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CBSE CLASS 10 · CHAPTER 7

Coordinate Geometry: MCQs with Answers

Practise 51 selected MCQs from official CBSE Basic and Standard board papers and sample papers. Open each explanation for the formula, symbol meanings, substitution and reasoning.

36 board-paper questions · 15 official sample-paper questions

Checked against NCERT Chapter 7 and the 2026–27 CBSE syllabus. Ordinary MCQs are presented here; Assertion–Reason and written case-study questions are kept separate.

Practise 11 Coordinate Geometry Assertion–Reason PYQs →

Coordinates and axes Distance formula Midpoints and diameters Internal section formula Equidistant points Triangles and quadrilaterals

Coordinates and axes

Question 1 · Basic

The distance of (3, 5) from the x-axis is k units. The value of k is:

CBSE 2023–24 official sample · Basic · Q6

A. 3B. 4C. 5D. 8

Show Answer & Explanation

Formula guide · Team TutorVertex

Perpendicular distances from the axes

Distance from x-axis = |y|; distance from y-axis = |x|.

What the symbols mean

x is the abscissa, or horizontal coordinate, of the point.
y is the ordinate, or vertical coordinate.
| | denotes absolute value: a distance cannot be negative.

Given here: P(3,5); required perpendicular distance from the x-axis.

Official CBSE answer · Solution by Team TutorVertex

Distance from the x-axis=|y|=|5|=5 units. Hence k=5.

Correct answer: C — 5

Question 2 · Basic

The mirror image of (−3, 5) in the x-axis is:

CBSE 2025–26 official sample · Basic · Q8

A. (3, 5)B. (3, −5)C. (−3, −5)D. (−3, 5)

Show Answer & Explanation

Formula guide · Team TutorVertex

Reflection in the x-axis

(x,y) ↦ (x,−y)

What the symbols mean

x is the abscissa and stays unchanged.
y is the ordinate and changes sign.
The original point and its image are equally far from the x-axis.

Given here: P(−3,5); reflection in the x-axis.

Official CBSE solution · typesetting normalised

Reflection in the x-axis preserves the abscissa and reverses the ordinate: (−3,5) ↦ (−3,−5).

Correct answer: C — (−3, −5)

TutorVertex illustration · added to explain the solution.

Question 3 · Standard

The shortest distance, in units, of (2, 3) from the y-axis is:

CBSE 2025–26 official sample · Standard · Q2

A. 2B. 3C. 5D. 1

Show Answer & Explanation

Formula guide · Team TutorVertex

Perpendicular distances from the axes

Distance from x-axis = |y|; distance from y-axis = |x|.

What the symbols mean

x is the abscissa, or horizontal coordinate, of the point.
y is the ordinate, or vertical coordinate.
| | denotes absolute value: a distance cannot be negative.

Given here: P(2,3); perpendicular distance from the y-axis.

Official CBSE solution · typesetting normalised

The shortest distance from (2,3) to the y-axis is its x-coordinate, 2 units.

Correct answer: A — 2

Distance formula

Question 4 · Standard

M lies on the y-axis, 4 units below the x-axis. The distance of M from Q(5, 1) is:

CBSE 2025 supplementary examination · Standard · 30/S/1 · Q8

A. √2 unitsB. √34 unitsC. √50 unitsD. √90 units

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: M=(0,−4), Q=(5,1).

Official CBSE answer · Solution by Team TutorVertex

MQ=√[(5−0)²+(1−(−4))²]=√(25+25)=√50 units.

Correct answer: C — √50 units

Question 5 · Standard

AB is a chord of a circle with centre O(2, 3). If A=(4, 3) and B=(x, 5), then x equals:

CBSE 2023 supplementary examination · Standard · 30/C/2 · Q2

A. 3B. 2C. 5D. 4

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: O(2,3), A(4,3), B(x,5); OA=OB because both are radii.

Official CBSE answer · Solution by Team TutorVertex

OA²=(4−2)²+(3−3)²=4. OB²=(x−2)²+(5−3)²=(x−2)²+4. Equating gives (x−2)²=0 ⇒ x=2.

Correct answer: B — 2

Question 6 · Basic

The distance of (3, 4) from the origin is:

CBSE 2024 board examination · Basic · 430/2/2 · Q17

A. 25B. 5C. 7D. 1

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: P(3,4), O(0,0).

Official CBSE answer · Solution by Team TutorVertex

OP=√(3²+4²)=√25=5 units.

Correct answer: B — 5

Question 7 · Standard

The endpoints of a diameter are (5, −2) and (5, 2). The radius of the circle is:

CBSE 2025 board examination · Standard · 30/2/1 · Q5

A. ±2B. ±4C. 4D. 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: Diameter endpoints A(5,−2), B(5,2).

Official CBSE answer · Solution by Team TutorVertex

AB=√[(5−5)²+(2−(−2))²]=4. Radius=AB/2=2 units. A radius is a positive length, so ±2 is not the answer.

