How to Use This Page
This page contains complete step-by-step solutions for all exercises in Samacheer Kalvi 9th Maths Chapter 5 (Coordinate Geometry). Work through each exercise on your own first, then compare your method and answer with the solutions below. Pay special attention to the working steps — the process matters as much as the final answer in exams.
Key Concepts (Quick Recap)
- Cartesian Plane: A plane formed by two perpendicular lines — the horizontal x-axis and the vertical y-axis. The origin (0,0) is where they intersect.
- Quadrants: The four regions of the cartesian plane — I (+,+), II (−,+), III (−,−), IV (+,−).
- Distance Formula: d = sqrt[(x₂−x₁)² + (y₂−y₁)²] — distance between (x₁,y₁) and (x₂,y₂).
- Section Formula (Internal Division): A point P dividing AB in ratio m:n has coordinates ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)).
- Midpoint Formula: M = ((x₁+x₂)/2, (y₁+y₂)/2) — the point that bisects a line segment.
- Centroid of a Triangle: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) — intersection of the three medians.
- Area of Triangle: Area = (1/2)|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.
- Collinearity Test: Three points are collinear if the area of the triangle they form equals zero.
Exercise 5.1 — Cartesian Plane and Plotting Points
Q1. Plot the following points on the cartesian plane and identify the quadrant: (3, 4)
Point (3, 4):
x-coordinate = 3 (positive)
y-coordinate = 4 (positive)
Since both coordinates are positive, the point lies in Quadrant I.
Move 3 units right along the x-axis and 4 units up along the y-axis.
Q2. Plot the point (−2, 5) and identify its quadrant.
Point (−2, 5):
x-coordinate = −2 (negative)
y-coordinate = 5 (positive)
Since x is negative and y is positive, the point lies in Quadrant II.
Move 2 units left along the x-axis and 5 units up along the y-axis.
Q3. Plot the point (−3, −4) and identify its quadrant.
Point (−3, −4):
x-coordinate = −3 (negative)
y-coordinate = −4 (negative)
Since both coordinates are negative, the point lies in Quadrant III.
Move 3 units left along the x-axis and 4 units down along the y-axis.
Q4. Plot the point (5, −2) and identify its quadrant.
Point (5, −2):
x-coordinate = 5 (positive)
y-coordinate = −2 (negative)
Since x is positive and y is negative, the point lies in Quadrant IV.
Move 5 units right along the x-axis and 2 units down along the y-axis.
Q5. Where does the point (0, 5) lie?
Point (0, 5):
x-coordinate = 0 → the point lies on the y-axis.
y-coordinate = 5 → the point is 5 units above the origin on the y-axis.
Any point with x = 0 lies on the y-axis.
Q6. Where does the point (−4, 0) lie?
Point (−4, 0):
x-coordinate = −4 → the point is 4 units to the left of the origin.
y-coordinate = 0 → the point lies on the x-axis.
Any point with y = 0 lies on the x-axis.
Q7. If a point has the form (x, y) where x > 0 and y < 0, in which quadrant does it lie?
Given: x > 0 (positive) and y < 0 (negative).
Positive x → right half of the plane.
Negative y → lower half of the plane.
The intersection is Quadrant IV.
Q8. In which quadrant does the point (−1, −6) lie?
Point (−1, −6):
x = −1 (negative)
y = −6 (negative)
Both negative → Quadrant III.
Q9. A point lies on the x-axis at a distance of 7 units from the origin. What are the possible coordinates?
A point on the x-axis has y = 0.
Distance from origin = |x| = 7.
So x = 7 or x = −7.
Possible coordinates: (7, 0) or (−7, 0).
Q10. Plot the points A(2, 3), B(−2, 3), C(−2, −3), D(2, −3). What shape do they form?
Plotting all four points:
A(2, 3) — Quadrant I
B(−2, 3) — Quadrant II
C(−2, −3) — Quadrant III
D(2, −3) — Quadrant IV
The four points are symmetric about both axes:
AB is horizontal (same y = 3, from x = −2 to x = 2), length = 4
BC is vertical (same x = −2, from y = 3 to y = −3), length = 6
CD is horizontal (same y = −3, from x = −2 to x = 2), length = 4
DA is vertical (same x = 2, from y = −3 to y = 3), length = 6
All angles are right angles (90°).
Opposite sides are equal.
