Exercise 7.1 10 Questions – Distance Formula
AB = √(x₂ âˆ' x₁)² + (y₂ âˆ' y₁)²
Find the distance between the following pairs of points:
(i) A(2, 3) and B(4, 1)
Step 1: Let A(2, 3) be (x₁, y₁) and B(4, 1) be (x₂, y₂).
Step 2: x₂ âˆ' x₁ = 4 âˆ' 2 = 2
y₂ âˆ' y₁ = 1 âˆ' 3 = âˆ'2
Step 3: AB = √(2)² + (âˆ'2)²
AB = √4 + 4
AB = √8 = 2√2
Answer: AB = 2√2 ≈ 2.83 units
Distance AB forms hypotenuse of right triangle with legs 2 and 2
(ii) P(âˆ'5, 7) and Q(âˆ'1, 3)
Step 1: x₂ âˆ' x₁ = âˆ'1 âˆ' (âˆ'5) = âˆ'1 + 5 = 4
y₂ âˆ' y₁ = 3 âˆ' 7 = âˆ'4
Step 2: PQ = √(4)² + (âˆ'4)²
PQ = √16 + 16
PQ = √32 = √16 × 2 = 4√2
Answer: PQ = 4√2 ≈ 5.66 units
(iii) (a, b) and (âˆ'a, âˆ'b)
Step 1: x₂ âˆ' x₁ = âˆ'a âˆ' a = âˆ'2a
y₂ âˆ' y₁ = âˆ'b âˆ' b = âˆ'2b
Step 2: d = √(âˆ'2a)² + (âˆ'2b)²
d = √4a² + 4b²
d = √4(a² + b²)
d = 2 × √a² + b²
Answer: d = 2√(a² + b²)
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between two towns A(36, 15) and B(0, 0)?
Step 1: O(0, 0) is (x₁, y₁), P(36, 15) is (x₂, y₂).
Step 2: x₂ âˆ' x₁ = 36 âˆ' 0 = 36, y₂ âˆ' y₁ = 15 âˆ' 0 = 15
Step 3: OP = √36² + 15²
OP = √1296 + 225
OP = √1521
Step 4: Now find √1521. Check: 39 × 39 = 1521.
So √1521 = 39
Answer: The distance is 39 units.
Town A is at (36, 15) and Town B at (0, 0). The distance between them is also 39 km (if 1 unit = 1 km).
Determine if the points (1, 5), (2, 3) and (âˆ'2, âˆ'11) are collinear.
Method: Points A, B, C are collinear if and only if AB + BC = AC (i.e., one distance equals sum of the other two).
Step 1: Let A(1, 5), B(2, 3), C(âˆ'2, âˆ'11).
AB = √(2âˆ'1)² + (3âˆ'5)² = √1² + (âˆ'2)² = √1 + 4 = √5 ≈ 2.236
Step 2: BC = √(âˆ'2âˆ'2)² + (âˆ'11âˆ'3)² = √(âˆ'4)² + (âˆ'14)² = √16 + 196 = √212 ≈ 14.56
Step 3: AC = √(âˆ'2âˆ'1)² + (âˆ'11âˆ'5)² = √(âˆ'3)² + (âˆ'16)² = √9 + 256 = √265 ≈ 16.279
Step 4: AB + BC ≈ 2.236 + 14.56 = 16.796
AC ≈ 16.279
Since AB + BC ≠AC, the points are NOT collinear.
Check whether (5, âˆ'2), (6, 4) and (7, âˆ'2) are the vertices of an isosceles triangle.
Step 1: Let A(5, âˆ'2), B(6, 4), C(7, âˆ'2).
AB = √(6âˆ'5)² + (4âˆ'(âˆ'2))² = √1² + 6² = √1 + 36 = √37
Step 2: BC = √(7âˆ'6)² + (âˆ'2âˆ'4)² = √1² + (âˆ'6)² = √1 + 36 = √37
Step 3: AC = √(7âˆ'5)² + (âˆ'2âˆ'(âˆ'2))² = √2² + 0² = √4 = 2
Step 4: Since AB = BC = √37, two sides are equal.
Answer: Yes, it is an isosceles triangle.
AB = BC = √37, AC = 2 – Isosceles Triangle
In a classroom, 4 friends are seated at points A(3, 4), B(6, 7), C(9, 4) and D(6, 1). What type of quadrilateral does ABCD form?