Correct answer: D — 2

Question 8 · Standard

The distance of (−6, 8) from the origin is:

CBSE 2023 board examination · Standard · 30/4/1 · Q13

A. 6B. −6C. 8D. 10

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: P(−6,8), O(0,0).

Official CBSE answer · Solution by Team TutorVertex

OP=√[(−6)²+8²]=√(36+64)=10 units.

Correct answer: D — 10

Question 9 · Standard

A point (x, y) is 5 units from the origin. How many such points lie in the third quadrant?

CBSE 2023–24 official sample · Standard · Q7

A. 0B. 1C. 2D. Infinitely many

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance from the origin

OP = √(x²+y²); OP=5 ⇒ x²+y²=25

What the symbols mean

O=(0,0) is the origin and P=(x,y) is a point on the plane.
The third quadrant has x<0 and y<0.
The equation describes a circle, including a continuous arc in the third quadrant.

Given here: OP=5; x<0 and y<0 in the third quadrant.

Official CBSE answer · Solution by Team TutorVertex

The distance condition gives x²+y²=25. The part of this circle in the third quadrant contains infinitely many points; the condition does not restrict x and y to integers.

Correct answer: D — Infinitely many

Question 10 · Basic

The distance between (a, b) and (−a, −b) is:

CBSE 2024–25 official sample · Basic · Q4

A. √(a² + b²)B. a² + b²C. 2√(a² + b²)D. 4√(a² + b²)

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: P(a,b), Q(−a,−b).

Official CBSE answer · Solution by Team TutorVertex

PQ=√[(−a−a)²+(−b−b)²]=√(4a²+4b²)=2√(a²+b²).

Correct answer: C — 2√(a² + b²)

Question 11 · Basic

The distance between (cos 30°, sin 30°) and (cos 60°, −sin 60°) is:

CBSE 2025–26 official sample · Basic · Q7

A. 0 unitsB. √3 unitsC. 1 unitD. √2 units

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: P=(√ $\frac{3}{2}$ , $\frac{1}{2}$ ), Q=( $\frac{1}{2}$ ,−√ $\frac{3}{2}$ ), using standard trigonometric values.

Official CBSE solution · typesetting normalised

Distance = √[((1−√3)/2)²+((1+√3)/2)²] = √2 units.

Correct answer: D — √2 units

Midpoints and diameters

Question 12 · Basic

If the midpoint of the line segment joining (3a, 4) and (−2, 2b) is (5, a), then:

CBSE 2024 supplementary examination · Basic · 430(B)/S · Q10

A. a = 2, b = 5B. a = 4, b = 2C. a = 3, b = 4D. a = 5, b = 8

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: Endpoints (3a,4), (−2,2b); midpoint (5,a).

Official CBSE answer · Solution by Team TutorVertex

(3a−2)/2=5 ⇒ 3a=12 ⇒ a=4. Also, (4+2b)/2=a=4 ⇒ 4+2b=8 ⇒ b=2.

Correct answer: B — a = 4, b = 2

Question 13 · Basic

The midpoint of A(3, −1) and B(7, 5) is (5a, 2b). The value of a + b is:

CBSE 2024 supplementary examination · Basic · 430/S/1 · Q3

A. 7B. 3C. 2D. 5

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: A(3,−1), B(7,5); midpoint (5a,2b).

Official CBSE answer · Solution by Team TutorVertex

The midpoint is ((3+7)/2,(−1+5)/2)=(5,2). Thus 5a=5 and 2b=2, so a=b=1 and a+b=2.

Correct answer: C — 2

Question 14 · Basic

If the origin is the midpoint of P(a, b) and Q(3, 3), then a + b equals:

CBSE 2024 supplementary examination · Basic · 430/S/2 · Q12

A. 0B. 3C. 6D. −6

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: Midpoint O(0,0); endpoints P(a,b), Q(3,3).

Official CBSE answer · Solution by Team TutorVertex

(a+3)/2=0 ⇒ a=−3. Similarly, (b+3)/2=0 ⇒ b=−3. Therefore a+b=−6.

Correct answer: D — −6

Question 15 · Basic

A circle has centre (2, −3). One endpoint of diameter AB is A(3, −10). The coordinates of B are:

CBSE 2024 supplementary examination · Basic · 430/S/3 · Q7

A. (4, 1)B. (−4, 1)C. (1, 4)D. (−1, −4)

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: Centre O(2,−3), diameter endpoint A(3,−10); let B=(x,y).

