Class 10 Maths · Original EZT practice
Coordinate Geometry (Class 10 Chapter 7): summary and practice
Coordinate Geometry measures the plane: the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) straight from Pythagoras, the section formula for a point dividing a segment in the ratio m:n, and the midpoint as its 1:1 special case. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Introduction
Every point on the plane carries an address - its abscissa x and ordinate y. Once points are numbers, geometry becomes arithmetic.
2. The distance formula
The distance between (x1, y1) and (x2, y2) is sqrt((x2 - x1)^2 + (y2 - y1)^2) - the Pythagoras theorem applied to the horizontal and vertical gaps. From the origin it shortens to sqrt(x^2 + y^2).
3. The section formula
The point dividing the segment from (x1, y1) to (x2, y2) in the ratio m:n is ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)). Note the cross-pairing: m multiplies the FAR endpoint x2.
4. The midpoint
Setting m = n = 1 in the section formula gives the midpoint ((x1 + x2)/2, (y1 + y2)/2) - just the average of the coordinates.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Distance between two points
Choose the method: Square the gaps, add, take the root.
EZT practice question
Find the distance between (2, 3) and (6, 6).
Show the worked solution
- Gaps: 6 - 2 = 4 and 6 - 3 = 3.
- Squares: 16 and 9.
- d = sqrt(16 + 9) = sqrt(25).
Answer: d = 5.
Check: A 3-4-5 right triangle hides in every such question - the hypotenuse 5 matches.
Watch for this: Do not stop at 25 - the formula ends with the square root.
Now try it yourself
Distance from the origin to (6, 8)?
Check your answer
10.
sqrt(36 + 64) = sqrt(100).
Find the midpoint
Choose the method: Average the x-coordinates and the y-coordinates.
EZT practice question
Midpoint of (1, 2) and (5, 8)?
Show the worked solution
- x: (1 + 5)/2 = 3.
- y: (2 + 8)/2 = 5.
- Combine.
Answer: (3, 5).
Check: The midpoint is equidistant from both ends: (3, 5) is sqrt(8) from each - equal, as required.
Watch for this: Average, do not add: (1 + 5) without the /2 doubles the answer.
Now try it yourself
Midpoint of (-2, 4) and (6, -2)?
Check your answer
(2, 1).
((-2 + 6)/2, (4 - 2)/2).
Apply the section formula
Choose the method: ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n)) - m pairs with the far end.
EZT practice question
Point dividing the segment from (-1, 7) to (4, -3) in the ratio 2:3?
Show the worked solution
- x: (2 x 4 + 3 x (-1))/5 = (8 - 3)/5 = 1.
- y: (2 x (-3) + 3 x 7)/5 = (-6 + 21)/5 = 3.
- Combine.
Answer: (1, 3).
Check: (1, 3) lies 2/5 of the way from (-1, 7) toward (4, -3): (-1 + 2, 7 - 4) - matches.
Watch for this: The ratio m:n pairs m with x2, the FAR endpoint - swapping them lands on the wrong side.
Now try it yourself
Point dividing (0, 0) to (10, 5) in the ratio 1:4?
Check your answer
(2, 1).
((1 x 10)/5, (1 x 5)/5).
Distance from the origin
Choose the method: sqrt(x^2 + y^2) - the formula with (0, 0) as one end.
EZT practice question
Which is farther from the origin: (3, 4) or (0, 6)?
Show the worked solution
- (3, 4): sqrt(9 + 16) = 5.
- (0, 6): sqrt(0 + 36) = 6.
- Compare.
Answer: (0, 6) is farther - 6 against 5.
Check: Both sit on circles about the origin; radius 6 beats radius 5.
Watch for this: Square BEFORE adding: sqrt(9 + 16) is not 3 + 4.
Now try it yourself
Distance of (-5, 12) from the origin?
Check your answer
13.
sqrt(25 + 144) = sqrt(169).
Test three points for collinearity
Choose the method: If AB + BC = AC, the points lie on one line.
EZT practice question
Are (1, 1), (2, 3) and (3, 5) collinear?
Show the worked solution
- AB = sqrt(1 + 4) = sqrt(5).
- BC = sqrt(1 + 4) = sqrt(5).
- AC = sqrt(4 + 16) = sqrt(20) = 2 sqrt(5).
Answer: Yes - AB + BC = AC, so the middle point sits on the segment.
Check: The slope from (1, 1) to (2, 3) is 2 and from (2, 3) to (3, 5) is also 2 - same line, as the distances said.
Watch for this: Collinearity needs the SUM test or equal slopes - near-equal distances alone prove nothing.
Now try it yourself
Are (0, 0), (1, 1) and (2, 3) collinear?
Check your answer
No.
sqrt(2) + sqrt(5) is not sqrt(13).
Find an unknown coordinate from a distance
Choose the method: Set up the distance equation and solve the square.
EZT practice question
The point (x, 2) is 5 units from (1, -2). Find x.
Show the worked solution
- (x - 1)^2 + (2 + 2)^2 = 25.
- (x - 1)^2 + 16 = 25, so (x - 1)^2 = 9.
- x - 1 = 3 or -3.
Answer: x = 4 or x = -2.
Check: Both work: (4, 2) and (-2, 2) each give sqrt(9 + 16) = 5.
Watch for this: A square has TWO roots - dropping x = -2 loses half the answer.
Now try it yourself
(3, y) is 5 units from the origin. Find y.
Check your answer
y = 4 or y = -4.
9 + y^2 = 25.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Distance formula?
Show answer and reason
sqrt((x2 - x1)^2 + (y2 - y1)^2).
Pythagoras on the coordinate gaps.
2. From the origin?
Show answer and reason
sqrt(x^2 + y^2).
The same formula with (0, 0).
3. Section formula?
Show answer and reason
((m x2 + n x1)/(m + n), same for y).
m pairs with the far endpoint.
4. Midpoint?
Show answer and reason
((x1 + x2)/2, (y1 + y2)/2).
The 1:1 case of the section formula.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Distance between (0, 0) and (6, 8).
- Midpoint of (-2, 4) and (6, -2).
- Point dividing (-1, 7) to (4, -3) in the ratio 2:3.
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Questions about this practice page
- What does the distance formula actually measure?
- The straight-line gap between two points. The horizontal and vertical differences form the legs of a right triangle, and the formula is the Pythagoras theorem solving for its hypotenuse.
- How does the midpoint formula follow from the section formula?
- A midpoint divides the segment in the ratio 1:1. Putting m = n = 1 into the section formula cancels the weights, leaving the plain average of the x-coordinates and of the y-coordinates.
- Why does m multiply x2 and not x1 in the section formula?
- The weight pulls toward the opposite end: a larger m moves the point closer to (x2, y2). Cross-pairing is what makes the 2:3 point land 2/5 of the way along, not 3/5.
- Is the area-of-a-triangle formula part of this chapter?
- Not in the Reprint 2026-27. The chapter runs 7.1 to 7.4 and ends with the section formula and summary - the triangle area section is not in this edition.
- Are these the NCERT exercise questions for Coordinate Geometry (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.