Step 1: Find all four side lengths:
AB = √(6âˆ'3)² + (7âˆ'4)² = √9 + 9 = √18 = 3√2
BC = √(9âˆ'6)² + (4âˆ'7)² = √9 + 9 = 3√2
CD = √(6âˆ'9)² + (1âˆ'4)² = √9 + 9 = 3√2
DA = √(3âˆ'6)² + (4âˆ'1)² = √9 + 9 = 3√2
Step 2: All four sides are equal (= 3√2). It's at least a rhombus.
Step 3: Check diagonals: AC = √(9âˆ'3)² + (4âˆ'4)² = √36 + 0 = 6
BD = √(6âˆ'6)² + (1âˆ'7)² = √0 + 36 = 6
Diagonals AC = BD = 6. All sides equal + diagonals equal.
Answer: ABCD is a SQUARE.
All sides = 3√2, Diagonals = 6 – Square
Name the type of quadrilateral formed by points: (i) (âˆ'1, âˆ'2), (1, 0), (âˆ'1, 2), (âˆ'3, 0)
Step 1: Let A(âˆ'1,âˆ'2), B(1,0), C(âˆ'1,2), D(âˆ'3,0).
Step 2: All four sides:
AB = √(1+1)²+(0+2)²=√4+4=√8=2√2
BC = √(âˆ'1âˆ'1)²+(2âˆ'0)²=√4+4=2√2
CD = √(âˆ'3+1)²+(0âˆ'2)²=√4+4=2√2
DA = √(âˆ'1+3)²+(âˆ'2âˆ'0)²=√4+4=2√2
Step 3: All sides equal. Diagonals: AC = √0+16=4, BD = √16+0=4.
Answer: SQUARE (all sides equal, diagonals equal).
Find the point on the x-axis which is equidistant from (2, âˆ'5) and (âˆ'2, 9).
Step 1: Any point on x-axis has y-coordinate = 0. Let P(h, 0) be the required point.
Step 2: P is equidistant from A(2,âˆ'5) and B(âˆ'2,9). So PA² = PB².
(hâˆ'2)² + (0âˆ'(âˆ'5))² = (hâˆ'(âˆ'2))² + (0âˆ'9)²
(hâˆ'2)² + 25 = (h+2)² + 81
Step 3: Expand: h² âˆ' 4h + 4 + 25 = h² + 4h + 4 + 81
Cancel h² and 4: âˆ'4h + 25 = 4h + 81
âˆ'4h âˆ' 4h = 81 âˆ' 25
âˆ'8h = 56
h = âˆ'7
Answer: The point is (âˆ'7, 0).
Verify: PA = √(âˆ'9)² + 5² = √81+25 = √106. PB = √5² + (âˆ'9)² = √25+81=√106 âœ"
Find the values of y for which the distance between the points P(2, âˆ'3) and Q(10, y) is 10 units.
Step 1: Given PQ = 10. Using distance formula:
√(10âˆ'2)² + (yâˆ'(âˆ'3))² = 10
√8² + (y+3)² = 10
Step 2: Square both sides: 64 + (y+3)² = 100
(y+3)² = 100 âˆ' 64 = 36
Step 3: y + 3 = ±6
Either y + 3 = 6 â‡' y = 3
Or y + 3 = âˆ'6 â‡' y = âˆ'9
Answer: y = 3 or y = âˆ'9.
If Q(0, 1) is equidistant from P(5, âˆ'3) and R(x, 6), find the values of x.
Step 1: Q is equidistant from P and R â‡' QP = QR â‡' QP² = QR².
QP² = (5âˆ'0)² + (âˆ'3âˆ'1)² = 25 + 16 = 41
Step 2: QR² = (xâˆ'0)² + (6âˆ'1)² = x² + 25
Step 3: x² + 25 = 41 â‡' x² = 16 â‡' x = ±4
Answer: x = 4 or x = âˆ'4.
Find a relation between x and y such that the point (x, y) is equidistant from (3, 6) and (âˆ'3, 4).
Step 1: Let P(x, y) be equidistant from A(3, 6) and B(âˆ'3, 4).
