    x₂ = 2×3 − 1 = 5, y₂ = 2×5 − 2 = 8. B(5,8).
19. Q19 (line through intersection): Equation of line through intersection of x+y−3=0 and x−y+1=0 and passing through (2,0).
    
    Intersection: x + y = 3, x − y = −1 → x=1, y=2.
    Line through (1,2),(2,0): Slope m = (0−2)/(2−1) = −2.
    y − 2 = −2(x − 1) → y = −2x + 4.
20. Q20 (transformations): Point (x,y) reflected in x-axis: new coordinates?
    
    Reflection: (x, −y). So (3, −2) → (3, 2).
21. Q21 (internal division fractional ratio): Divide (0,0) and (9,12) in ratio 3:1 (internal).
    
    x = (3×9 + 1×0)/4 = 6.75; y = (3×12 + 1×0)/4 = 9. So (6.75, 9).
22. Q22 (find intersection): Intersection of y = 2x + 1 and y = −x + 4.
    
    2x + 1 = −x + 4 → 3x = 3 → x = 1; y = 2×1 + 1 = 3. Intersection (1,3).
23. Q23 (area using coordinates): Triangle vertices (−1,0), (2,3), (4,−2). Find area.
    
    Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
    = ½ |(−1)(3 − (−2)) + 2((−2) − 0) + 4(0 − 3)|
    = ½ |−5 −4 −12| = ½(21) = 10.5.
24. Q24 (find unknown coordinate): Area of triangle with (0,0), (x,2), (3,0) is 6, find x.
    
    Area = ½ |0(2−0) + x(0−0) + 3(0−2)|
    = ½ |−6| = 3. So, given area 6, not possible by this setup; area is fixed at 3.
25. Q25 (challenge): Equation of perpendicular bisector of segment joining (1,2) and (5,6).
    
    Midpoint: (3, 4). Slope = (6−2)/(5−1) = 1;
    Perpendicular slope = −1.
    y − 4 = −1(x − 3) → x + y − 7 = 0.

---

15 FAQs (Quick Answers)

1. Q1: What is the distance formula?
   d = √[(x₂ − x₁)² + (y₂ − y₁)²]
2. Q2: How to find midpoint?
   Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
3. Q3: Slope of vertical line?
   Undefined (x₂ = x₁ gives division by zero). Vertical line x = constant.
4. Q4: Slope of horizontal line?
   Zero (y₂ = y₁). Horizontal line y = constant.
5. Q5: When are three points collinear?
   If slopes between pairwise points are equal, or area determinant equals zero.
6. Q6: How to find area of triangle by coordinates?
   Use ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.
7. Q7: What is section formula used for?
   To find coordinates of point dividing a line in a given ratio (internal/external).
8. Q8: How to test if two lines are parallel?
   If slopes are equal, or a₁/a₂ = b₁/b₂.
9. Q9: How to test if lines coincide?
   If a₁/a₂ = b₁/b₂ = c₁/c₂, lines coincide.
10. Q10: Slope-intercept vs point-slope forms?
    Slope-intercept y=mx+c if slope/intercept known. Point-slope y−y₁=m(x−x₁) for point and slope.
11. Q11: Equation of perpendicular bisector?
    Find midpoint, perpendicular slope to segment, use point-slope form.
12. Q12: Distance from point to line?
    Distance = |ax₀ + by₀ + c| / √(a² + b²).
13. Q13: When to use coordinate geometry in advanced math?
    In advanced topics, e.g., area by integration, centroids, Class 10 focuses on algebraic formulas.
14. Q14: Quick tip to avoid mistakes?
    Write formula first, substitute carefully, check signs, verify by plugging back.
15. Q15: Where to practice more problems?
    Find more at www.fuzymathacademy.com.

---

FUZY Math Academy – Online Math Tuition for Classes 5 to 12

Personalized Online Coaching | AI-Powered LMS | 24/7 AI Math Solver. Learn from expert teachers and get step-by-step NCERT solutions for Classes 5–12.

Read Full Post

Visit www.fuzymathacademy.com for more courses and free demo classes.