Official CBSE answer · Solution by Team TutorVertex

The centre is the midpoint of a diameter. (3+x)/2=2 ⇒ x=1; (−10+y)/2=−3 ⇒ y=4. Thus B=(1,4).

Correct answer: C — (1, 4)

TutorVertex illustration · added to explain the solution.

Question 16 · Standard

A(−4, 5) and C(8, 2) are opposite vertices of parallelogram ABCD. Its diagonals intersect at P(a, b). Which relation holds?

CBSE 2025 supplementary examination · Standard · 30/S/1 · Q15

A. b = a − 1.5B. b = a + 1.5C. b = a − 4.5D. b = a + 4.5

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: Opposite vertices A(−4,5), C(8,2); diagonal intersection P(a,b).

Official CBSE answer · Solution by Team TutorVertex

Diagonals of a parallelogram bisect each other. P=((−4+8)/2,(5+2)/2)=(2,3.5). Hence b=a+1.5.

Correct answer: B — b = a + 1.5

Question 17 · Basic

If (a, 6) is the midpoint of A(−7, 7) and B(−3, 5), the value of a is:

CBSE 2024 board examination · Basic · 430(B) · Q9

A. 5B. −4C. 2D. −5

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: A(−7,7), B(−3,5); midpoint (a,6).

Official CBSE answer · Solution by Team TutorVertex

a=(−7+(−3))/2= $\frac{−10}{2}$ =−5. The ordinate also agrees: (7+5)/2=6.

Correct answer: D — −5

Question 18 · Basic

C(1, −1) is the midpoint of A(4, x) and B(−2, 4). The value of x is:

CBSE 2024 board examination · Basic · 430/4/1 · Q2

A. 5B. −5C. 6D. −6

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: A(4,x), B(−2,4), midpoint C(1,−1).

Official CBSE answer · Solution by Team TutorVertex

Using the ordinate, (x+4)/2=−1 ⇒ x+4=−2 ⇒ x=−6. The abscissa agrees: (4−2)/2=1.

Correct answer: D — −6

Question 19 · Basic

The midpoint of the line segment joining (−6, −4) and (0, 4) is:

CBSE 2024 board examination · Basic · 430/4/3 · Q17

A. (−6, 0)B. (−3, 0)C. (−6, 8)D. (−6, 4)

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: A(−6,−4), B(0,4).

Official CBSE answer · Solution by Team TutorVertex

M=((−6+0)/2,(−4+4)/2)=(−3,0).

Correct answer: B — (−3, 0)

Question 20 · Basic

The midpoint of A(−2, 8) and B(−6, 4) is:

CBSE 2024 board examination · Basic · 430/5/1 · Q15

A. (2, 6)B. (−4, 12)C. (−4, 6)D. (4, −6)

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: A(−2,8), B(−6,4).

Official CBSE answer · Solution by Team TutorVertex

M=((−2−6)/2,(8+4)/2)=(−4,6).

Correct answer: C — (−4, 6)

Question 21 · Standard

PQ is a diameter of a circle with centre O(2, −4). If P=(−4, 5), then Q is:

CBSE 2024 supplementary examination · Standard · 30/S/1 · Q8

A. (−3, 4.5)B. (−1, 0.5)C. (4, −5)D. (8, −13)

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: O(2,−4), P(−4,5); let Q=(x,y).

Official CBSE answer · Solution by Team TutorVertex

The centre is the midpoint of PQ. (−4+x)/2=2 ⇒ x=8; (5+y)/2=−4 ⇒ y=−13. Thus Q=(8,−13).

Correct answer: D — (8, −13)

Question 22 · Basic

If (a, 2b) is the midpoint of (3, 5) and (−1, −1), then (a, b) is:

CBSE 2025 board examination · Basic · 430/3/3 · Q11

A. (1, 2)B. (2, 2)C. (2, 1)D. (1, 1)

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: Endpoints (3,5), (−1,−1); midpoint (a,2b).

Official CBSE answer · Solution by Team TutorVertex

The midpoint is ((3−1)/2,(5−1)/2)=(1,2). Thus a=1 and 2b=2 ⇒ b=1. Hence (a,b)=(1,1).

Correct answer: D — (1, 1)

Question 23 · Standard

If the midpoint of (a, 4) and (2, 2b) is (2, 6), then a + b equals:

CBSE 2025 board examination · Standard · 30/3/2 · Q15

A. 6B. 7C. 8D. 16

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: Endpoints (a,4), (2,2b); midpoint (2,6).

Official CBSE answer · Solution by Team TutorVertex

(a+2)/2=2 ⇒ a=2. Also, (4+2b)/2=6 ⇒ b=4. Hence a+b=6.

