Important Questions Class 10 Maths Chapter 7: Coordinate Geometry With Solutions

Coordinate Geometry is the study of points, lines, and figures using ordered pairs on a plane.
It connects algebra with geometry by showing the exact position of a point as (x, y).

Algebra gives every point a fixed address in Coordinate Geometry. CBSE 2026 tests Important Questions Class 10 Maths Chapter 7 through Distance Formula, Section Formula, midpoint use, collinearity, triangle type, and missing-coordinate questions. The NCERT Reprint 2026-27 syllabus includes two exercises and 20 NCERT questions. Area of Triangle does not form part of Class 10 Chapter 7 for CBSE 2026. Students should focus on formula selection, sign accuracy, and step-by-step substitution to score well in board-style questions.

Key Takeaways

• Coordinate Geometry: Chapter 7 focuses on distance, section, midpoint, collinearity, and missing-coordinate questions.
• Formula Choice: Use Distance Formula for length and Section Formula for ratio-based division.
• Exam Trap: In Section Formula, m1 multiplies the second point and m2 multiplies the first point.
• CBSE 2026 Update: Area of Triangle does not form part of Class 10 Chapter 7 for CBSE 2026.

Important Questions Class 10 Maths Chapter 7 Structure 2026

Important Questions Class 10 Maths Chapter 7: Key Formulas

These formulas cover Chapter 7 Maths Class 10 important questions in CBSE 2026. Students must write the formula before substitution.

Q1. What Is The Distance Formula In Class 10 Coordinate Geometry?

The Distance Formula finds the distance between two points on a coordinate plane. For P(x1, y1) and Q(x2, y2), the distance is √[(x2 - x1)² + (y2 - y1)²].

1. Given Data:
   P(x1, y1), Q(x2, y2)
2. Formula Used:
   PQ = √[(x2 - x1)² + (y2 - y1)²]
3. Final Result:
   Distance between two points = PQ

Q2. What Is The Section Formula In Class 10 Coordinate Geometry?

The Section Formula finds the point that divides a line segment internally in a given ratio. If P divides AB in m1 : m2, use weighted coordinates.

1. Given Data:
   A(x1, y1), B(x2, y2), ratio m1 : m2
2. Formula Used:
   P = ((m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2))
3. Final Result:
   Point of division = P(x, y)

Q3. What Is The Midpoint Formula Class 10?

The midpoint formula class 10 finds the middle point of a line segment. It comes from the Section Formula when the ratio is 1 : 1.

1. Given Data:
   A(x1, y1), B(x2, y2)
2. Formula Used:
   M = ((x1 + x2)/2, (y1 + y2)/2)
3. Final Result:
   Midpoint = M(x, y)

Class 10 Coordinate Geometry MCQ With Answers

These class 10 coordinate geometry MCQ questions follow the CBSE 2026 board exam pattern. Each question tests direct formula use.

Q1. What Is The Distance Of Point P(3, 4) From The Origin?

The distance of P(3, 4) from the origin is 5 units. Use OP = √(x² + y²) for distance from (0, 0).

1. Given Data:
   P(3, 4), O(0, 0)
2. Formula Used:
   OP = √(x² + y²)
3. Calculation:
   OP = √(3² + 4²)
   OP = √(9 + 16)
   OP = √25 = 5
4. Final Result:
   Answer: (C) 5 units

Q2. What Is The Distance Between A(0, 6) And B(0, -2)?

The distance between A(0, 6) and B(0, -2) is 8 units. Both points lie on the y-axis.

1. Given Data:
   A(0, 6), B(0, -2)
2. Formula Used:
   AB = |y2 - y1|
3. Calculation:
   AB = |-2 - 6|
   AB = |-8|
   AB = 8
4. Final Result:
   Answer: (C) 8 units

Q3. What Is The Distance Between (0, 5) And (-5, 0)?

The distance between (0, 5) and (-5, 0) is 5√2 units. Use the Distance Formula.

