Coordinate Geometry - 1 Mark Problems
Important Formulas:
• Distance between (x₁, y₁) and (x₂, y₂): √[(x₂-x₁)² + (y₂-y₁)²]
• Midpoint of (x₁, y₁) and (x₂, y₂): [(x₁+x₂)/2, (y₁+y₂)/2]
• Centroid of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
[(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
• Slope of line through (x₁, y₁) and (x₂, y₂): (y₂-y₁)/(x₂-x₁)
• If C is center and A is one endpoint of diameter, then other endpoint B:
B = (2×C_x - A_x, 2×C_y - A_y)
• Midpoint of (x₁, y₁) and (x₂, y₂): [(x₁+x₂)/2, (y₁+y₂)/2]
• Centroid of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
[(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
• Slope of line through (x₁, y₁) and (x₂, y₂): (y₂-y₁)/(x₂-x₁)
• If C is center and A is one endpoint of diameter, then other endpoint B:
B = (2×C_x - A_x, 2×C_y - A_y)
1
Find the centroid of a triangle, whose vertices are (3, 4), (–7, –2) and (10, –5).
(M'15)
Step 1: Use centroid formula
Centroid = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
Step 2: Substitute values
x-coordinate = (3 + (-7) + 10)/3 = (3 - 7 + 10)/3 = 6/3 = 2
y-coordinate = (4 + (-2) + (-5))/3 = (4 - 2 - 5)/3 = -3/3 = -1
∴ The centroid is (2, -1).
2
Find the distance between the points (0, 0) and (a, b).
(J'15)
Step 1: Use distance formula
Distance = √[(x₂-x₁)² + (y₂-y₁)²]
Step 2: Substitute values
Distance = √[(a-0)² + (b-0)²] = √[a² + b²]
∴ The distance is √(a² + b²).
3
Find the midpoint of the line segment joining the points (–5, 5) and (5, –5).
(M'16)
Step 1: Use midpoint formula
Midpoint = [(x₁+x₂)/2, (y₁+y₂)/2]
Step 2: Substitute values
x-coordinate = (-5 + 5)/2 = 0/2 = 0
y-coordinate = (5 + (-5))/2 = 0/2 = 0
∴ The midpoint is (0, 0).
4
If the slope of the line passing through the two points (2, 5) and (5, 8) is represented by tan θ (where 0° < θ < 90°) in trigonometry, then find angle 'θ'.
(J'16)
Step 1: Find the slope
Slope = (y₂-y₁)/(x₂-x₁) = (8-5)/(5-2) = 3/3 = 1
Step 2: Relate to tan θ
tan θ = 1
Step 3: Find θ
θ = tan⁻¹(1) = 45°
∴ The angle θ is 45°.
5
A(0, 3), B(k, 0) and AB = 5. Find the positive value of k.
(M'17)
Step 1: Use distance formula
AB = √[(k-0)² + (0-3)²] = √[k² + 9]
Step 2: Set AB = 5
√[k² + 9] = 5
Step 3: Square both sides
k² + 9 = 25
k² = 16
k = ±4
Step 4: Take positive value
k = 4
∴ The positive value of k is 4.
6
Find the distance between the points (1, 5) and (5, 8).
(M'18)
Step 1: Use distance formula
Distance = √[(x₂-x₁)² + (y₂-y₁)²]
Step 2: Substitute values
Distance = √[(5-1)² + (8-5)²] = √[4² + 3²] = √[16 + 9] = √25
∴ The distance is 5 units.
7
What is the other end of the diameter of the circle, whose centre is (1, 2) and one end point of the diameter is (3, 4)?
(J'18)
Step 1: Use center formula
Center is the midpoint of the diameter
If A(x₁, y₁) and B(x₂, y₂) are endpoints, center = [(x₁+x₂)/2, (y₁+y₂)/2]
Step 2: Let other endpoint be (x, y)
Center = [(3+x)/2, (4+y)/2] = (1, 2)
Step 3: Solve for x and y
(3+x)/2 = 1 ⇒ 3+x = 2 ⇒ x = -1
(4+y)/2 = 2 ⇒ 4+y = 4 ⇒ y = 0
∴ The other endpoint is (-1, 0).