ABCD forms a rectangle.
Exercise 5.2 — Distance Formula
Q1. Find the distance between the points A(4, 1) and B(1, 5).
Using distance formula: d = sqrt[(x₂ − x₁)² + (y₂ − y₁)²]
x₁ = 4, y₁ = 1, x₂ = 1, y₂ = 5
d = sqrt[(1 − 4)² + (5 − 1)²]
= sqrt[(−3)² + (4)²]
= sqrt[9 + 16]
= sqrt[25]
= 5 units
Q2. Find the distance between the points P(6, 2) and Q(2, 6).
x₁ = 6, y₁ = 2, x₂ = 2, y₂ = 6
d = sqrt[(2 − 6)² + (6 − 2)²]
= sqrt[(−4)² + (4)²]
= sqrt[16 + 16]
= sqrt[32]
= 4 sqrt(2) units
Q3. Find the distance between (−1, 2) and (3, −1).
x₁ = −1, y₁ = 2, x₂ = 3, y₂ = −1
d = sqrt[(3 − (−1))² + (−1 − 2)²]
= sqrt[(3 + 1)² + (−3)²]
= sqrt[16 + 9]
= sqrt[25]
= 5 units
Q4. Find the distance of the point (5, 12) from the origin.
Distance from origin (0, 0) to (5, 12):
d = sqrt[(5 − 0)² + (12 − 0)²]
= sqrt[25 + 144]
= sqrt[169]
= 13 units
Q5. Find the distance of the point (3, −4) from the origin.
Distance from origin (0, 0) to (3, −4):
d = sqrt[(3)² + (−4)²]
= sqrt[9 + 16]
= sqrt[25]
= 5 units
Q6. Show that the points A(1, 1), B(5, 4), C(2, 8) form a right-angled triangle.
Step 1: Find the three side lengths.
AB = sqrt[(5−1)² + (4−1)²] = sqrt[16 + 9] = sqrt[25] = 5
BC = sqrt[(2−5)² + (8−4)²] = sqrt[9 + 16] = sqrt[25] = 5
AC = sqrt[(2−1)² + (8−1)²] = sqrt[1 + 49] = sqrt[50] = 5sqrt(2)
Step 2: Check Pythagoras theorem.
AB² + BC² = 25 + 25 = 50
AC² = 50
Since AB² + BC² = AC², the triangle is right-angled at B.
Q7. Show that the points A(0, 0), B(3, 0), C(3, 4) form a right triangle.
Step 1: Find all three sides.
AB = sqrt[(3−0)² + (0−0)²] = sqrt[9 + 0] = 3
BC = sqrt[(3−3)² + (4−0)²] = sqrt[0 + 16] = 4
AC = sqrt[(3−0)² + (4−0)²] = sqrt[9 + 16] = 5
Step 2: Check Pythagoras theorem.
AB² + BC² = 9 + 16 = 25
AC² = 25
Since AB² + BC² = AC², triangle ABC is right-angled at B.
Q8. Show that the points (1, 2), (3, 4), (5, 6) are collinear.
Using the area of triangle formula to test collinearity:
Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
x₁ = 1, y₁ = 2; x₂ = 3, y₂ = 4; x₃ = 5, y₃ = 6
Area = (1/2)|1(4 − 6) + 3(6 − 2) + 5(2 − 4)|
= (1/2)|1(−2) + 3(4) + 5(−2)|
= (1/2)|−2 + 12 − 10|
= (1/2)|0|
= 0
Since area = 0, the three points are collinear.
Q9. Find a point on the x-axis which is equidistant from (2, −5) and (−2, 9).
Let the point on the x-axis be (a, 0).
Since it is equidistant from (2, −5) and (−2, 9):
Distance from (a, 0) to (2, −5) = Distance from (a, 0) to (−2, 9)
sqrt[(2 − a)² + (−5 − 0)²] = sqrt[(−2 − a)² + (9 − 0)²]
Squaring both sides:
(2 − a)² + 25 = (−2 − a)² + 81
Expand:
4 − 4a + a² + 25 = 4 + 4a + a² + 81
Simplify:
29 − 4a = 85 + 4a
29 − 85 = 4a + 4a
−56 = 8a
a = −7
The point is (−7, 0).