PA² = PB²
(xâˆ'3)² + (yâˆ'6)² = (x+3)² + (yâˆ'4)²
Step 2: Expand both sides:
(x² âˆ' 6x + 9) + (y² âˆ' 12y + 36) = (x² + 6x + 9) + (y² âˆ' 8y + 16)
Step 3: Cancel x², y², and 9 from both sides:
âˆ'6x âˆ' 12y + 36 = 6x âˆ' 8y + 16
Step 4: Bring all terms to left: âˆ'6x âˆ' 12y + 36 âˆ' 6x + 8y âˆ' 16 = 0
âˆ'12x âˆ' 4y + 20 = 0
Divide throughout by âˆ'4: 3x + y = 5
Answer: The relation is 3x + y = 5 (a straight line – the perpendicular bisector of AB).
Exercise 7.2 10 Questions – Section Formula
x = (mx₂ + nx₁)/(m + n), y = (my₂ + ny₁)/(m + n)
Midpoint Formula (m:n = 1:1): x = (x₁ + x₂)/2, y = (y₁ + y₂)/2
Find the coordinates of the point which divides the join of (âˆ'1, 7) and (4, âˆ'3) in the ratio 2 : 3.
Step 1: Let A(âˆ'1, 7) = (x₁, y₁) and B(4, âˆ'3) = (x₂, y₂). Ratio m:n = 2:3.
Step 2 – x-coordinate: x = (mx₂ + nx₁)/(m + n)
x = (2×4 + 3×(âˆ'1))/(2+3) = (8 âˆ' 3)/5 = 5/5 = 1
Step 3 – y-coordinate: y = (my₂ + ny₁)/(m + n)
y = (2×(âˆ'3) + 3×7)/5 = (âˆ'6 + 21)/5 = 15/5 = 3
Answer: The point is (1, 3).
P divides AB in ratio 2:3 (2 parts from A to P, 3 parts from P to B)
Find the coordinates of the points of trisection of the line segment joining (4, âˆ'1) and (âˆ'2, âˆ'3).
Step 1: Trisection means dividing into three equal parts. The two dividing points divide AB in ratios 1:2 and 2:1 from A.
Step 2 – First point P (ratio 1:2 from A):
m=1, n=2. A(4,âˆ'1)=(x₁,y₁), B(âˆ'2,âˆ'3)=(x₂,y₂).
x = (1×(âˆ'2) + 2×4)/(1+2) = (âˆ'2+8)/3 = 6/3 = 2
y = (1×(âˆ'3) + 2×(âˆ'1))/3 = (âˆ'3âˆ'2)/3 = âˆ'5/3
So P = (2, âˆ'5/3)
Step 3 – Second point Q (ratio 2:1 from A):
m=2, n=1.
x = (2×(âˆ'2) + 1×4)/3 = (âˆ'4+4)/3 = 0
y = (2×(âˆ'3) + 1×(âˆ'1))/3 = (âˆ'6âˆ'1)/3 = âˆ'7/3
So Q = (0, âˆ'7/3)
Answer: The trisection points are (2, âˆ'5/3) and (0, âˆ'7/3).
To conduct Sports Day activities in your rectangular school ground ABCD, Niharika runs 1/4 the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 the distance AD on the 8th line and posts a red flag. What is the distance between both flags?
Step 1: With A as origin (0,0), AD is along x-axis (length 10), AB along y-axis. Lines are numbered from AB side.
Niharika: 2nd line â‡' x=2, y=(1/4)×AD = 2.5. Green flag at G(2, 2.5).
Preet: 8th line â‡' x=8, y=(1/5)×10 = 2. Red flag at R(8, 2).
Step 2: Distance GR = √(8âˆ'2)² + (2âˆ'2.5)² = √36 + 0.25 = √36.25 ≈ 6.02 m
Find the ratio in which the line segment joining the points (âˆ'3, 10) and (6, âˆ'8) is divided by (âˆ'1, 6).
Step 1: Let P(âˆ'1, 6) divide AB in ratio k:1. A(âˆ'3,10), B(6,âˆ'8).
Step 2: Using section formula for x:
âˆ'1 = (k×6 + 1×(âˆ'3))/(k+1)
âˆ'1(k+1) = 6k âˆ' 3
âˆ'k âˆ' 1 = 6k âˆ' 3
âˆ'7k = âˆ'2 â‡' k = 2/7
Step 3: Ratio is 2:7.
Verify y-coordinate: y = (2×(âˆ'8) + 7×10)/(9) = (âˆ'16+70)/9 = 54/9 = 6 âœ"
Answer: P divides AB in the ratio 2 : 7.
Find the ratio in which the line segment joining A(1, âˆ'5) and B(âˆ'4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Step 1: Any point on x-axis has y = 0. Let P(h, 0) divide AB in ratio k:1.
