15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
How far apart are the points \((1, -2)\) and \((6, 10)\)?
Why: d = √(5² + 12²) = √169.
\(d = \sqrt{(6 - 1)^2 + (10 + 2)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\). (Adding the differences, \(5 + 12 = 17\), is a common slip.)
The distance of the point \((-8, 15)\) from the origin is
Why: √(64 + 225) = √289.
\(\sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\).
The mid-point of the segment joining \((-3, 8)\) and \((7, -2)\) is
Why: Average the coordinates: ((−3 + 7)/2, (8 − 2)/2).
\(\left(\dfrac{-3 + 7}{2}, \dfrac{8 - 2}{2}\right) = (2, 3)\).
The point dividing the segment joining \((1, 4)\) and \((10, -2)\) internally in the ratio \(2 : 1\) is
Why: Section formula with m : n = 2 : 1: ((2·10 + 1·1)/3, (2·(−2) + 1·4)/3).
\(\left(\dfrac{2(10) + 1(1)}{3}, \dfrac{2(-2) + 1(4)}{3}\right) = (7, 0)\).
The point on the \(y\)-axis that is equidistant from \((2, 3)\) and \((-4, 1)\) is
Why: Take (0, y): 4 + (y − 3)² = 16 + (y − 1)² gives y = −1.
\(2^2 + (y - 3)^2 = 4^2 + (y - 1)^2 \Rightarrow 13 - 6y = 17 - 2y \Rightarrow y = -1\). The point is \((0, -1)\).
The \(x\)-axis divides the segment joining \((3, -6)\) and \((7, 4)\) in the ratio
Why: On the x-axis y = 0: (4k − 6)/(k + 1) = 0 gives k = 3/2.
Let the ratio be \(k : 1\). \(\dfrac{4k - 6}{k + 1} = 0 \Rightarrow k = \dfrac32\), i.e. \(3 : 2\).
If \((3, b)\) is the mid-point of the segment joining \((a, 5)\) and \((4, -1)\), then \(a + b\) equals
Why: (a + 4)/2 = 3 gives a = 2; b = (5 − 1)/2 = 2.
\(\dfrac{a + 4}{2} = 3 \Rightarrow a = 2\); \(b = \dfrac{5 + (-1)}{2} = 2\). So \(a + b = 4\).
The points \(A(1, 1)\), \(B(4, 5)\) and \(C(7, 1)\) are the vertices of
Why: AB = BC = 5 but AC = 6.
\(AB = \sqrt{9 + 16} = 5\), \(BC = \sqrt{9 + 16} = 5\), \(AC = 6\). Two equal sides, not three: isosceles.
If the distance between \((p, 3)\) and \((5, -1)\) is \(5\) units, then \(p\) is
Why: (p − 5)² + 16 = 25 gives p − 5 = ±3.
\((p - 5)^2 + 4^2 = 25 \Rightarrow (p - 5)^2 = 9 \Rightarrow p = 8\) or \(p = 2\).
The centre of a circle is \((2, -3)\) and one end of a diameter is \((-1, 1)\). The other end of that diameter is
Why: The centre is the mid-point of the diameter: other end = 2 × centre − known end.
\((x, y) = (2 \times 2 - (-1),\ 2 \times (-3) - 1) = (5, -7)\).
The distance of the point \((6, -9)\) from the \(x\)-axis is
Why: Distance from the x-axis is |y|.
The foot of the perpendicular is \((6, 0)\), so the distance is \(|-9| = 9\).
For the points \(A(-1, -2)\), \(B(1, 0)\) and \(C(3, 2)\), which statement is true?
Why: AB = BC = 2√2 and AC = 4√2 = AB + BC, and ((−1 + 3)/2, (−2 + 2)/2) = (1, 0).
Mid-point of \(AC\) \(= \left(\dfrac{-1 + 3}{2}, \dfrac{-2 + 2}{2}\right) = (1, 0) = B\). (Also \(AB + BC = 2\sqrt2 + 2\sqrt2 = 4\sqrt2 = AC\), so the points are collinear: no triangle.)
\(P\) lies on the segment joining \(A(-2, 5)\) and \(B(4, -1)\) with \(\dfrac{AP}{PB} = \dfrac12\). The coordinates of \(P\) are
Why: Section formula with m : n = 1 : 2.
\(P = \left(\dfrac{1(4) + 2(-2)}{3}, \dfrac{1(-1) + 2(5)}{3}\right) = (0, 3)\).
A right-angled triangle has its corners at the origin, at \((5, 0)\) and at \((0, 12)\). How long is its boundary?
Why: Sides 5, 12 and √(25 + 144) = 13.
Sides: \(5\), \(12\) and \(\sqrt{5^2 + 12^2} = \sqrt{169} = 13\). Perimeter \(= 30\).
The points \(P\) and \(Q\) trisect the segment joining \(A(2, 3)\) and \(B(6, 7)\), with \(P\) nearer to \(A\). Then \(P\) is
Why: P divides AB in the ratio 1 : 2.
\(P = \left(\dfrac{1(6) + 2(2)}{3}, \dfrac{1(7) + 2(3)}{3}\right) = \left(\dfrac{10}{3}, \dfrac{13}{3}\right)\).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A delivery company plots its service area on a grid in which \(1\) unit is \(1\) km. Its drone warehouse is at \(W(-2, 1)\), and two customers live at \(H_1(6, 7)\) and \(H_2(4, -7)\). Drones fly in straight lines.
A rectangular park is drawn on a grid in which \(1\) unit is \(10\) m. Its corners are \(P(1, 1)\), \(Q(9, 1)\), \(R(9, 7)\) and \(S(1, 7)\). A fountain is to be built where the diagonals \(PR\) and \(QS\) cross.
This free set is separate from the chapter's question bank. On the Coordinate Geometry chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 40 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.