Correct answer: A — 6

Question 24 · Standard

Which point lies on the perpendicular bisector of the segment joining A(−3, −4) and B(3, 4)?

CBSE 2026–27 official sample · Standard · Q7

A. (0, 0)B. (0, 3)C. (3, 0)D. (0, 4)

Show Answer & Explanation

Formula guide · Team TutorVertex

Midpoint formula

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M is the midpoint; it divides AB in the ratio 1:1.
For a diameter, the centre of the circle is its midpoint.

Given here: A(−3,−4), B(3,4).

Official CBSE solution · typesetting normalised

The midpoint of AB lies on its perpendicular bisector. The midpoint is ((−3+3)/2,(−4+4)/2)=(0,0).

Correct answer: A — (0, 0)

Internal section formula

Question 25 · Basic

The y-axis divides the line segment joining (1, −5) and (−4, 5) in the ratio:

CBSE 2026 second board examination · Basic · 430/7/1 · Q2

A. 4 : 1B. 1 : 1C. 1 : 5D. 1 : 4

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(1,−5), B(−4,5); P lies on the y-axis.

Official CBSE answer · Solution by Team TutorVertex

Let AP:PB=m:n. Since xₚ=0, (−4m+n)/(m+n)=0 ⇒ n=4m. Thus m:n=1:4.

Correct answer: D — 1 : 4

Question 26 · Standard

The y-axis intersects the line segment A(−2, −3) to B(3, 7) internally in the ratio:

CBSE 2024 supplementary examination · Standard · 30/S/2 · Q8

A. 3 : 2B. 2 : 3C. 3 : 7D. 7 : 3

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(−2,−3), B(3,7); P lies on the y-axis.

Official CBSE answer · Solution by Team TutorVertex

Let AP:PB=m:n. The abscissa is (3m−2n)/(m+n)=0 ⇒ 3m=2n. Therefore m:n=2:3.

Correct answer: B — 2 : 3

TutorVertex illustration · added to explain the solution.

Question 27 · Standard

M(2, −3) divides the segment from P(−5, 11) to Q internally, with PM : MQ = 7 : 2. The coordinates of Q are:

CBSE 2024 supplementary examination · Standard · 30/S/3 · Q8

A. (4, −7)B. (27.5, −52)C. (−7, 4)D. (

\frac{4}{9}

\frac{1}{9}

)

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: P(−5,11), M(2,−3), PM:MQ=7:2; let Q=(x,y).

Official CBSE answer · Solution by Team TutorVertex

2=(7x+2(−5))/9 ⇒ 18=7x−10 ⇒ x=4. Also, −3=(7y+2(11))/9 ⇒ −27=7y+22 ⇒ y=−7. Thus Q=(4,−7).

Correct answer: A — (4, −7)

Question 28 · Standard

P divides the line segment joining A(4, −5) and B(1, 2) in the ratio 5 : 2. The coordinates of P are:

CBSE 2024 board examination · Standard · 30/2/1 · Q2

A. (

\frac{5}{2}

\frac{−3}{2}

)B. (

\frac{11}{7}

, 0)C. (

\frac{13}{7}

, 0)D. (0,

\frac{13}{7}

)

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(4,−5), B(1,2); AP:PB=5:2.

Official CBSE answer · Solution by Team TutorVertex

xₚ=(5×1+2×4)/7= $\frac{13}{7}$ . yₚ=(5×2+2×(−5))/7=0. Thus P=( $\frac{13}{7}$ ,0).

Correct answer: C — ( $\frac{13}{7}$ , 0)

Question 29 · Basic

The x-axis divides the segment from A(−8, 4) to B(−6, −2) in the ratio:

CBSE 2026 board examination · Basic · 430(B) · Q7

A. 5 : 1B. 3 : 1C. 2 : 1D. 1 : 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(−8,4), B(−6,−2); P lies on the x-axis.

Official CBSE answer · Solution by Team TutorVertex

Let AP:PB=m:n. Its ordinate is (−2m+4n)/(m+n)=0 ⇒ m=2n. Therefore AP:PB=2:1.

Correct answer: C — 2 : 1

Question 30 · Standard

The y-axis divides the segment from P(−4, −2) to Q(10, 4) in the ratio:

CBSE 2026 board examination · Standard · 30/5/1 · Q2

A. 2 : 5B. 1 : 2C. 2 : 1D. 5 : 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: P(−4,−2), Q(10,4); R lies on the y-axis.

Official CBSE answer · Solution by Team TutorVertex

Let PR:RQ=m:n. Since xᵣ=0, (10m−4n)/(m+n)=0 ⇒ 10m=4n. Therefore m:n=2:5.