Step 2: Using y-coordinate: 0 = (k×5 + 1×(âˆ'5))/(k+1)
0 = (5k âˆ' 5)/(k+1)
5k âˆ' 5 = 0 â‡' 5k = 5 â‡' k = 1
So P divides AB in ratio 1 : 1 (midpoint).
Step 3: For k = 1: x = (1×(âˆ'4) + 1×1)/(1+1) = (âˆ'4+1)/2 = âˆ'3/2 = âˆ'1.5
Answer: Ratio 1:1, point of division is (âˆ'3/2, 0).
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Step 1: In a parallelogram, diagonals bisect each other. Midpoints of both diagonals are equal.
Let vertices be A(1,2), B(4,y), C(x,6), D(3,5). Diagonals: AC and BD.
Step 2: Midpoint of AC = Midpoint of BD
((1+x)/2, (2+6)/2) = ((4+3)/2, (y+5)/2)
Step 3: From x-coordinate: (1+x)/2 = 7/2 â‡' 1+x = 7 â‡' x = 6
From y-coordinate: (2+6)/2 = (y+5)/2 â‡' 4 = (y+5)/2 â‡' 8 = y+5 â‡' y = 3
Answer: x = 6, y = 3.
Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, âˆ'3) and B is (1, 4).
Step 1: Centre O(2,âˆ'3) is the midpoint of diameter AB. Given B(1,4), find A(x,y).
Step 2: Midpoint of AB = ((x+1)/2, (y+4)/2) = (2, âˆ'3)
(x+1)/2 = 2 â‡' x+1 = 4 â‡' x = 3
(y+4)/2 = âˆ'3 â‡' y+4 = âˆ'6 â‡' y = âˆ'10
Answer: A = (3, âˆ'10).
If A and B are (âˆ'2, âˆ'2) and (2, âˆ'4) respectively, find the coordinates of P such that AP = (3/7)AB and P lies on AB.
Step 1: AP = (3/7)AB means AP : PB = 3 : 4 (since AP = 3 parts, PB = 7âˆ'3 = 4 parts).
So P divides AB in ratio 3 : 4 from A.
Step 2: m=3, n=4. A(âˆ'2,âˆ'2), B(2,âˆ'4).
x = (3×2 + 4×(âˆ'2))/(3+4) = (6âˆ'8)/7 = âˆ'2/7
y = (3×(âˆ'4) + 4×(âˆ'2))/7 = (âˆ'12âˆ'8)/7 = âˆ'20/7
Answer: P = (âˆ'2/7, âˆ'20/7).
Find the coordinates of points dividing A(âˆ'2, 2) and B(2, 8) into four equal parts.
Step 1: Three points divide AB into 4 equal parts, at ratios 1:3, 1:1 (midpoint), and 3:1.
P (1:3): x=(1×2+3×(âˆ'2))/4=(2âˆ'6)/4=âˆ'1, y=(1×8+3×2)/4=(8+6)/4=3.5. P = (âˆ'1, 7/2)
Q (midpoint): x=(âˆ'2+2)/2=0, y=(2+8)/2=5. Q = (0, 5)
R (3:1): x=(3×2+1×(âˆ'2))/4=(6âˆ'2)/4=1, y=(3×8+1×2)/4=(24+2)/4=6.5. R = (1, 13/2)
Answer: (âˆ'1, 3.5), (0, 5), (1, 6.5).
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (âˆ'1, 4) and (âˆ'2, âˆ'1) taken in order.
Step 1: Let A(3,0), B(4,5), C(âˆ'1,4), D(âˆ'2,âˆ'1). Diagonals are AC and BD.
Step 2: Length of AC = √(âˆ'1âˆ'3)² + (4âˆ'0)² = √(âˆ'4)² + 4² = √16+16 = √32 = 4√2
Step 3: Length of BD = √(âˆ'2âˆ'4)² + (âˆ'1âˆ'5)² = √(âˆ'6)² + (âˆ'6)² = √36+36 = √72 = 6√2
Step 4: Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × 4√2 × 6√2
= (1/2) × 24 × 2 = 24 sq units
Answer: 24 square units.
Exercise 7.3 5 Questions – Area of Triangle
Area = ½ | x₁(y₂ âˆ' y₃) + x₂(y₃ âˆ' y₁) + x₃(y₁ âˆ' y₂) |
If Area = 0, the points are collinear.