8
Find the centroid of a ∆PQR, when vertices are P(1, 1), Q(2, 2), R(–3, –3).
(M'19)
Step 1: Use centroid formula
Centroid = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
Step 2: Substitute values
x-coordinate = (1 + 2 + (-3))/3 = (1 + 2 - 3)/3 = 0/3 = 0
y-coordinate = (1 + 2 + (-3))/3 = (1 + 2 - 3)/3 = 0/3 = 0
∴ The centroid is (0, 0).
9
Determine 'x' so that 2 is the slope of the line passing through A(–2, 4) and B(x, –2).
(J'19)
Step 1: Use slope formula
Slope = (y₂-y₁)/(x₂-x₁)
Step 2: Substitute values
2 = (-2 - 4)/(x - (-2)) = (-6)/(x + 2)
Step 3: Solve for x
2 = -6/(x + 2)
2(x + 2) = -6
2x + 4 = -6
2x = -10
x = -5
∴ The value of x is -5.
10
Find the distance between the points (0, 0) and (sin θ, cos θ), where (0° ≤ θ ≤ 90°).
(May 2022)
Step 1: Use distance formula
Distance = √[(x₂-x₁)² + (y₂-y₁)²]
Step 2: Substitute values
Distance = √[(sin θ - 0)² + (cos θ - 0)²] = √[sin²θ + cos²θ]
Step 3: Use trigonometric identity
sin²θ + cos²θ = 1
Distance = √1 = 1
∴ The distance is always 1 unit, regardless of θ.
11
Find the distance between the points (sec θ, 0) and (0, tan θ) when θ = 45°.
(Aug 22)
Step 1: Find values when θ = 45°
sec 45° = √2
tan 45° = 1
Step 2: Points become
Point 1: (√2, 0)
Point 2: (0, 1)
Step 3: Use distance formula
Distance = √[(0-√2)² + (1-0)²] = √[2 + 1] = √3
∴ The distance is √3 units.
12
Find the centroid of the triangle whose vertices are (2, 3), (-4, 7) and (2, –4).
(Apr 23)
Step 1: Use centroid formula
Centroid = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
Step 2: Substitute values
x-coordinate = (2 + (-4) + 2)/3 = (2 - 4 + 2)/3 = 0/3 = 0
y-coordinate = (3 + 7 + (-4))/3 = (3 + 7 - 4)/3 = 6/3 = 2
∴ The centroid is (0, 2).
Coordinate Geometry - 2 Mark Problems
Important Formulas:
• Distance between (x₁, y₁) and (x₂, y₂): √[(x₂-x₁)² + (y₂-y₁)²]
• Section formula (dividing in ratio m:n):
[(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
• Centroid of triangle: [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
• Slope of line: m = (y₂-y₁)/(x₂-x₁)
• Angle with X-axis: θ = tan⁻¹(m)
• Area of triangle: ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|
• For collinearity: Area = 0 or slopes equal
• Section formula (dividing in ratio m:n):
[(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
• Centroid of triangle: [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
• Slope of line: m = (y₂-y₁)/(x₂-x₁)
• Angle with X-axis: θ = tan⁻¹(m)
• Area of triangle: ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|
• For collinearity: Area = 0 or slopes equal
1
Show that the points A(4, 2), B(7, 5) and C(9, 7) are collinear.
(M'15)
Method 1: Using Area of Triangle
Step 1: Area formula
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Step 2: Substitute values
Area = ½|4(5-7) + 7(7-2) + 9(2-5)|
= ½|4(-2) + 7(5) + 9(-3)|
= ½|-8 + 35 - 27|
= ½|0| = 0
Method 2: Using Slope
Step 1: Calculate slopes
Slope of AB = (5-2)/(7-4) = 3/3 = 1
Slope of BC = (7-5)/(9-7) = 2/2 = 1
Slope of AC = (7-2)/(9-4) = 5/5 = 1
Step 2: Compare slopes
Since all slopes are equal (m = 1), the points are collinear.
∴ Points A, B, and C are collinear.
2
A(3, 6), B(3, 2) and C(8, 2) are the vertices of a rectangle ABCD. Plot these points on a graph paper. From this find the coordinates of vertex D, so that ABCD will be a rectangle.