1. Given Data:
   A(0, 5), B(-5, 0)
2. Formula Used:
   AB = √[(x2 - x1)² + (y2 - y1)²]
3. Calculation:
   AB = √[(-5 - 0)² + (0 - 5)²]
   AB = √(25 + 25)
   AB = √50 = 5√2
4. Final Result:
   Answer: (B) 5√2 units

Q4. What Is The Midpoint Of A(2, 3) And B(4, 7)?

The midpoint of A(2, 3) and B(4, 7) is (3, 5). Use the midpoint formula for equal division.

1. Given Data:
   A(2, 3), B(4, 7)
2. Formula Used:
   M = ((x1 + x2)/2, (y1 + y2)/2)
3. Calculation:
   M = ((2 + 4)/2, (3 + 7)/2)
   M = (3, 5)
4. Final Result:
   Answer: (A) (3, 5)

Q5. What Is The X-Coordinate When P Divides A(1, 2) And B(4, 5) In Ratio 2 : 1?

The x-coordinate of P is 3. The ratio 2 : 1 means the first part multiplies the second endpoint.

1. Given Data:
   A(1, 2), B(4, 5), ratio 2 : 1
2. Formula Used:
   x = (m1x2 + m2x1)/(m1 + m2)
3. Calculation:
   x = (2 × 4 + 1 × 1)/(2 + 1)
   x = 9/3
   x = 3
4. Final Result:
   Answer: (B) 3

Q6. What Is The Distance Of P(-6, 8) From The Origin?

The distance of P(-6, 8) from the origin is 10 units. The signs become positive after squaring.

1. Given Data:
   P(-6, 8), O(0, 0)
2. Formula Used:
   OP = √(x² + y²)
3. Calculation:
   OP = √[(-6)² + 8²]
   OP = √(36 + 64)
   OP = √100 = 10
4. Final Result:
   Answer: (C) 10 units

Q7. If The Midpoint Of (x, 4) And (3, y) Is (5, 3), Find x And y.

The values are x = 7 and y = 2. Compare both coordinates using the midpoint formula.

1. Given Data:
   A(x, 4), B(3, y), midpoint (5, 3)
2. Formula Used:
   M = ((x1 + x2)/2, (y1 + y2)/2)
3. Calculation:
   (x + 3)/2 = 5
   x + 3 = 10
   x = 7
4. Calculation:
   (4 + y)/2 = 3
   4 + y = 6
   y = 2
5. Final Result:
   Answer: (A) x = 7, y = 2

Q8. In What Ratio Does The Y-Axis Divide A(5, -6) And B(-1, -4)?

The y-axis divides the segment in the ratio 5 : 1. A point on the y-axis has x-coordinate 0.

1. Given Data:
   A(5, -6), B(-1, -4), x = 0
2. Formula Used:
   x = (m1x2 + m2x1)/(m1 + m2)
3. Calculation:
   Let ratio = k : 1
   (-k + 5)/(k + 1) = 0
   k = 5
4. Final Result:
   Answer: (B) 5 : 1

Q9. If A(1, 2), B(3, k), And C(5, -4) Are Collinear, Find k.

The value of k is -1. For collinear points, slopes of consecutive line segments remain equal.

1. Given Data:
   A(1, 2), B(3, k), C(5, -4)
2. Formula Used:
   Slope of AB = Slope of BC
3. Calculation:
   (k - 2)/(3 - 1) = (-4 - k)/(5 - 3)
   (k - 2)/2 = (-4 - k)/2
   k - 2 = -4 - k
   2k = -2
4. Final Result:
   Answer: (A) k = -1

Q10. What Is The Length Of Diagonal AB In Rectangle AOBC With A(0, 3), O(0, 0), And B(5, 0)?

The length of diagonal AB is √34 units. Apply the Distance Formula between A and B.