Verification:
Distance to (2, −5): sqrt[(2−(−7))² + (−5)²] = sqrt[81 + 25] = sqrt[106]
Distance to (−2, 9): sqrt[(−2−(−7))² + 9²] = sqrt[25 + 81] = sqrt[106]
Both distances are equal. ✓
Q10. Determine whether the triangle with vertices A(2, 3), B(4, 7), C(6, 11) is right-angled, isosceles, or scalene.
Step 1: Find the three side lengths.
AB = sqrt[(4−2)² + (7−3)²] = sqrt[4 + 16] = sqrt[20] = 2sqrt(5)
BC = sqrt[(6−4)² + (11−7)²] = sqrt[4 + 16] = sqrt[20] = 2sqrt(5)
AC = sqrt[(6−2)² + (11−3)²] = sqrt[16 + 64] = sqrt[80] = 4sqrt(5)
Step 2: Check the type.
AB = BC = 2sqrt(5) → two sides are equal.
The triangle is isosceles.
Step 3: Check if right-angled.
AB² + BC² = 20 + 20 = 40
AC² = 80
AB² + BC² ≠ AC² → Not right-angled.
Q11. The distance between the points (a, 2) and (3, 4) is sqrt(8). Find the value of a.
d = sqrt[(3 − a)² + (4 − 2)²] = sqrt(8)
Squaring both sides:
(3 − a)² + 4 = 8
(3 − a)² = 4
3 − a = ±2
Case 1: 3 − a = 2 → a = 1
Case 2: 3 − a = −2 → a = 5
Verification:
If a = 1: d = sqrt[(3−1)² + 4] = sqrt[4 + 4] = sqrt(8) ✓
If a = 5: d = sqrt[(3−5)² + 4] = sqrt[4 + 4] = sqrt(8) ✓
Q12. Find the distance between the points (−5, 7) and (−1, 3).
x₁ = −5, y₁ = 7, x₂ = −1, y₂ = 3
d = sqrt[(−1 − (−5))² + (3 − 7)²]
= sqrt[(4)² + (−4)²]
= sqrt[16 + 16]
= sqrt[32]
= 4sqrt(2) units
Q13. Show that the points A(4, 3), B(6, 7), C(8, 3) form an isosceles triangle.
Find the three sides:
AB = sqrt[(6−4)² + (7−3)²] = sqrt[4 + 16] = sqrt[20] = 2sqrt(5)
BC = sqrt[(8−6)² + (3−7)²] = sqrt[4 + 16] = sqrt[20] = 2sqrt(5)
AC = sqrt[(8−4)² + (3−3)²] = sqrt[16 + 0] = 4
AB = BC = 2sqrt(5), but AC = 4.
Since AB = BC, the triangle is isosceles.
Q14. Show that the points A(2, −2), B(8, 6), C(5, 2) are collinear.
Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
x₁ = 2, y₁ = −2; x₂ = 8, y₂ = 6; x₃ = 5, y₃ = 2
Area = (1/2)|2(6 − 2) + 8(2 − (−2)) + 5((−2) − 6)|
= (1/2)|2(4) + 8(4) + 5(−8)|
= (1/2)|8 + 32 − 40|
= (1/2)|0|
= 0
Since area = 0, the points are collinear.
Q15. Find the perimeter of the triangle with vertices A(0, 0), B(4, 0), C(0, 3).
AB = sqrt[(4−0)² + (0−0)²] = sqrt[16] = 4
BC = sqrt[(0−4)² + (3−0)²] = sqrt[16 + 9] = sqrt[25] = 5
CA = sqrt[(0−0)² + (0−3)²] = sqrt[9] = 3
Perimeter = AB + BC + CA = 4 + 5 + 3 = 12 units
Exercise 5.3 — Section Formula, Midpoint, and Centroid
Q1. Find the midpoint of the line segment joining A(2, 4) and B(6, 8).
Midpoint formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
x₁ = 2, y₁ = 4, x₂ = 6, y₂ = 8
M = ((2 + 6)/2, (4 + 8)/2)
= (8/2, 12/2)
= (4, 6)
Q2. Find the midpoint of the segment joining (−3, 5) and (7, −1).
M = ((−3 + 7)/2, (5 + (−1))/2)
= (4/2, 4/2)
= (2, 2)
Q3. Find the point that divides the line segment joining A(2, −3) and B(5, 6) in the ratio 2:1 internally.