(J'15)
Step 1: Plot the given points
A(3, 6) - Top left
B(3, 2) - Bottom left
C(8, 2) - Bottom right
Step 2: Identify rectangle properties
In a rectangle, opposite sides are equal and parallel
AB is vertical (x=3), BC is horizontal (y=2)
Step 3: Find coordinates of D
Since AB is parallel to CD, D has x-coordinate = 8
Since AD is parallel to BC, D has y-coordinate = 6
∴ The coordinates of vertex D are (8, 6).
3
Show that the points A(-3, 3), B(0, 0), C(3, -3) are collinear.
(M'16)
Method: Using Area of Triangle
Step 1: Area formula
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Step 2: Substitute values
Area = ½|-3(0-(-3)) + 0(-3-3) + 3(3-0)|
= ½|-3(3) + 0(-6) + 3(3)|
= ½|-9 + 0 + 9|
= ½|0| = 0
∴ Since area = 0, points A, B, and C are collinear.
4
The distance between the points (8, x) and (x, 8) is 2√2 units, then find the value of x.
(J'16)
Step 1: Distance formula
Distance = √[(x₂-x₁)² + (y₂-y₁)²]
Step 2: Substitute values
2√2 = √[(x-8)² + (8-x)²]
Step 3: Square both sides
(2√2)² = (x-8)² + (8-x)²
8 = (x-8)² + (8-x)²
Step 4: Simplify
Since (x-8)² = (8-x)², we have:
8 = 2(x-8)²
(x-8)² = 4
x-8 = ±2
Step 5: Solve for x
x-8 = 2 ⇒ x = 10
x-8 = -2 ⇒ x = 6
∴ The values of x are 6 and 10.
5
Two vertices of a triangle are (3, 2), (-2, 1) and its centroid is (5/3, -1/3). Find the third vertex of the triangle.
(M'17)
Step 1: Centroid formula
Centroid = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3]
Step 2: Let third vertex be (x, y)
(5/3, -1/3) = [(3 + (-2) + x)/3, (2 + 1 + y)/3]
(5/3, -1/3) = [(1 + x)/3, (3 + y)/3]
Step 3: Equate coordinates
(1 + x)/3 = 5/3 ⇒ 1 + x = 5 ⇒ x = 4
(3 + y)/3 = -1/3 ⇒ 3 + y = -1 ⇒ y = -4
∴ The third vertex is (4, -4).
6
Find the angle made by the line joining (5, 3) and (-1, -3) with the positive direction of X-axis.
(M'17)
Step 1: Find slope
Slope (m) = (y₂-y₁)/(x₂-x₁) = (-3-3)/(-1-5) = (-6)/(-6) = 1
Step 2: Relate slope to angle
m = tan θ, where θ is the angle with positive X-axis
tan θ = 1
Step 3: Find θ
θ = tan⁻¹(1) = 45°
∴ The line makes an angle of 45° with the positive direction of X-axis.
7
Determine 'x', if the slope of the line joining the two points (4, x), (7, 2) is 8/3.
(J'17)
Step 1: Slope formula
Slope = (y₂-y₁)/(x₂-x₁)
Step 2: Substitute values
8/3 = (2 - x)/(7 - 4)
8/3 = (2 - x)/3
Step 3: Cross multiply
8/3 = (2 - x)/3
8 = 2 - x
x = 2 - 8 = -6
∴ The value of x is -6.
8
In the diagram on a lunar eclipse, the positions of the sun, earth and moon are shown by (-4, 6), (k, -2), and (5, -6) respectively, then find the value of k.
(M'18)
Step 1: Understanding the problem
During a lunar eclipse, the sun, earth, and moon are collinear
So the points (-4, 6), (k, -2), and (5, -6) are collinear
Step 2: Condition for collinearity
Area of triangle = 0
½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)| = 0
Step 3: Substitute values
½|-4(-2-(-6)) + k(-6-6) + 5(6-(-2))| = 0
½|-4(4) + k(-12) + 5(8)| = 0
½|-16 - 12k + 40| = 0
½|24 - 12k| = 0
Step 4: Solve for k
|24 - 12k| = 0
24 - 12k = 0
12k = 24
k = 2
∴ The value of k is 2.