1. Given Data:
   A(0, 3), B(5, 0)
2. Formula Used:
   AB = √[(x2 - x1)² + (y2 - y1)²]
3. Calculation:
   AB = √[(5 - 0)² + (0 - 3)²]
   AB = √(25 + 9)
   AB = √34
4. Final Result:
   Answer: (C) √34 units

Distance Formula Class 10 Important Questions With Step-By-Step Solutions

These distance formula class 10 important questions test length, triangle type, and unknown coordinates. The Distance Formula comes from Pythagoras theorem.

Q1. Do P(3, 2), Q(-2, -3), And R(2, 3) Form A Triangle?

Yes, the points form a right-angled scalene triangle. The side lengths satisfy the triangle inequality and Pythagoras theorem.

1. Given Data:
   P(3, 2), Q(-2, -3), R(2, 3)
2. Formula Used:
   Distance Formula
3. Calculation:
   PQ = √[(-2 - 3)² + (-3 - 2)²] = 5√2
   QR = √[(2 + 2)² + (3 + 3)²] = 2√13
   PR = √[(2 - 3)² + (3 - 2)²] = √2
4. Check:
   PQ² = 50
   QR² = 52
   PR² = 2
   PQ² + PR² = QR²
5. Final Result:
   Right-angled scalene triangle

Q2. Show That A(1, 7), B(4, 2), C(-1, -1), And D(-4, 4) Form A Square.

The four points form a square. All four sides measure √34, and both diagonals measure √68.

1. Given Data:
   A(1, 7), B(4, 2), C(-1, -1), D(-4, 4)
2. Formula Used:
   Distance Formula
3. Calculation:
   AB = √[(4 - 1)² + (2 - 7)²] = √34
   BC = √[(-1 - 4)² + (-1 - 2)²] = √34
   CD = √[(-4 + 1)² + (4 + 1)²] = √34
   DA = √[(1 + 4)² + (7 - 4)²] = √34
4. Diagonal Check:
   AC = √[(-1 - 1)² + (-1 - 7)²] = √68
   BD = √[(-4 - 4)² + (4 - 2)²] = √68
5. Final Result:
   ABCD is a square

Q3. Name The Type Of Triangle Formed By S(-5, 6), T(-4, -2), And U(7, 5).

The points form a scalene triangle. All three sides have different lengths.

1. Given Data:
   S(-5, 6), T(-4, -2), U(7, 5)
2. Formula Used:
   Distance Formula
3. Calculation:
   ST = √[(-4 + 5)² + (-2 - 6)²] = √65
   TU = √[(7 + 4)² + (5 + 2)²] = √170
   SU = √[(7 + 5)² + (5 - 6)²] = √145
4. Final Result:
   Scalene triangle

Q4. Find y If The Distance Between P(2, -3) And Q(10, y) Is 10 Units.

The possible values of y are 3 or -9. The given distance creates a quadratic equation in y.

1. Given Data:
   P(2, -3), Q(10, y), PQ = 10
2. Formula Used:
   Distance Formula
3. Calculation:
   √[(10 - 2)² + (y + 3)²] = 10
   64 + (y + 3)² = 100
   (y + 3)² = 36
   y + 3 = ±6
4. Final Result:
   y = 3 or y = -9

Q5. Find The Point On The X-Axis Equidistant From (2, -5) And (-2, 9).

The required point is (-7, 0). A point on the x-axis has y-coordinate 0.

1. Given Data:
   Point on x-axis = (x, 0)
   A(2, -5), B(-2, 9)
2. Formula Used:
   PA² = PB²
3. Calculation:
   (x - 2)² + (0 + 5)² = (x + 2)² + (0 - 9)²
   x² - 4x + 29 = x² + 4x + 85
   -8x = 56
   x = -7
4. Final Result:
   Required point = (-7, 0)

Q6. Find A Relation Between x And y If (x, y) Is Equidistant From (7, 1) And (3, 5).

The required relation is x - y = 2. Equidistant points have equal squared distances.

1. Given Data:
   P(x, y), A(7, 1), B(3, 5)
2. Formula Used:
   PA² = PB²
3. Calculation:
   (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
   x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² - 10y + 25
   -8x + 8y = -16
4. Final Result:
   x - y = 2

Collinear Points Class 10 Questions With Board Exam Pattern

These collinear points class 10 questions test whether three points lie on one straight line. Students can use distances or slopes.