Section formula: P = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))
m = 2, n = 1
A = (2, −3), B = (5, 6)
P = ((2×5 + 1×2)/(2+1), (2×6 + 1×(−3))/(2+1))
= ((10 + 2)/3, (12 − 3)/3)
= (12/3, 9/3)
= (4, 3)
Q4. Find the point that divides the segment joining P(1, 6) and Q(5, −2) in the ratio 3:1 internally.
m = 3, n = 1
P = (1, 6), Q = (5, −2)
R = ((3×5 + 1×1)/(3+1), (3×(−2) + 1×6)/(3+1))
= ((15 + 1)/4, (−6 + 6)/4)
= (16/4, 0/4)
= (4, 0)
Q5. Find the coordinates of the point which divides the line segment joining A(−2, 3) and B(4, −5) in the ratio 1:2 internally.
m = 1, n = 2
A = (−2, 3), B = (4, −5)
P = ((1×4 + 2×(−2))/(1+2), (1×(−5) + 2×3)/(1+2))
= ((4 − 4)/3, (−5 + 6)/3)
= (0/3, 1/3)
= (0, 1/3)
Q6. Find the centroid of the triangle with vertices A(1, 2), B(3, 4), C(5, 6).
Centroid formula: G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)
G = ((1 + 3 + 5)/3, (2 + 4 + 6)/3)
= (9/3, 12/3)
= (3, 4)
Q7. Find the centroid of the triangle with vertices (2, −4), (6, 2), and (−2, 6).
G = ((2 + 6 + (−2))/3, (−4 + 2 + 6)/3)
= (6/3, 4/3)
= (2, 4/3)
Q8. The midpoint of a line segment is (3, 4). If one endpoint is (1, 2), find the other endpoint.
Midpoint formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Given: M = (3, 4), endpoint A = (1, 2), endpoint B = (x, y)
(1 + x)/2 = 3 → 1 + x = 6 → x = 5
(2 + y)/2 = 4 → 2 + y = 8 → y = 6
The other endpoint is (5, 6).
Q9. The midpoint of a segment is (5, −1). If one endpoint is (−3, 7), find the other endpoint.
((−3 + x)/2, (7 + y)/2) = (5, −1)
(−3 + x)/2 = 5 → −3 + x = 10 → x = 13
(7 + y)/2 = −1 → 7 + y = −2 → y = −9
The other endpoint is (13, −9).
Q10. Find the ratio in which the point (4, 3) divides the line segment joining A(2, 1) and B(8, 7).
Let the ratio be m:n.
Using section formula for x-coordinate:
(mx₂ + nx₁)/(m + n) = 4
(8m + 2n)/(m + n) = 4
8m + 2n = 4m + 4n
4m = 2n
m/n = 2/4 = 1/2
So the ratio is 1:2.
Verification with y-coordinate:
(my₂ + ny₁)/(m + n) = (1×7 + 2×1)/(1 + 2) = (7 + 2)/3 = 9/3 = 3 ✓
Q11. Find the ratio in which the x-axis divides the line segment joining (2, −3) and (5, 6).
The x-axis divides the segment at a point (x, 0).
Let the ratio be m:n.
Using the y-coordinate of the section formula:
(my₂ + ny₁)/(m + n) = 0
(6m + (−3)n)/(m + n) = 0
6m − 3n = 0
6m = 3n
m/n = 3/6 = 1/2
The x-axis divides the segment in the ratio 1:2.
Finding the x-coordinate of division:
x = (1×5 + 2×2)/(1 + 2) = (5 + 4)/3 = 9/3 = 3
Point of division: (3, 0)
Q12. Find the centroid of the triangle with vertices A(0, 0), B(6, 0), C(0, 8).
G = ((0 + 6 + 0)/3, (0 + 0 + 8)/3)
= (6/3, 8/3)
= (2, 8/3)
Q13. The point P divides the line segment joining A(−1, 3) and B(9, 8) in the ratio 3:2. Find the coordinates of P.
m = 3, n = 2
A = (−1, 3), B = (9, 8)
P = ((3×9 + 2×(−1))/(3+2), (3×8 + 2×3)/(3+2))
= ((27 − 2)/5, (24 + 6)/5)
= (25/5, 30/5)
= (5, 6)
Q14. If the centroid of a triangle with vertices (a, b), (c, d), (e, f) is (0, 0), find the relation between the coordinates.