9
Find the coordinates of the point which divides the segment joining (2, 3) and (-4, 0) in 1:2.
(J'18)
Step 1: Section formula
Point dividing in ratio m:n = [(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
Step 2: Substitute values
m = 1, n = 2
(x₁, y₁) = (2, 3), (x₂, y₂) = (-4, 0)
Step 3: Calculate coordinates
x = (1×(-4) + 2×2)/(1+2) = (-4 + 4)/3 = 0/3 = 0
y = (1×0 + 2×3)/(1+2) = (0 + 6)/3 = 6/3 = 2
∴ The point is (0, 2).
10
Akhila says, "points A(1, 3), B(2, 2), C(5, 1) are collinear". Do you agree with Akhila? Why?
(M'19)
Step 1: Check using area method
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Step 2: Substitute values
Area = ½|1(2-1) + 2(1-3) + 5(3-2)|
= ½|1(1) + 2(-2) + 5(1)|
= ½|1 - 4 + 5|
= ½|2| = 1
Step 3: Interpret result
Since area ≠ 0, the points are not collinear
∴ No, I don't agree with Akhila. The points are not collinear as the area of triangle formed is 1 (not zero).
11
If the area of the triangle formed by joining the points A(x, y), B(3, 2) and C(-2, 4) is 10 square units, then show that 2x + 5y + 4 = 0.
(Jun'23)
Step 1: Area formula
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Step 2: Substitute values
10 = ½|x(2-4) + 3(4-y) + (-2)(y-2)|
10 = ½|x(-2) + 3(4-y) - 2(y-2)|
10 = ½|-2x + 12 - 3y - 2y + 4|
10 = ½|-2x - 5y + 16|
Step 3: Remove absolute value
|-2x - 5y + 16| = 20
-2x - 5y + 16 = ±20
Step 4: Consider both cases
Case 1: -2x - 5y + 16 = 20
⇒ -2x - 5y = 4
⇒ 2x + 5y = -4
⇒ 2x + 5y + 4 = 0 ✓
Step 5: Verify second case
Case 2: -2x - 5y + 16 = -20
⇒ -2x - 5y = -36
⇒ 2x + 5y = 36
This gives a different equation
∴ For the given condition, we get 2x + 5y + 4 = 0, as required.
Coordinate Geometry - 4 Mark Problems
Important Formulas:
• Distance formula: √[(x₂-x₁)² + (y₂-y₁)²]
• Section formula (internal): [(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
• Area of triangle: ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
• Area of quadrilateral: Divide into triangles and sum areas
• For parallelogram: Diagonals bisect each other
• For rectangle: All angles 90°, diagonals equal
• For square: All sides equal, diagonals equal and perpendicular
• For rhombus: All sides equal, diagonals perpendicular
• Section formula (internal): [(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
• Area of triangle: ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
• Area of quadrilateral: Divide into triangles and sum areas
• For parallelogram: Diagonals bisect each other
• For rectangle: All angles 90°, diagonals equal
• For square: All sides equal, diagonals equal and perpendicular
• For rhombus: All sides equal, diagonals perpendicular
1
Name the type of quadrilateral formed by joining the points A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) on a graph paper. Justify your answer.
(M'15)
Step 1: Calculate distances between points
AB = √[(1-(-1))² + (0-(-2))²] = √[(2)² + (2)²] = √8 = 2√2
BC = √[(-1-1)² + (2-0)²] = √[(-2)² + (2)²] = √8 = 2√2
CD = √[(-3-(-1))² + (0-2)²] = √[(-2)² + (-2)²] = √8 = 2√2
DA = √[(-1-(-3))² + (-2-0)²] = √[(2)² + (-2)²] = √8 = 2√2
Step 2: Check diagonals
AC = √[(-1-(-1))² + (2-(-2))²] = √[0 + 16] = 4
BD = √[(-3-1)² + (0-0)²] = √[(-4)² + 0] = 4
Step 3: Analyze the shape
All sides equal: AB = BC = CD = DA = 2√2
Diagonals equal: AC = BD = 4
In a square, all sides are equal and diagonals are equal.
∴ The quadrilateral is a square as all sides are equal and diagonals are equal.
2
If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, then find the area of the quadrilateral ABCD.
