🔵 Section A — Multiple Choice 10 × 1 = 10 marks
Q1 · MCQ
The point (−3, 5) lies in which quadrant?
I
II
III
IV
Q2 · MCQ
The distance between the points (0, 0) and (6, 8) is:
14
10
√14
2√7
Q3 · MCQ
The midpoint of the segment joining (2, 3) and (4, 7) is:
(3, 5)
(6, 10)
(1, 2)
(3, 4)
Q4 · MCQ
The distance of the point (−4, −3) from the origin is:
7
5
√7
1
Q5 · MCQ
A point on the y-axis is of the form:
(x, 0)
(0, y)
(x, x)
(y, y)
Q6 · MCQ
The point dividing (1, 2) and (3, 4) in ratio 1 : 1 is the:
Midpoint (2, 3)
(4, 6)
(1, 1)
(3, 4)
Q7 · MCQ
If the area of a triangle formed by three points is 0, the points are:
Vertices of a square
Collinear
Equidistant
On the y-axis
Q8 · MCQ
The centroid of a triangle with vertices (0,0), (3,0), (0,3) is:
(1, 1)
(3, 3)
(1.5, 1.5)
(0, 0)
Q9 · MCQ
In what ratio does the x-axis divide the segment joining (2, −3) and (5, 6)?
1 : 2
2 : 1
1 : 3
3 : 1
Q10 · MCQ
The distance between (a, 0) and (0, b) is:
a + b
√(a² + b²)
a² + b²
|a − b|
🟣 Section B — Assertion & Reason 3 × 1 = 3 marks
(a) Both A & R true, R explains A · (b) Both true, R doesn't explain A · (c) A true, R false · (d) A false, R true
Q11 · A–R
Assertion: The points (1,1), (2,2), (3,3) are collinear.Reason: Three points are collinear if the area of the triangle they form is zero.
Show Answer
(a) Both correct and R is the exact reason. Area = ½|1(2−3)+2(3−1)+3(1−2)| = ½|−1+4−3| = 0, so they are collinear.
Q12 · A–R
Assertion: The distance of (3, 4) from the origin is 5.Reason: Distance from origin of (x, y) is x + y.
Show Answer
(c) Assertion is true (√(9+16)=5) but the Reason is false — the correct formula is √(x²+y²) , not x + y.
Q13 · A–R
Assertion: The midpoint of (−2, 3) and (4, −1) is (1, 1).Reason: Midpoint = ( (x₁+x₂)/2 , (y₁+y₂)/2 ).
Show Answer
(a) Both true and R explains A: ((−2+4)/2, (3−1)/2) = (1, 1).
🟢 Section C — Short Answer 12 × 2 = 24 marks
Q14 · Short
Find the distance between the points (2, 3) and (4, 1).
Show Answer
d = √[(4−2)²+(1−3)²] = √(4+4) = 2√2 units.
Q15 · Short
Show that (1, 5), (2, 3) and (−2, −11) are not collinear.
Show Answer
Area = ½|1(3−(−11)) + 2((−11)−5) + (−2)(5−3)| = ½|14 − 32 − 4| = ½(22) = 11 ≠ 0 . Since area ≠ 0, they are not collinear.
Q16 · Short
Find the coordinates of the point which divides (4, −3) and (8, 5) in the ratio 3 : 1 internally.
Show Answer
x = (3·8+1·4)/4 = 28/4 = 7; y = (3·5+1·(−3))/4 = 12/4 = 3. Point = (7, 3) .
Q17 · Short
The midpoint of segment AB is (2, 5). If A = (−1, 4), find B.
Show Answer
B = 2M − A = (2·2−(−1), 2·5−4) = (5, 6) .
Q18 · Short
Find the area of the triangle with vertices (1, −1), (−4, 6) and (−3, −5).
Show Answer
Area = ½|1(6−(−5)) + (−4)((−5)−(−1)) + (−3)((−1)−6)| = ½|11 + 16 + 21| = ½(48) = 24 sq. units .