Centroid = ((a + c + e)/3, (b + d + f)/3) = (0, 0)
Therefore:
(a + c + e)/3 = 0 → a + c + e = 0
(b + d + f)/3 = 0 → b + d + f = 0
The sum of x-coordinates = 0 and the sum of y-coordinates = 0.
Q15. A line segment is divided into two parts in the ratio 2:3. If the segment joins (1, 2) and (6, 7), find the point of division.
m = 2, n = 3
A = (1, 2), B = (6, 7)
P = ((2×6 + 3×1)/(2+3), (2×7 + 3×2)/(2+3))
= ((12 + 3)/5, (14 + 6)/5)
= (15/5, 20/5)
= (3, 4)
Exercise 5.4 — Area of Triangle and Collinearity
Q1. Find the area of the triangle whose vertices are A(1, 2), B(3, 4), C(5, 3).
Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
x₁ = 1, y₁ = 2; x₂ = 3, y₂ = 4; x₃ = 5, y₃ = 3
Area = (1/2)|1(4 − 3) + 3(3 − 2) + 5(2 − 4)|
= (1/2)|1(1) + 3(1) + 5(−2)|
= (1/2)|1 + 3 − 10|
= (1/2)|−6|
= (1/2)(6)
= 3 sq. units
Q2. Find the area of the triangle with vertices (0, 0), (4, 0), (0, 3).
x₁ = 0, y₁ = 0; x₂ = 4, y₂ = 0; x₃ = 0, y₃ = 3
Area = (1/2)|0(0 − 3) + 4(3 − 0) + 0(0 − 0)|
= (1/2)|0 + 12 + 0|
= (1/2)(12)
= 6 sq. units
Verification: This is a right triangle with legs 4 and 3.
Area = (1/2)(4)(3) = 6 sq. units ✓
Q3. Find the area of the triangle with vertices A(2, 3), B(4, 7), C(6, 5).
Area = (1/2)|2(7 − 5) + 4(5 − 3) + 6(3 − 7)|
= (1/2)|2(2) + 4(2) + 6(−4)|
= (1/2)|4 + 8 − 24|
= (1/2)|−12|
= (1/2)(12)
= 6 sq. units
Q4. Show that the points A(1, 1), B(3, 3), C(5, 5) are collinear.
Area = (1/2)|1(3 − 5) + 3(5 − 1) + 5(1 − 3)|
= (1/2)|1(−2) + 3(4) + 5(−2)|
= (1/2)|−2 + 12 − 10|
= (1/2)|0|
= 0
Since area = 0, the three points are collinear.
Q5. Show that the points (2, 5), (5, 7), (8, 9) are collinear.
Area = (1/2)|2(7 − 9) + 5(9 − 5) + 8(5 − 7)|
= (1/2)|2(−2) + 5(4) + 8(−2)|
= (1/2)|−4 + 20 − 16|
= (1/2)|0|
= 0
Since area = 0, the points are collinear.
Q6. Find the area of the triangle formed by the points (1, −1), (2, 3), and (4, 1).
Area = (1/2)|1(3 − 1) + 2(1 − (−1)) + 4((−1) − 3)|
= (1/2)|1(2) + 2(2) + 4(−4)|
= (1/2)|2 + 4 − 16|
= (1/2)|−10|
= (1/2)(10)
= 5 sq. units
Q7. Find the area of the triangle with vertices (−2, 3), (4, 3), and (1, 7).
Area = (1/2)|(−2)(3 − 7) + 4(7 − 3) + 1(3 − 3)|
= (1/2)|(−2)(−4) + 4(4) + 1(0)|
= (1/2)|8 + 16 + 0|
= (1/2)(24)
= 12 sq. units
Q8. Determine whether the points (a, b), (c, d), and (a + c − b, b + d − a) are collinear.
Let P₁ = (a, b), P₂ = (c, d), P₃ = (a + c − b, b + d − a)
Area = (1/2)|a(d − (b + d − a)) + c((b + d − a) − b) + (a + c − b)(b − d)|
Simplify each term:
First term: a(d − b − d + a) = a(a − b) = a² − ab
Second term: c(b + d − a − b) = c(d − a) = cd − ac
Third term: (a + c − b)(b − d)
= (a + c − b)(b − d) = ab − ad + bc − cd − b² + bd
Sum = (a² − ab) + (cd − ac) + (ab − ad + bc − cd − b² + bd)
= a² − ab + cd − ac + ab − ad + bc − cd − b² + bd
= a² − ac − ad + bc + bd − b²
This simplifies further but in general does NOT equal 0.
