Worksheet Solutions: Section and Mid-Point Formula | Mathematics Class 10 ICSE PDF Download

Section A: Very Short Questions (1 mark each)

Q1. Find the midpoint of the line joining (2, 5) and (8, -3).
Solution:
Midpoint = ((2+8)/2, (5+(-3))/2) = (10/2, 2/2) = (5, 1).

Q2. Find the coordinates of the centroid of triangle with vertices (0, 0), (6, 0), (0, 6).
Solution:
x = (0+6+0)/3 = 6/3 = 2
y = (0+0+6)/3 = 6/3 = 2
Centroid = (2, 2).

Q3. Point P(4, 2) is the midpoint of line AB. If A = (6, -1), find B.
Solution:
(6+x)/2 = 4 → 6+x = 8 → x = 2
(-1+y)/2 = 2 → y-1 = 4 → y = 5
B = (2, 5).

Q4. In which ratio does point (7, 2) divide the line joining (5, 6) and (9, 0)?
Solution:
x = (m1×9 + m2×5)/(m1+m2) = 7
7(m1+m2) = 9m1 + 5m2
7m1+7m2 = 9m1+5m2 → 2m1 = 2m2 → m1:m2 = 1:1
Ratio = 1:1.

Q5. Find the point on y-axis equidistant from (2, -3) and (-2, 5).
Solution:
Point = (0, y)
(0-2)²+(y+3)² = (0+2)²+(y-5)²
4+(y+3)² = 4+(y-5)²
(y+3)² = (y-5)²
y+3 = ±(y-5)
Case 1: y+3 = y-5 → 3=-5 (not possible)
Case 2: y+3 = -(y-5) → y+3 = -y+5 → 2y=2 → y=1
Point = (0, 1).

Section B: Short Answer Questions (2–3 marks each)

Q6. Find the coordinates of the point dividing the line joining (3, -4) and (7, 6) in the ratio 3:1.
Solution:
x = (3×7 + 1×3)/(3+1) = (21+3)/4 = 24/4 = 6
y = (3×6 + 1×(-4))/(3+1) = (18-4)/4 = 14/4 = 7/2
Point = (6, 7/2).

Q7. A(-6, 3) and B(6, -1). Find the points of trisection of AB.
Solution:
For P(1:2):
x = (1×6+2×(-6))/3 = (6-12)/3 = -2
y = (1×-1+2×3)/3 = (-1+6)/3 = 5/3
P = (-2, 5/3)

For Q(2:1):
x = (2×6+1×(-6))/3 = (12-6)/3 = 2
y = (2×-1+1×3)/3 = (-2+3)/3 = 1/3
Q = (2, 1/3).

Q8. The centroid of triangle is (2, -1). Two vertices are (1, 3) and (3, -5). Find the third vertex.
Solution:
(4+x)/3 = 2 → x = 2
(-2+y)/3 = -1 → y = -1
Third vertex = (2, -1).

Q9. Show that the points (2, 3), (4, -1), (6, 3) are collinear using section formula.
Solution:
Check if (4, -1) lies on line joining (2, 3) and (6, 3).
Line AB: (2,3) to (6,3) → slope = 0 (horizontal line, y=3).
But point (4,-1) does not satisfy y=3.
Hence, points are not collinear.

Q10. Find the ratio in which the x-axis divides the line joining (2, -4) and (6, 8). Also find the coordinates.
Solution:
y = (k×8+1×(-4))/(k+1) = 0
8k-4=0 → k=1/2
Ratio = 1:2
x = (1×6+2×2)/(1+2) = (6+4)/3 = 10/3
Point = (10/3, 0).

Section C: Long Answer Questions (4–5 marks each)

Q11. Find the coordinates of the centroid of triangle A(2, -2), B(6, 4), C(-4, 2). Verify by showing that centroid divides median in 2:1 ratio.
Solution:
Centroid = ((2+6+(-4))/3, (-2+4+2)/3) = (4/3, 4/3).
Midpoint of BC = ((6+(-4))/2, (4+2)/2) = (1, 3).
Median from A(2, -2) to (1, 3).
Check 2:1 ratio:
x = (2×1+1×2)/3 = 4/3, y=(2×3+1×-2)/3=4/3.
Centroid = (4/3, 4/3).

Q12. Find the points of trisection of the line joining (7, -2) and (-2, 4). Verify that the distance between them is one-third of total length.
Solution:
For P(1:2):
x = (1×(-2)+2×7)/3 = 12/3=4
y = (1×4+2×-2)/3 = 0
P=(4, 0)

For Q(2:1):
x = (2×(-2)+1×7)/3 = 1
y = (2×4+1×-2)/3 = 2
Q=(1, 2)

PQ=√((4-1)²+(0-2)²)=√(9+4)=√13
AB=√((7-(-2))²+(-2-4)²)=√(9²+(-6)²)=√117=3√13
PQ = (1/3)AB.

Q13. A(1, 2), B(5, -2), C(-3, -4). Find centroid and verify that medians intersect at same point.
Solution:
Centroid = ((1+5+(-3))/3, (2+(-2)+(-4))/3) = (1, -4/3).

Median 1: Midpoint of BC=((5+(-3))/2, (-2+(-4))/2)=(1, -3).
Median from A(1, 2) to (1, -3).
Check: y=(2×-3+2)/3 = -4/3.

Median 2: Midpoint of AC=((1+(-3))/2,(2+(-4))/2)=(-1,-1).
Median from B(5, -2) to (-1, -1).
Check: x=(2×-1+5)/3=1.
Centroid verified at (1, -4/3).