CBSE Class 10 Mathematics Standard 2025: question 23
Set in the CBSE Class 10 board exam of 2025, paper 30/6/1, section B, question 23, for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q23 · 2 marks · Very short answerCoordinate Geometry
Prove that abscissa of a point P which is equidistant from points with coordinates A(7, 1) and B(3, 5) is 2 more than its ordinate.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/6/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Take P as and equate its distances from A and B.
Hint 2
Square both distances; the and terms cancel, leaving a linear relation.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- ½ mark
Let be equidistant from A(7, 1) and B(3, 5). Then .
- ½ mark
[distance formula]
- ½ mark
- ½ mark
. So the abscissa of P is 2 more than its ordinate. Hence proved.
Answer: (proved)
Where the marks are
- Writes (½ mark)
- Applies distance formula correctly (½ mark)
- Expands and simplifies (½ mark)
- Concludes (½ mark)
Revise the chapter
- Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
- Topic: Distance Formula
More questions from Coordinate Geometry
- Sample paper 2024-25 (Standard), Q15 · 1 markThe point on the x-axis nearest to the point (-4,-5) is
- Sample paper 2024-25 (Standard), Q24 · 2 marksFind the point(s) on the x-axis which is at a distance of √(41) units from the point (8,-5).
- Sample paper 2024-25 (Standard), Q25 · 2 marksShow that the points A(-5,6), B(3,0) and C(9,8) are the vertices of an isosceles triangle.
The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.