Correct answer: A — 2 : 5

Question 31 · Standard

P(−1, 5) and Q(5, 2) are two vertices of triangle PQR. The point dividing PQ in the ratio 2 : 1 is:

CBSE 2025 board examination · Standard · 30/3/1 · Q7

A. (3, −3)B. (5, 5)C. (3, 3)D. (5, 1)

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: P(−1,5), Q(5,2); PX:XQ=2:1.

Official CBSE answer · Solution by Team TutorVertex

X=((2×5+1×(−1))/3,(2×2+1×5)/3)=( $\frac{9}{3}$ , $\frac{9}{3}$ )=(3,3).

Correct answer: C — (3, 3)

Question 32 · Standard

The x-axis divides the segment joining A(−6, 5) and B(−4, −1) in the ratio:

CBSE 2023 board examination · Standard · 30(B) · Q7

A. 1 : 5B. 1 : 7C. 5 : 1D. 7 : 1

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(−6,5), B(−4,−1); P lies on the x-axis.

Official CBSE answer · Solution by Team TutorVertex

Let AP:PB=m:n. Since yₚ=0, (−m+5n)/(m+n)=0 ⇒ m=5n. Therefore AP:PB=5:1.

Correct answer: C — 5 : 1

Question 33 · Basic

In what ratio does the y-axis divide the segment joining (5, −6) and (−1, −4)?

CBSE 2022–23 official sample · Basic · Q8

A. 1 : 5B. 5 : 1C. 1 : 1D. 1 : 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(5,−6), B(−1,−4); P lies on the y-axis.

Official CBSE answer · Solution by Team TutorVertex

Let AP:PB=m:n. Its abscissa is (−m+5n)/(m+n)=0 ⇒ m=5n. Thus m:n=5:1.

Correct answer: B — 5 : 1

Question 34 · Standard

In what ratio does the x-axis divide the segment joining (2, −3) and (5, 6)?

CBSE 2023–24 official sample · Standard · Q6

A. 1 : 2B. 2 : 1C. 2 : 5D. 5 : 2

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(2,−3), B(5,6); P lies on the x-axis.

Official CBSE answer · Solution by Team TutorVertex

Let AP:PB=m:n. Its ordinate is (6m−3n)/(m+n)=0 ⇒ n=2m. Therefore m:n=1:2.

Correct answer: A — 1 : 2

Question 35 · Standard

A point on the x-axis divides A(2, −3) to B(5, 6) in the ratio 1 : 2. The point is:

CBSE 2024–25 official sample · Standard · Q17

A. (4, 0)B. (

\frac{7}{2}

\frac{3}{2}

)C. (3, 0)D. (0, 3)

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(2,−3), B(5,6); AP:PB=1:2.

Official CBSE answer · Solution by Team TutorVertex

P=((1×5+2×2)/3,(1×6+2×(−3))/3)=(3,0).

Correct answer: C — (3, 0)

Question 36 · Basic

P(5, 2) divides the segment joining A(8, 5) and B(4, y) in the ratio 3 : 1. The value of y is:

CBSE 2026–27 official sample · Basic · Q8

A. 4B. 3C. 2D. 1

Show Answer & Explanation

Formula guide · Team TutorVertex

Internal section formula

P = ( $\frac{mx₂ + nx₁}{m + n}$ , $\frac{my₂ + ny₁}{m + n}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P lies between A and B, with AP:PB = m:n; m and n are positive ratio terms.
m multiplies the coordinates of B; n multiplies those of A.
A point on the x-axis has y=0; a point on the y-axis has x=0.

Given here: A(8,5), B(4,y), P(5,2); AP:PB=3:1.

Official CBSE solution · typesetting normalised

Using the section formula, (3y+5)/(3+1)=2 ⇒ 3y+5=8 ⇒ y=1.

Correct answer: D — 1

Equidistant points

Question 37 · Basic

The coordinates of a point on the x-axis, equidistant from (−2, 5) and (2, −3), are:

CBSE 2024 supplementary examination · Basic · 430(B)/S · Q9

A. (−5, 0)B. (−3, 0)C. (−4, 0)D. (−2, 0)

Show Answer & Explanation

Formula guide · Team TutorVertex

Equal distances

PA = PB ⇒ PA² = PB²; AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P is the point at equal distances from A and B.
On the x-axis use P(x,0); on the y-axis use P(0,y).
A point on the perpendicular bisector of AB is equidistant from A and B.

Given here: A(−2, 5), B(2, −3); P lies on the x-axis.

Official CBSE answer · Solution by Team TutorVertex

Let P=(x,0). PA²=PB² gives (x+2)²+25=(x−2)²+9. Thus 8x=−16, so x=−2 and P=(−2,0).

Correct answer: D — (−2, 0)

Question 38 · Basic

P(x, y) is equidistant from (0, 5) and (5, 0). Which statement is true?