Q1. Are A(3, 1), B(6, 4), And C(8, 6) Collinear?

Yes, the points are collinear. The sum of the two smaller distances equals the largest distance.

1. Given Data:
   A(3, 1), B(6, 4), C(8, 6)
2. Formula Used:
   AB + BC = AC
3. Calculation:
   AB = √[(6 - 3)² + (4 - 1)²] = 3√2
   BC = √[(8 - 6)² + (6 - 4)²] = 2√2
   AC = √[(8 - 3)² + (6 - 1)²] = 5√2
4. Check:
   AB + BC = 3√2 + 2√2 = 5√2
5. Final Result:
   The points are collinear

Q2. Determine Whether (1, 5), (2, 3), And (-2, -11) Are Collinear.

The points are not collinear. The distances do not satisfy the straight-line distance condition.

1. Given Data:
   A(1, 5), B(2, 3), C(-2, -11)
2. Formula Used:
   AB + BC = AC
3. Calculation:
   AB = √[(2 - 1)² + (3 - 5)²] = √5
   BC = √[(-2 - 2)² + (-11 - 3)²] = √212
   AC = √[(-2 - 1)² + (-11 - 5)²] = √265
4. Check:
   √5 + √212 ≠ √265
5. Final Result:
   The points are not collinear

Q3. Why Are Collinear Points Repeatedly Asked In CBSE Boards?

Collinearity questions repeat because they test both Distance Formula and slope logic. They also check calculation accuracy.

1. Given Data:
   Three coordinate points
2. Formula Used:
   AB + BC = AC
   or
   Slope of AB = Slope of BC
3. Board Exam Pattern:
   MCQ questions use slope comparison.
   Short-answer questions often use distance comparison.
4. Final Result:
   Collinearity remains a high-frequency Chapter 7 pattern

Section Formula Class 10 Important Questions With Solutions

These section formula class 10 important questions cover internal division, trisection, axis division, and ratio finding. The order of ratio matters.

Q1. Find The Point Dividing (4, -3) And (8, 5) In Ratio 3 : 1 Internally.

The required point is (7, 3). Use the Section Formula with m1 = 3 and m2 = 1.

1. Given Data:
   A(4, -3), B(8, 5), ratio 3 : 1
2. Formula Used:
   P = ((m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2))
3. Calculation:
   x = (3 × 8 + 1 × 4)/4 = 7
   y = (3 × 5 + 1 × (-3))/4 = 3
4. Final Result:
   Point = (7, 3)

Q2. Find The Trisection Points Of The Segment Joining A(2, -2) And B(-7, 4).

The trisection points are (-1, 0) and (-4, 2). They divide the line segment in ratios 1 : 2 and 2 : 1.

1. Given Data:
   A(2, -2), B(-7, 4)
2. Formula Used:
   Section Formula
3. Calculation For P:
   P divides AB in 1 : 2
   P = ((1 × -7 + 2 × 2)/3, (1 × 4 + 2 × -2)/3)
   P = (-3/3, 0/3)
   P = (-1, 0)
4. Calculation For Q:
   Q divides AB in 2 : 1
   Q = ((2 × -7 + 1 × 2)/3, (2 × 4 + 1 × -2)/3)
   Q = (-12/3, 6/3)
   Q = (-4, 2)
5. Final Result:
   Trisection points = (-1, 0) and (-4, 2)

Q3. Find The Coordinates Of A If AB Is The Diameter Of A Circle With Centre (2, -3) And B(1, 4).

The coordinate of A is (3, -10). The centre of a circle bisects its diameter.

1. Given Data:
   Centre = (2, -3)
   B(1, 4), A(x, y)
2. Formula Used:
   Centre = Midpoint of AB
3. Calculation:
   (x + 1)/2 = 2
   x + 1 = 4
   x = 3
4. Calculation:
   (y + 4)/2 = -3
   y + 4 = -6
   y = -10
5. Final Result:
   A = (3, -10)

Section Formula Important Questions Class 10 On Finding Ratio

These section formula important questions class 10 focus on axis division and ratio calculation. Use x = 0 for the y-axis and y = 0 for the x-axis.