Q19 · Short
Find a point on the x-axis equidistant from (2, −5) and (−2, 9).
Show Answer
Let P(x, 0). PA² = PB²: (x−2)²+25 = (x+2)²+81 → −4x+25 = 4x+81 → 8x = −56 → x = −7. Point = (−7, 0) .
Q20 · Short
In what ratio does the point (−4, 6) divide the segment joining A(−6, 10) and B(3, −8)?
Show Answer
Let ratio k:1. x: (3k−6)/(k+1) = −4 → 3k−6 = −4k−4 → 7k = 2 → k = 2/7. Ratio = 2 : 7 .
Q21 · Short
Find the centroid of the triangle with vertices (−1, 3), (2, −4) and (5, 7).
Show Answer
G = ((−1+2+5)/3, (3−4+7)/3) = (6/3, 6/3) = (2, 2) .
Q22 · Short
Name the quadrant / axis for the points (0, −4), (5, 0) and (−2, −6).
Show Answer
(0, −4) → on the negative y-axis ; (5, 0) → on the positive x-axis ; (−2, −6) → Quadrant III .
Q23 · Short
Find the value of y for which the distance between (2, −3) and (10, y) is 10 units.
Show Answer
100 = (10−2)² + (y+3)² = 64 + (y+3)² → (y+3)² = 36 → y+3 = ±6 → y = 3 or y = −9 .
Q24 · Short
Find the coordinates of the points of trisection of the segment joining (2, −2) and (−7, 4).
Show Answer
1:2 → ((1·−7+2·2)/3, (1·4+2·−2)/3) = (−1, 0). 2:1 → ((2·−7+1·2)/3, (2·4+1·−2)/3) = (−4, 2). Points: (−1, 0) and (−4, 2) .
Q25 · Short
Find the value of k if the points (2, 3), (4, k) and (6, −3) are collinear.
Show Answer
Area = 0: 2(k+3) + 4(−3−3) + 6(3−k) = 0 → 2k+6 −24 +18 −6k = 0 → −4k = 0 → k = 0 .
🔴 Section D — Long Answer 5 × 2 = 10 marks
Q26 · Long
Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square .
Show Answer
Sides: AB = √(9+25)=√34, BC = √(25+9)=√34, CD = √(9+25)=√34, DA = √(25+9)=√34 — all equal. Diagonals AC = √(4+64)=√68, BD = √(64+4)=√68 — equal. All sides equal and diagonals equal ⇒ square .
Q27 · Long
Find the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by (−1, 6).
Show Answer
Let ratio k:1. x: (6k−3)/(k+1) = −1 → 6k−3 = −k−1 → 7k = 2 → k = 2/7. Ratio = 2 : 7 . (Check with y: (−8·2/7+10)/(2/7+1) = 6 ✓)
Q28 · Long
Find the area of the quadrilateral whose vertices, in order, are (−4, −2), (−3, −5), (3, −2) and (2, 3).
Show Answer
Split by diagonal AC. Δ ABC: ½|−4(−5+2) −3(−2+2) +3(−2+5)| = ½|12+0+9| = 10.5. Δ ACD: ½|−4(−2−3) +3(3+2) +2(−2+2)| = ½|20+15+0| = 17.5. Total = 28 sq. units .
Q29 · Long
Find the point on the y-axis which is equidistant from the points (5, −2) and (−3, 2).
Show Answer
Let P(0, y). PA² = PB²: 25+(y+2)² = 9+(y−2)² → 25 + y²+4y+4 = 9 + y²−4y+4 → 8y = −16 → y = −2. Point = (0, −2) .
Q30 · Long
Two vertices of a triangle are (1, 2) and (3, 5) and its centroid is (2, 3). Find the third vertex.
Show Answer
Centroid = average: (1+3+x)/3 = 2 → x = 2; (2+5+y)/3 = 3 → y = 2. Third vertex = (2, 2) .