CBSE 2026 second board examination · Basic · 430/7/1 · Q6

A. y = xB. x + y = 0C. x = 5yD. x + y = 5

Show Answer & Explanation

Formula guide · Team TutorVertex

Equal distances

PA = PB ⇒ PA² = PB²; AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P is the point at equal distances from A and B.
On the x-axis use P(x,0); on the y-axis use P(0,y).
A point on the perpendicular bisector of AB is equidistant from A and B.

Given here: A(0,5), B(5,0); PA=PB.

Official CBSE answer · Solution by Team TutorVertex

PA²=PB² gives x²+(y−5)²=(x−5)²+y². On expanding and cancelling common terms, −10y=−10x. Hence y=x.

Correct answer: A — y = x

Question 39 · Standard

The point on the x-axis equidistant from (5, −3) and (4, 2) is:

CBSE 2024 supplementary examination · Standard · 30/S/2 · Q2

A. (4.5, 0)B. (7, 0)C. (0.5, 0)D. (−7, 0)

Show Answer & Explanation

Formula guide · Team TutorVertex

Equal distances

PA = PB ⇒ PA² = PB²; AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P is the point at equal distances from A and B.
On the x-axis use P(x,0); on the y-axis use P(0,y).
A point on the perpendicular bisector of AB is equidistant from A and B.

Given here: A(5,−3), B(4,2); P=(x,0).

Official CBSE answer · Solution by Team TutorVertex

(x−5)²+9=(x−4)²+4 ⇒ −10x+34=−8x+20 ⇒ 2x=14 ⇒ x=7. Therefore P=(7,0).

Correct answer: B — (7, 0)

Question 40 · Standard

The perpendicular bisector of the segment joining A(−1, 3) and B(2, 4) meets the y-axis at:

CBSE 2024 board examination · Standard · 30(B) · Q7

A. (0, 5)B. (0, −5)C. (0, 4)D. (0, −4)

Show Answer & Explanation

Formula guide · Team TutorVertex

Equal distances

PA = PB ⇒ PA² = PB²; AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P is the point at equal distances from A and B.
On the x-axis use P(x,0); on the y-axis use P(0,y).
A point on the perpendicular bisector of AB is equidistant from A and B.

Given here: A(−1,3), B(2,4); P=(0,y) lies on their perpendicular bisector.

Official CBSE answer · Solution by Team TutorVertex

A point on the perpendicular bisector is equidistant from the endpoints. PA²=PB² ⇒ 1+(y−3)²=4+(y−4)² ⇒ −6y+10=−8y+20 ⇒ y=5. Hence P=(0,5).

Correct answer: A — (0, 5)

Question 41 · Standard

The point P on the y-axis equidistant from (−2, 7) and (3, 6) is:

CBSE 2023 board examination · Standard · 30(B) · Q6

A. (0, 1)B. (4, 0)C. (0, 4)D. (0, −1)

Show Answer & Explanation

Formula guide · Team TutorVertex

Equal distances

PA = PB ⇒ PA² = PB²; AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
P is the point at equal distances from A and B.
On the x-axis use P(x,0); on the y-axis use P(0,y).
A point on the perpendicular bisector of AB is equidistant from A and B.

Given here: A(−2,7), B(3,6); P=(0,y).

Official CBSE answer · Solution by Team TutorVertex

4+(y−7)²=9+(y−6)² ⇒ −14y+53=−12y+45 ⇒ 2y=8 ⇒ y=4. Thus P=(0,4).

Correct answer: C — (0, 4)

Triangles and quadrilaterals

Question 42 · Standard

The points (−2, −2), (6, −2) and (2, 1) are vertices of:

CBSE 2024 supplementary examination · Standard · 30(B) · Q18

A. a right-angled triangleB. an isosceles triangleC. an isosceles right triangleD. a scalene triangle

Show Answer & Explanation

Formula guide · Team TutorVertex

Side lengths and triangle type

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
An isosceles triangle has at least two equal side lengths.
For squared side lengths u≤v≤w, a right angle requires u+v=w.
Compare all three side lengths to distinguish isosceles, equilateral and scalene triangles.

Given here: A(−2,−2), B(6,−2), C(2,1).

Official CBSE answer · Solution by Team TutorVertex

AB²=8²=64; AC²=4²+3²=25; BC²=(−4)²+3²=25. Thus AC=BC=5, so the triangle is isosceles. It is not right-angled, because 25+25≠64.

Correct answer: B — an isosceles triangle

TutorVertex illustration · added to explain the solution.