Q1. In What Ratio Does (-4, 6) Divide A(-6, 10) And B(3, -8)?

The point (-4, 6) divides the segment in the ratio 2 : 7. Use the x-coordinate to find the ratio.

1. Given Data:
   A(-6, 10), B(3, -8), point (-4, 6)
2. Formula Used:
   x = (m1x2 + m2x1)/(m1 + m2)
3. Calculation:
   Let ratio = k : 1
   (3k - 6)/(k + 1) = -4
   3k - 6 = -4k - 4
   7k = 2
4. Final Result:
   Ratio = 2 : 7

Q2. Find The Ratio In Which The Y-Axis Divides (5, -6) And (-1, -4).

The y-axis divides the segment in the ratio 5 : 1 at (0, -13/3).

1. Given Data:
   A(5, -6), B(-1, -4)
   Point on y-axis has x = 0
2. Formula Used:
   Section Formula
3. Calculation:
   Let ratio = k : 1
   (-k + 5)/(k + 1) = 0
   k = 5
4. Point Of Division:
   y = [5 × (-4) + 1 × (-6)]/6
   y = -26/6
   y = -13/3
5. Final Result:
   Ratio = 5 : 1, point = (0, -13/3)

Q3. Find The Ratio In Which The X-Axis Divides A(1, -5) And B(-4, 5).

The x-axis divides the segment in the ratio 1 : 1 at (-3/2, 0).

1. Given Data:
   A(1, -5), B(-4, 5)
   Point on x-axis has y = 0
2. Formula Used:
   Section Formula
3. Calculation:
   Let ratio = k : 1
   (5k - 5)/(k + 1) = 0
   k = 1
4. Point Of Division:
   x = [1 × (-4) + 1 × 1]/2
   x = -3/2
5. Final Result:
   Ratio = 1 : 1, point = (-3/2, 0)

Midpoint Formula Class 10 And Parallelogram Questions

The midpoint formula class 10 problems often use parallelogram properties. In a parallelogram, diagonals bisect each other.

Q1. If A(6, 1), B(8, 2), C(9, 4), And D(p, 3) Form A Parallelogram, Find p.

The value of p is 7. In a parallelogram, the midpoints of both diagonals remain equal.

1. Given Data:
   A(6, 1), B(8, 2), C(9, 4), D(p, 3)
2. Formula Used:
   Midpoint of AC = Midpoint of BD
3. Calculation:
   (6 + 9)/2 = (8 + p)/2
   15 = 8 + p
   p = 7
4. Final Result:
   p = 7

Q2. If A(-2, 1), B(a, 0), C(4, b), And D(1, 2) Form A Parallelogram, Find a And b.

The values are a = 1 and b = 1. Equate the midpoints of both diagonals.

1. Given Data:
   A(-2, 1), B(a, 0), C(4, b), D(1, 2)
2. Formula Used:
   Midpoint of AC = Midpoint of BD
3. Calculation:
   (-2 + 4)/2 = (a + 1)/2
   a = 1
4. Calculation:
   (1 + b)/2 = (0 + 2)/2
   b = 1
5. Final Result:
   a = 1, b = 1

Q3. Find The Fourth Vertex Of A Parallelogram With Vertices (-1, 0), (3, 1), And (2, 2) In Order.

The fourth vertex is (-2, 1). Use equal midpoints of diagonals for vertices taken in order.

1. Given Data:
   A(-1, 0), B(3, 1), C(2, 2), D(x, y)
2. Formula Used:
   Midpoint of AC = Midpoint of BD
3. Calculation:
   (-1 + 2)/2 = (3 + x)/2
   x = -2
4. Calculation:
   (0 + 2)/2 = (1 + y)/2
   y = 1
5. Final Result:
   D = (-2, 1)

Coordinate Geometry Class 10 Extra Questions With Solutions

These coordinate geometry class 10 extra questions combine formulas with line conditions and equal division. They match CBSE 2026 school exam practice.