The collinearity depends on specific values of a, b, c, d.
For a specific numerical example, check using the area formula.
Q9. The area of a triangle is 15 sq. units. If two vertices are (1, 3) and (5, 7), find the third vertex if it lies on the line y = x.
Let the third vertex be (a, a) since it lies on y = x.
A(1, 3), B(5, 7), C(a, a)
Area = (1/2)|1(7 − a) + 5(a − 3) + a(3 − 7)| = 15
(1/2)|7 − a + 5a − 15 − 4a| = 15
(1/2)|7 − 15 − a + 5a − 4a| = 15
(1/2)|−8 + 0a| = 15
(1/2)(8) = 15
4 = 15
This is not possible! This means no point on y = x gives area = 15 with these two vertices.
Let me recheck:
(1/2)|7 − a + 5a − 15 + a(3 − 7)|
= (1/2)|7 − a + 5a − 15 + 3a − 7a|
= (1/2)|7 − 15 + (−a + 5a + 3a − 7a)|
= (1/2)|−8 + 0|
= (1/2)(8) = 4
The area is always 4, regardless of the value of a.
So it is impossible for the area to be 15.
This means no such point exists on y = x.
Q10. Find the area of the quadrilateral with vertices A(1, 1), B(5, 1), C(5, 4), D(1, 4) taken in order.
Divide the quadrilateral into two triangles using diagonal AC.
Triangle ABC:
Area = (1/2)|1(1 − 4) + 5(4 − 1) + 5(1 − 1)|
= (1/2)|1(−3) + 5(3) + 5(0)|
= (1/2)|−3 + 15|
= (1/2)(12) = 6
Triangle ACD:
Area = (1/2)|1(4 − 4) + 5(4 − 1) + 1(1 − 4)|
= (1/2)|0 + 15 − 3|
= (1/2)(12) = 6
Total area = 6 + 6 = 12 sq. units
Verification: This is a rectangle with length 4 and width 3.
Area = 4 × 3 = 12 sq. units ✓
Q11. For what value of k are the points (k, 2), (3, 4), (7, 6) collinear?
For collinearity, area must be 0:
(1/2)|k(4 − 6) + 3(6 − 2) + 7(2 − 4)| = 0
(1/2)|k(−2) + 3(4) + 7(−2)| = 0
(1/2)|−2k + 12 − 14| = 0
(1/2)|−2k − 2| = 0
−2k − 2 = 0
−2k = 2
k = −1
Q12. Find the area of the triangle with vertices (3, 8), (−7, 2), and (1, −4).
Area = (1/2)|3(2 − (−4)) + (−7)((−4) − 8) + 1(8 − 2)|
= (1/2)|3(6) + (−7)(−12) + 1(6)|
= (1/2)|18 + 84 + 6|
= (1/2)(108)
= 54 sq. units
Q13. If the area of the triangle formed by the points (a, 0), (0, b), and (1, 1) is zero, find the relation between a and b.
Area = (1/2)|a(b − 1) + 0(1 − 0) + 1(0 − b)| = 0
(1/2)|ab − a + 0 − b| = 0
ab − a − b = 0
ab = a + b
ab − a − b + 1 = 1
(a − 1)(b − 1) = 1
Exercise 5.5 — Word Problems and Applications
Q1. A farmer has a triangular field with vertices at A(0, 0), B(40, 0), and C(20, 30). Find the area of the field.
Area = (1/2)|0(0 − 30) + 40(30 − 0) + 20(0 − 0)|
= (1/2)|0 + 1200 + 0|
= (1/2)(1200)
= 600 sq. units
If 1 unit = 1 m, then Area = 600 m².
Q2. Three friends live at coordinates A(0, 0), B(6, 0), C(3, 4). Find the centroid of the triangle. If they want to meet at a point equidistant from all three, where should they meet?
Centroid:
G = ((0 + 6 + 3)/3, (0 + 0 + 4)/3)
= (9/3, 4/3)
= (3, 4/3)
The centroid divides each median in the ratio 2:1 and is the center of mass.
It is NOT equidistant from all three vertices (that would be the circumcenter).