Question 43 · Standard

The fourth vertex D of parallelogram ABCD, whose vertices are A(−2, 3), B(6, 7) and C(8, 3), is:

CBSE 2024 board examination · Standard · 30/4/1 · Q16

A. (0, 1)B. (0, −1)C. (−1, 0)D. (1, 0)

Show Answer & Explanation

Formula guide · Team TutorVertex

Diagonals and midpoints

Midpoint of AC = midpoint of BD; M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M denotes a midpoint. In a parallelogram, the two diagonals bisect each other.
The stated vertices are taken in order; pair opposite vertices when forming the diagonals.

Given here: Consecutive vertices A(−2,3), B(6,7), C(8,3); let D=(x,y).

Official CBSE answer · Solution by Team TutorVertex

The diagonals bisect each other. Midpoint of AC is ((−2+8)/2,(3+3)/2)=(3,3). Therefore (6+x)/2=3 and (7+y)/2=3, giving D=(0,−1).

Correct answer: B — (0, −1)

TutorVertex illustration · added to explain the solution.

Question 44 · Basic

The fourth vertex D of parallelogram ABCD, with A(−4, 1), B(4, 5) and C(6, 1), is:

CBSE 2026 board examination · Basic · 430(B) · Q6

A. (−2, −3)B. (3, −2)C. (0, −1)D. (0, 1)

Show Answer & Explanation

Formula guide · Team TutorVertex

Diagonals and midpoints

Midpoint of AC = midpoint of BD; M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M denotes a midpoint. In a parallelogram, the two diagonals bisect each other.
The stated vertices are taken in order; pair opposite vertices when forming the diagonals.

Given here: Consecutive vertices A(−4,1), B(4,5), C(6,1).

Official CBSE answer · Solution by Team TutorVertex

Midpoint of AC=((−4+6)/2,(1+1)/2)=(1,1). Let D=(x,y). Since the midpoint of BD is also (1,1), (4+x)/2=1 and (5+y)/2=1. Hence D=(−2,−3).

Correct answer: A — (−2, −3)

Question 45 · Standard

The points (−5, 0), (5, 0) and (0, 4) are vertices of:

CBSE 2025 board examination · Standard · 30/2/1 · Q6

A. a right-angled triangleB. an isosceles triangleC. an equilateral triangleD. a scalene triangle

Show Answer & Explanation

Formula guide · Team TutorVertex

Side lengths and triangle type

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
An isosceles triangle has at least two equal side lengths.
For squared side lengths u≤v≤w, a right angle requires u+v=w.
Compare all three side lengths to distinguish isosceles, equilateral and scalene triangles.

Given here: A(−5,0), B(5,0), C(0,4).

Official CBSE answer · Solution by Team TutorVertex

AC²=5²+4²=41 and BC²=(−5)²+4²=41, while AB²=10²=100. Exactly two side lengths are equal, so the triangle is isosceles.

Correct answer: B — an isosceles triangle

Question 46 · Standard

The perimeter of the triangle with vertices (0, 0), (2, 0) and (0, 2) is:

CBSE 2025 board examination · Standard · 30/2/2 · Q3

A. 4 unitsB. 6 unitsC. 6√2 unitsD. (4 + 2√2) units

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance and perimeter

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]; perimeter = AB + BC + CA

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
A, B and C are the triangle’s vertices.
AB, BC and CA are side lengths; their sum is the perimeter.

Given here: A(0,0), B(2,0), C(0,2).

Official CBSE answer · Solution by Team TutorVertex

AB=2 and AC=2. BC=√[(0−2)²+(2−0)²]=√8=2√2. Perimeter=AB+BC+CA=4+2√2 units.

Correct answer: D — (4 + 2√2) units

Question 47 · Standard

AOBC is a rectangle with A(0, 2), O(0, 0) and B(4, 0). The square of the length of a diagonal is:

CBSE 2025 board examination · Standard · 30/3/1 · Q16

A. 36B. 20C. 16D. 4

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance formula

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
AB is the non-negative distance between A and B, measured in units.
For distance from the origin O(0,0), OP = √(x² + y²).

Given here: Rectangle AOBC; opposite vertices A(0,2), B(4,0).

Official CBSE answer · Solution by Team TutorVertex

AB²=(4−0)²+(0−2)²=16+4=20 square units. Both diagonals of a rectangle have equal length.

Correct answer: B — 20

TutorVertex illustration · added to explain the solution.

Question 48 · Basic

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is:

CBSE 2022–23 official sample · Basic · Q6

A. 5 unitsB. 12 unitsC. 11 unitsD. (7 + √5) units

Show Answer & Explanation

Formula guide · Team TutorVertex

Distance and perimeter

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]; perimeter = AB + BC + CA

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
A, B and C are the triangle’s vertices.
AB, BC and CA are side lengths; their sum is the perimeter.

Given here: A(0,4), B(0,0), C(3,0).