Q1. Point P Divides A(3, -5) And B(-4, 8) Such That AP : PB = k : 1. If P Lies On x + y = 0, Find k.

The value of k is 1/2. The point divides the segment in the ratio 1 : 2.

1. Given Data:
   A(3, -5), B(-4, 8), ratio k : 1
   P lies on x + y = 0
2. Formula Used:
   Section Formula
3. Calculation:
   P = ((-4k + 3)/(k + 1), (8k - 5)/(k + 1))
   (-4k + 3 + 8k - 5)/(k + 1) = 0
   4k - 2 = 0
   k = 1/2
4. Final Result:
   k = 1/2, so AP : PB = 1 : 2

Q2. If Q(0, 1) Is Equidistant From P(5, -3) And R(x, 6), Find x And QR.

The value of x is 4 or -4. The distance QR equals √41.

1. Given Data:
   Q(0, 1), P(5, -3), R(x, 6)
2. Formula Used:
   QP² = QR²
3. Calculation:
   QP² = (5 - 0)² + (-3 - 1)² = 41
   QR² = (x - 0)² + (6 - 1)²
   QR² = x² + 25
   x² + 25 = 41
4. Final Result:
   x = 4 or -4, QR = √41

Q3. Find The Points That Divide A(-2, 2) And B(2, 8) Into Four Equal Parts.

The required points are (-1, 7/2), (0, 5), and (1, 13/2). They divide the segment into four equal parts.

1. Given Data:
   A(-2, 2), B(2, 8)
2. Formula Used:
   Section Formula and Midpoint Formula
3. First Point:
   P1 divides AB in 1 : 3
   P1 = ((1 × 2 + 3 × -2)/4, (1 × 8 + 3 × 2)/4)
   P1 = (-1, 7/2)
4. Second Point:
   P2 = ((-2 + 2)/2, (2 + 8)/2)
   P2 = (0, 5)
5. Third Point:
   P3 divides AB in 3 : 1
   P3 = ((3 × 2 + 1 × -2)/4, (3 × 8 + 1 × 2)/4)
   P3 = (1, 13/2)
6. Final Result:
   (-1, 7/2), (0, 5), and (1, 13/2)

Q4. Find The Ratio In Which (-1, 6) Divides The Segment Joining (-3, 10) And (6, -8).

The ratio is 2 : 7. Use the x-coordinate in the Section Formula.

1. Given Data:
   A(-3, 10), B(6, -8), point (-1, 6)
2. Formula Used:
   x = (m1x2 + m2x1)/(m1 + m2)
3. Calculation:
   Let ratio = k : 1
   (6k - 3)/(k + 1) = -1
   6k - 3 = -k - 1
   7k = 2
4. Final Result:
   Ratio = 2 : 7

Q5. Find The Fourth Vertex Of A Parallelogram With Three Vertices (-1, 0), (3, 1), And (2, 2).

The fourth vertex is (-2, 1). Equal midpoints of diagonals give the missing point.

1. Given Data:
   A(-1, 0), B(3, 1), C(2, 2), D(x, y)
2. Formula Used:
   Midpoint of AC = Midpoint of BD
3. Calculation:
   (-1 + 2)/2 = (3 + x)/2
   x = -2
4. Calculation:
   (0 + 2)/2 = (1 + y)/2
   y = 1
5. Final Result:
   D = (-2, 1)

Coordinate Geometry Class 10 Short Answer Questions

These coordinate geometry class 10 short answer questions match 2-mark and 3-mark patterns. Write formulas clearly and show substitutions.

Q1. Find The Distance Between (a, b) And (-a, -b).

The distance between (a, b) and (-a, -b) is 2√(a² + b²). Treat a and b as variables.