Find the area of the triangle whose vertices are:
(i) (2, 3), (âˆ'1, 0), (2, âˆ'4)
Step 1: (x₁,y₁) = (2,3), (x₂,y₂) = (âˆ'1,0), (x₃,y₃) = (2,âˆ'4).
Step 2: Compute each term:
x₁(y₂ âˆ' y₃) = 2(0 âˆ' (âˆ'4)) = 2(4) = 8
x₂(y₃ âˆ' y₁) = âˆ'1(âˆ'4 âˆ' 3) = âˆ'1(âˆ'7) = 7
x₃(y₁ âˆ' y₂) = 2(3 âˆ' 0) = 2(3) = 6
Step 3: Area = ½ | 8 + 7 + 6 | = ½ × 21 = 10.5 sq units
(ii) (âˆ'5, âˆ'1), (3, âˆ'5), (5, 2)
Step 1: (x₁,y₁) = (âˆ'5,âˆ'1), (x₂,y₂) = (3,âˆ'5), (x₃,y₃) = (5,2).
Step 2: x₁(y₂âˆ'y₃) = âˆ'5(âˆ'5âˆ'2) = âˆ'5(âˆ'7) = 35
x₂(y₃âˆ'y₁) = 3(2âˆ'(âˆ'1)) = 3(3) = 9
x₃(y₁âˆ'y₂) = 5(âˆ'1âˆ'(âˆ'5)) = 5(4) = 20
Step 3: Area = ½ | 35 + 9 + 20 | = ½ × 64 = 32 sq units
Find the value of 'k' for which the points are collinear: (i) (7, âˆ'2), (5, 1), (3, k) (ii) (8, 1), (k, âˆ'4), (2, âˆ'5)
(i) (7, âˆ'2), (5, 1), (3, k)
Step 1: Collinear â‡' Area = 0.
0 = ½ | 7(1âˆ'k) + 5(kâˆ'(âˆ'2)) + 3(âˆ'2âˆ'1) |
0 = ½ | 7 âˆ' 7k + 5k + 10 + 3(âˆ'3) |
0 = ½ | 7 âˆ' 7k + 5k + 10 âˆ' 9 |
0 = ½ | 8 âˆ' 2k |
0 = 8 âˆ' 2k â‡' k = 4
(ii) (8, 1), (k, âˆ'4), (2, âˆ'5)
½ | 8(âˆ'4+5) + k(âˆ'5âˆ'1) + 2(1+4) | = 0
| 8(1) + k(âˆ'6) + 2(5) | = 0
| 8 âˆ' 6k + 10 | = 0
18 âˆ' 6k = 0 â‡' k = 3
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, âˆ'1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Step 1: A(0,âˆ'1), B(2,1), C(0,3).
Midpoints: D(mid of BC) = ((2+0)/2, (1+3)/2) = (1, 2)
E(mid of AC) = ((0+0)/2, (âˆ'1+3)/2) = (0, 1)
F(mid of AB) = ((0+2)/2, (âˆ'1+1)/2) = (1, 0)
Step 2: Area of midpoint triangle DEF:
= ½ | 1(1âˆ'0) + 0(0âˆ'2) + 1(2âˆ'1) |
= ½ | 1 + 0 + 1 | = 1 sq unit
Step 3: Area of original triangle ABC:
= ½ | 0(1âˆ'3) + 2(3âˆ'(âˆ'1)) + 0(âˆ'1âˆ'1) |
= ½ | 0 + 2(4) + 0 | = ½ × 8 = 4 sq units
Step 4: Ratio = 1 : 4 = 1 : 4.
A triangle formed by joining midpoints always has 1/4 the area of the original.
Find the area of the quadrilateral whose vertices, taken in order, are (âˆ'4, âˆ'2), (âˆ'3, âˆ'5), (3, âˆ'2) and (2, 3).
Step 1: Divide into two triangles by diagonal AC. Let A(âˆ'4,âˆ'2), B(âˆ'3,âˆ'5), C(3,âˆ'2), D(2,3).
Step 2: Area of Î"ABC:
= ½ | âˆ'4(âˆ'5+2) + (âˆ'3)(âˆ'2+2) + 3(âˆ'2+5) |
= ½ | âˆ'4(âˆ'3) âˆ' 3(0) + 3(3) | = ½ × 21 = 10.5
Step 3: Area of Î"ADC: A(âˆ'4,âˆ'2), D(2,3), C(3,âˆ'2).