However, the centroid is the meeting point that minimizes total distance.
Q3. The vertices of a triangle are A(0, 0), B(6, 0), C(0, 8). Find the length of the median from A to BC and find the centroid.
Step 1: Find midpoint of BC (this is point D).
D = ((6 + 0)/2, (0 + 8)/2) = (3, 4)
Step 2: Length of median from A to D.
AD = sqrt[(3 − 0)² + (4 − 0)²] = sqrt[9 + 16] = sqrt[25] = 5 units
Step 3: Centroid.
G = ((0 + 6 + 0)/3, (0 + 0 + 8)/3) = (2, 8/3)
Q4. A park is in the shape of a triangle with vertices at A(2, 1), B(6, 1), C(4, 5). Find the area of the park. If 1 unit = 10 m, what is the area in square metres?
Area = (1/2)|2(1 − 5) + 6(5 − 1) + 4(1 − 1)|
= (1/2)|2(−4) + 6(4) + 4(0)|
= (1/2)|−8 + 24 + 0|
= (1/2)(16)
= 8 sq. units
If 1 unit = 10 m, then 1 sq. unit = 100 m².
Area = 8 × 100 = 800 m²
Q5. Prove that the points A(4, 0), B(0, 3), C(0, 0) form a right triangle and find its area.
Step 1: Find side lengths.
AB = sqrt[(0−4)² + (3−0)²] = sqrt[16 + 9] = sqrt[25] = 5
BC = sqrt[(0−0)² + (0−3)²] = sqrt[9] = 3
CA = sqrt[(4−0)² + (0−0)²] = sqrt[16] = 4
Step 2: Check Pythagoras.
BC² + CA² = 9 + 16 = 25 = AB² = 25 ✓
Right-angled at C.
Step 3: Area = (1/2) × BC × CA = (1/2)(3)(4) = 6 sq. units.
Verification using coordinate formula:
Area = (1/2)|0(3 − 0) + 0(0 − 0) + 4(0 − 3)|
= (1/2)|0 + 0 − 12|
= (1/2)(12) = 6 ✓
Q6. The end points of a diameter of a circle are (4, 0) and (0, 6). Find the centre and radius of the circle.
Step 1: The centre of the circle is the midpoint of the diameter.
Centre = ((4 + 0)/2, (0 + 6)/2) = (2, 3)
Step 2: The radius is half the length of the diameter.
Diameter = sqrt[(0 − 4)² + (6 − 0)²] = sqrt[16 + 36] = sqrt[52] = 2sqrt(13)
Radius = sqrt(52)/2 = sqrt(13) units
Alternative: Radius = distance from centre to an endpoint.
r = sqrt[(4 − 2)² + (0 − 3)²] = sqrt[4 + 9] = sqrt[13]
Q7. Show that the quadrilateral with vertices A(1, 2), B(3, 6), C(5, 2), D(3, −2) is a rhombus.
Step 1: Find all four side lengths.
AB = sqrt[(3−1)² + (6−2)²] = sqrt[4 + 16] = sqrt[20] = 2sqrt(5)
BC = sqrt[(5−3)² + (2−6)²] = sqrt[4 + 16] = sqrt[20] = 2sqrt(5)
CD = sqrt[(3−5)² + (−2−2)²] = sqrt[4 + 16] = sqrt[20) = 2sqrt(5)
DA = sqrt[(1−3)² + (2−(−2))²] = sqrt[4 + 16] = sqrt[20) = 2sqrt(5)
All four sides are equal (2sqrt(5)).
Step 2: Check diagonals are NOT equal.
AC = sqrt[(5−1)² + (2−2)²] = sqrt[16] = 4
BD = sqrt[(3−3)² + (−2−6)²] = sqrt[64] = 8
Diagonals are not equal (AC ≠ BD), so it is NOT a square.
Since all sides are equal and diagonals are unequal, it is a rhombus.
Q8. Show that the quadrilateral with vertices (1, 1), (2, 4), (4, 3), (3, 0) is a parallelogram.
Step 1: Find all four side lengths.
AB = sqrt[(2−1)² + (4−1)²] = sqrt[1 + 9] = sqrt[10]
BC = sqrt[(4−2)² + (3−4)²] = sqrt[4 + 1] = sqrt[5]
CD = sqrt[(3−4)² + (0−3)²] = sqrt[1 + 9] = sqrt[10]
DA = sqrt[(1−3)² + (1−0)²] = sqrt[4 + 1] = sqrt[5]
AB = CD = sqrt(10) and BC = DA = sqrt(5).