1. Given Data:
   P(a, b), Q(-a, -b)
2. Formula Used:
   PQ = √[(x2 - x1)² + (y2 - y1)²]
3. Calculation:
   PQ = √[(-a - a)² + (-b - b)²]
   PQ = √(4a² + 4b²)
   PQ = 2√(a² + b²)
4. Final Result:
   2√(a² + b²)

Q2. Check Whether (5, -2), (6, 4), And (7, -2) Form An Isosceles Triangle.

Yes, the points form an isosceles triangle. Two sides have equal length √37.

1. Given Data:
   A(5, -2), B(6, 4), C(7, -2)
2. Formula Used:
   Distance Formula
3. Calculation:
   AB = √[(6 - 5)² + (4 + 2)²] = √37
   BC = √[(7 - 6)² + (-2 - 4)²] = √37
   CA = √[(7 - 5)² + (-2 + 2)²] = 2
4. Final Result:
   The points form an isosceles triangle

Q3. Find k If A(2, 3), B(4, k), And C(6, -3) Are Collinear.

The value of k is 0. Equal slopes confirm collinearity.

1. Given Data:
   A(2, 3), B(4, k), C(6, -3)
2. Formula Used:
   Slope of AB = Slope of BC
3. Calculation:
   (k - 3)/(4 - 2) = (-3 - k)/(6 - 4)
   (k - 3)/2 = (-3 - k)/2
   k - 3 = -3 - k
   2k = 0
4. Final Result:
   k = 0

Q4. Find x And y If (1, 2), (4, y), (x, 6), And (3, 5) Form A Parallelogram.

The values are x = 6 and y = 3. Equal midpoints of diagonals give both values.

1. Given Data:
   A(1, 2), B(4, y), C(x, 6), D(3, 5)
2. Formula Used:
   Midpoint of AC = Midpoint of BD
3. Calculation:
   (1 + x)/2 = (4 + 3)/2
   1 + x = 7
   x = 6
4. Calculation:
   (2 + 6)/2 = (y + 5)/2
   8 = y + 5
   y = 3
5. Final Result:
   x = 6, y = 3

Q5. Find The Ratio In Which The Line Segment Joining A(1, -5) And B(-4, 5) Is Divided By The X-Axis.

The x-axis divides the segment in the ratio 1 : 1 at (-3/2, 0).

1. Given Data:
   A(1, -5), B(-4, 5)
   Point on x-axis has y = 0
2. Formula Used:
   Section Formula
3. Calculation:
   Let ratio = k : 1
   (5k - 5)/(k + 1) = 0
   k = 1
4. Point Of Division:
   x = (-4 + 1)/2
   x = -3/2
5. Final Result:
   Ratio = 1 : 1, point = (-3/2, 0)

Class 10 Maths Chapter 7 Questions And Answers: Most Repeated Variations

These class 10 maths chapter 7 questions and answers target repeated CBSE 2026 patterns. Students should practise each variation with exact formula steps.

Q1. What Are The Most Repeated Variations In Coordinate Geometry Important Questions?

The most repeated variations include distance from origin, triangle type, collinearity, axis division, trisection, and parallelogram missing vertex.

1. Distance Formula:
   Find distance between two points.
   Identify triangle type using side lengths.
2. Section Formula:
   Find coordinates in a given ratio.
   Find ratio when the point lies on an axis.
3. Midpoint Formula:
   Find missing vertex of a parallelogram.
   Find endpoint when midpoint and one endpoint appear.
4. Final Result:
   These six patterns cover most Chapter 7 board-style questions

Q2. How Should Students Solve Important Questions Class 10 Maths Chapter 7 With Solutions?

Students should identify the formula first, substitute coordinates carefully, and simplify without skipping signs. Sign errors cause most wrong answers.

1. Read The Condition:
   Length, ratio, midpoint, collinearity, or parallelogram.
2. Choose The Formula:
   Distance Formula, Section Formula, or Midpoint Formula.
3. Substitute Coordinates:
   Keep negative signs inside brackets.
4. Final Result:
   Correct formula selection decides the solution path

Useful Important Questions Class 10 Maths Links