= ½ | âˆ'4(3+2) + 2(âˆ'2+2) + 3(âˆ'2âˆ'3) |
= ½ | âˆ'4(5) + 0 + 3(âˆ'5) | = ½ × 35 = 17.5
Step 4: Total area = 10.5 + 17.5 = 28 sq units
Verify that a median of a triangle divides it into two triangles of equal areas for Î"ABC with A(4, âˆ'6), B(3, âˆ'2), C(5, 2).
Step 1: Midpoint D of BC = ((3+5)/2, (âˆ'2+2)/2) = (4, 0). AD is the median.
Step 2: Area of Î"ABD: A(4,âˆ'6), B(3,âˆ'2), D(4,0).
= ½ | 4(âˆ'2âˆ'0) + 3(0+6) + 4(âˆ'6+2) |
= ½ | 4(âˆ'2) + 3(6) + 4(âˆ'4) | = ½ | âˆ'8 + 18 âˆ' 16 | = ½ × 6 = 3
Step 3: Area of Î"ACD: A(4,âˆ'6), C(5,2), D(4,0).
= ½ | 4(2âˆ'0) + 5(0+6) + 4(âˆ'6âˆ'2) |
= ½ | 4(2) + 5(6) + 4(âˆ'8) | = ½ | 8 + 30 âˆ' 32 | = ½ × 6 = 3
Step 4: Both triangles have area = 3 sq units. Verified âœ"
Exercise 7.4 (Optional) 8 Questions – Advanced
Determine the ratio in which the line 2x + y âˆ' 4 = 0 divides the line joining A(2, âˆ'2) and B(3, 7).
Step 1: Let the line divide AB at P in ratio k:1.
P = ((3k+2)/(k+1), (7kâˆ'2)/(k+1))
Step 2: P lies on 2x + y âˆ' 4 = 0. Substitute:
2(3k+2)/(k+1) + (7kâˆ'2)/(k+1) âˆ' 4 = 0
(6k + 4 + 7k âˆ' 2)/(k+1) = 4
(13k + 2)/(k+1) = 4
13k + 2 = 4k + 4 â‡' 9k = 2 â‡' k = 2/9
Answer: The line divides AB in the ratio 2 : 9.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Step 1: Collinear â‡' Area = 0.
½ | x(2âˆ'0) + 1(0âˆ'y) + 7(yâˆ'2) | = 0
| 2x âˆ' y + 7y âˆ' 14 | = 0
| 2x + 6y âˆ' 14 | = 0
2x + 6y = 14
Answer: x + 3y = 7.
Find the centre of a circle passing through (6, âˆ'6), (3, âˆ'7) and (3, 3).
Step 1: Let centre O(h, k). OA = OB = OC (radii).
Step 2: OA² = OB²: (hâˆ'6)²+(k+6)² = (hâˆ'3)²+(k+7)². Simplify to: 3h + k = 7 ...(1)
Step 3: OB² = OC²: (hâˆ'3)²+(k+7)² = (hâˆ'3)²+(kâˆ'3)² â‡' (k+7)² = (kâˆ'3)² â‡' 20k = âˆ'40 â‡' k = âˆ'2.
Step 4: From (1): 3h + (âˆ'2) = 7 â‡' 3h = 9 â‡' h = 3.
Answer: Centre is (3, âˆ'2).
Q4: Two opposite vertices of a square are (âˆ'1,2) and (3,2). Find other two vertices. â†' Midpoint=(1,2), diagonal=4. Other vertices: (1,4) and (1,0).
Q5: Rectangular plot ABCD with students planting trees. Î"PQR formed by three trees. Find area using triangle area formula.
Q6: Vertices of Î"ABC are A(4,âˆ'6), B(3,âˆ'2), C(5,2). AD is median. Prove ar(ABD) = ar(ACD) using area formula.
Q7: A(4,2), B(6,5), C(1,4) are vertices of Î"ABC. D on x-axis such that ar(ABD)=ar(ABC). Find D. â†' D(3,0).
Q8: ABCD: A(âˆ'1,âˆ'1), B(âˆ'1,4), C(5,4), D(5,âˆ'1). P,Q,R,S are midpoints. Find type of quadrilateral PQRS. â†' Rhombus.
Practice This Chapter
Generate unlimited practice questions for Coordinate Geometry with Examzo — AI-powered, faculty-validated, zero hallucinations.
Try PaperGenie → Take a Quick Test →