Opposite sides are equal → parallelogram.
Step 2: Alternatively, check midpoints of diagonals.
Midpoint of AC = ((1+4)/2, (1+3)/2) = (5/2, 2)
Midpoint of BD = ((2+3)/2, (4+0)/2) = (5/2, 2)
Same midpoint → diagonals bisect each other → parallelogram ✓
Q9. Find the area of the triangle formed by the points (0, 0), (a, 0), and (0, b).
Area = (1/2)|0(0 − b) + a(b − 0) + 0(0 − 0)|
= (1/2)|0 + ab + 0|
= (1/2)ab
= ab/2 sq. units
This is a right triangle with legs a and b along the axes.
Q10. The vertices of a triangle are (1, 2), (h, 3), and (4, k). Find the values of h and k if the centroid is (2, 3).
Centroid = ((1 + h + 4)/3, (2 + 3 + k)/3) = (2, 3)
From x-coordinate:
(5 + h)/3 = 2
5 + h = 6
h = 1
From y-coordinate:
(5 + k)/3 = 3
5 + k = 9
k = 4
Common Mistakes
1. Forgetting the square root in the distance formula — Writing d = (x₂−x₁)² + (y₂−y₁)² instead of d = sqrt[(x₂−x₁)² + (y₂−y₁)²]. Always apply the square root.
2. Sign errors in the distance formula — When coordinates are negative, e.g., distance between (−1, 3) and (2, −5), forgetting parentheses: (2−(−1))² = 9, not (2−1)² = 1. Always write (x₂ − x₁)², not (x₁ − x₂)² — both work, but be consistent.
3. Confusing section formula ratio order — Swapping m and n gives a different point. If the ratio is AP:PB = m:n, the formula is P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)). Verify by checking that P is closer to the point whose weight is larger.
| 4. Missing absolute value in the area formula — The area formula uses | ... | because area is always positive. Forgetting the absolute value can give a negative answer, which is wrong for area. |
|---|
5. Confusing midpoint with section formula — Midpoint is a special case of the section formula with m = n = 1. Don't divide by m+n = 2 and then divide by 2 again; midpoint = ((x₁+x₂)/2, (y₁+y₂)/2) directly.
6. Assuming collinear means distance equals zero — Three collinear points do NOT have distance zero between them. Collinearity means the area of the triangle formed is zero, not that any two points coincide.
7. Wrong quadrant identification — Point (−3, 5) is in Quadrant II (not IV), and (3, −5) is in Quadrant IV (not II). Remember: Q-I is (+,+), Q-II is (−,+), Q-III is (−,−), Q-IV is (+,−).
Exam Checklist
- [ ] Can plot points and identify quadrants correctly
- [ ] Know the distance formula and can apply it to find distances between two points
- [ ] Can find distance from a point to the origin (a special case of the distance formula)
- [ ] Can verify if three points form a right triangle using the distance formula and Pythagoras
- [ ] Can verify if three points are collinear using the area formula (= 0)
- [ ] Know the section formula for internal division in ratio m:n
- [ ] Can find the midpoint (special case: ratio 1:1)
- [ ] Can find the centroid of a triangle using ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
- [ ] Can find an unknown endpoint given the midpoint and one endpoint
- [ ] Can find the ratio in which a point divides a line segment
- [ ] Can find the area of a triangle given three vertices
- [ ] Can use the area formula to test collinearity (area = 0 means collinear)
- [ ] Can find the ratio in which the x-axis or y-axis divides a line segment
Internal Links
- [Class 9 Maths — Coordinate Geometry Chapter Guide](/school/tamil-nadu-state-board/class/9/maths/coordinate-geometry)
- [All Class 9 Maths Solutions](/school/tamil-nadu-state-board/class/9/maths/solutions)
- [Class 9 Maths Subject Guide](/school/tamil-nadu-state-board/class/9/maths)
Frequently Asked Questions
Use the distance formula: d = sqrt[(x2-x1)^2 + (y2-y1)^2]. For example, the distance between (1,2) and (4,6) is sqrt[(4-1)^2 + (6-2)^2] = sqrt[9+16] = sqrt[25] = 5 units.