Official CBSE answer · Solution by Team TutorVertex

AB=4, BC=3 and AC=√[(3−0)²+(0−4)²]=5. Perimeter=4+3+5=12 units.

Correct answer: B — 12 units

TutorVertex illustration · added to explain the solution.

Question 49 · Standard

P(3, 4), Q(−2, 3) and R(−3, −2) are consecutive vertices of parallelogram PQRS. The fourth vertex S is:

CBSE 2022–23 official sample · Standard · Q5

A. (−2, −1)B. (−2, −3)C. (2, −1)D. (1, 2)

Show Answer & Explanation

Formula guide · Team TutorVertex

Diagonals and midpoints

Midpoint of AC = midpoint of BD; M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M denotes a midpoint. In a parallelogram, the two diagonals bisect each other.
The stated vertices are taken in order; pair opposite vertices when forming the diagonals.

Given here: P(3,4), Q(−2,3), R(−3,−2).

Official CBSE answer · Solution by Team TutorVertex

The diagonals PR and QS bisect each other. Midpoint of PR=(0,1). Let S=(x,y). Then (−2+x)/2=0 and (3+y)/2=1, giving S=(2,−1).

Correct answer: C — (2, −1)

Question 50 · Standard

A(9, 0), B(9, −6), C(−9, 0) and D(−9, 6) are vertices of a:

CBSE 2025–26 official sample · Standard · Q16

A. SquareB. RectangleC. ParallelogramD. Trapezium

Show Answer & Explanation

Formula guide · Team TutorVertex

Diagonals and midpoints

Midpoint of AC = midpoint of BD; M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ )

What the symbols mean

x₁, y₁ are the coordinates of the first endpoint A.
x₂, y₂ are the coordinates of the second endpoint B.
M denotes a midpoint. In a parallelogram, the two diagonals bisect each other.
The stated vertices are taken in order; pair opposite vertices when forming the diagonals.

Given here: Vertices A(9,0), B(9,−6), C(−9,0), D(−9,6), taken in order.

Official CBSE solution · typesetting normalised

Plotting the four points shows that the quadrilateral is a parallelogram.

Correct answer: C — Parallelogram

TutorVertex illustration · added to explain the solution.

Question 51 · Standard

PM is a median of triangle PQR, with P(5, −6), Q(6, 4) and R(0, 0). The length of PM is:

CBSE 2026–27 official sample · Standard · Q14

A. 2√17 unitsB. 2√15 unitsC. √101 unitsD. 10 units

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Formula guide · Team TutorVertex

Midpoint and median

M = ( $\frac{x₁ + x₂}{2}$ , $\frac{y₁ + y₂}{2}$ ); PM = √[(xₘ−xₚ)²+(yₘ−yₚ)²]

What the symbols mean

M is the midpoint of QR, because PM is a median.
xₘ,yₘ are M’s coordinates; xₚ,yₚ are P’s coordinates.
Find M first, then apply the distance formula to P and M.

Given here: P(5,−6), Q(6,4), R(0,0); M is the midpoint of QR.

Official CBSE solution · typesetting normalised

M=((6+0)/2,(4+0)/2)=(3,2). PM=√[(5−3)²+(−6−2)²]=√68=2√17 units.

Correct answer: A — 2√17 units

TutorVertex illustration · added to explain the solution.

Common questions

Which formulas are covered?

Distance, midpoint and internal section formulas, with applications to axes, equidistant points, triangles, diameters and parallelograms. External division and the coordinate determinant formula for triangle area are not added as current chapter topics.

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Where the corresponding official marking scheme supplies working, the explanation is labelled as an official CBSE solution. When it supplies only an answer, Team TutorVertex provides the working. Formula guides, symbol definitions and supplementary illustrations are by Team TutorVertex.

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The included MCQs are text-only in their cited papers. Selected explanations contain extra TutorVertex coordinate illustrations. These are teaching aids and are labelled separately from the official question.

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The 2022 papers in the checked CBSE archive use the written-answer format. They are not converted into invented multiple-choice questions. This bank includes official sample-paper MCQs from 2022–23.

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Source links are collected here. Each question above identifies its year/session, level, paper code and question number.

How this question bank is prepared

This is a selected collection of 36 board-examination MCQs from 2023–2026 and 15 official sample-paper MCQs from sessions 2022–23 to 2026–27. Answers are checked against corresponding official marking schemes and independently checked mathematically. Mathematical data and choice order are retained; wording is lightly edited for clarity. Repeated questions across sets are represented selectively.

The available CBSE archive collections are screened with text extraction and OCR where needed. This is not an exhaustive transcription of every question or a claim of complete translation of every language version. Unclear or ambiguous questions are excluded. NCERT is used to check chapter relevance, notation and methods.