(J'15)
Step 1: Divide quadrilateral into two triangles
We can divide ABCD into triangles ABC and ACD
Step 2: Area of triangle ABC
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
= ½|(-5)(-5-(-6)) + (-4)(-6-7) + (-1)(7-(-5))|
= ½|(-5)(1) + (-4)(-13) + (-1)(12)|
= ½|-5 + 52 - 12| = ½|35| = 17.5
Step 3: Area of triangle ACD
Area = ½|(-5)(-6-5) + (-1)(5-7) + (4)(7-(-6))|
= ½|(-5)(-11) + (-1)(-2) + (4)(13)|
= ½|55 + 2 + 52| = ½|109| = 54.5
Step 4: Total area
Area of ABCD = Area(ABC) + Area(ACD)
= 17.5 + 54.5 = 72 sq. units
∴ The area of quadrilateral ABCD is 72 square units.
3
Find the co-ordinates of the points of trisection of the line segment joining the points (-3, 3) and (3, -3).
(M'16)
Step 1: Understand trisection
Trisection means dividing into three equal parts
We need to find two points that divide the segment in ratios 1:2 and 2:1
Step 2: First point (dividing in ratio 1:2)
Using section formula: [(mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n)]
m = 1, n = 2
x = (1×3 + 2×(-3))/(1+2) = (3 - 6)/3 = -3/3 = -1
y = (1×(-3) + 2×3)/(1+2) = (-3 + 6)/3 = 3/3 = 1
First point: (-1, 1)
Step 3: Second point (dividing in ratio 2:1)
m = 2, n = 1
x = (2×3 + 1×(-3))/(2+1) = (6 - 3)/3 = 3/3 = 1
y = (2×(-3) + 1×3)/(2+1) = (-6 + 3)/3 = -3/3 = -1
Second point: (1, -1)
∴ The points of trisection are (-1, 1) and (1, -1).
4
If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, then find the values of a and b.
(J'16)
Step 1: Condition for collinearity
Area of triangle = 0
½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)| = 0
Step 2: Substitute values
½|(-3)(b-(-5)) + a(-5-9) + 4(9-b)| = 0
½|(-3)(b+5) + a(-14) + 4(9-b)| = 0
|-3b - 15 - 14a + 36 - 4b| = 0
|-14a - 7b + 21| = 0
Step 3: Simplify
-14a - 7b + 21 = 0
Divide by -7: 2a + b - 3 = 0
2a + b = 3 ...(1)
Step 4: Use given condition
a + b = 1 ...(2)
Step 5: Solve equations (1) and (2)
Subtract (2) from (1): (2a+b) - (a+b) = 3-1
a = 2
Substitute in (2): 2 + b = 1 ⇒ b = -1
∴ The values are a = 2 and b = -1.
5
The points C and D on the line segment joining A(-4, 7) and B(5, 13) such that AC = CD = DB. Then find the co-ordinates of point C and D.
(M'17)
Step 1: Understand the division
AC = CD = DB means the segment is divided into three equal parts
C divides AB in ratio 1:2
D divides AB in ratio 2:1
Step 2: Coordinates of C (dividing in ratio 1:2)
Using section formula:
x = (1×5 + 2×(-4))/(1+2) = (5 - 8)/3 = -3/3 = -1
y = (1×13 + 2×7)/(1+2) = (13 + 14)/3 = 27/3 = 9
C = (-1, 9)
Step 3: Coordinates of D (dividing in ratio 2:1)
x = (2×5 + 1×(-4))/(2+1) = (10 - 4)/3 = 6/3 = 2
y = (2×13 + 1×7)/(2+1) = (26 + 7)/3 = 33/3 = 11
D = (2, 11)
∴ The coordinates are C(-1, 9) and D(2, 11).
6
The area of the triangle is 18 sq. units, whose vertices are (3, 4), (-3, -2) and (p, -1); then find the value of 'p'.
(J'17)
Step 1: Area formula
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Step 2: Substitute values
18 = ½|3(-2-(-1)) + (-3)(-1-4) + p(4-(-2))|
18 = ½|3(-1) + (-3)(-5) + p(6)|
18 = ½|-3 + 15 + 6p|
18 = ½|12 + 6p|
Step 3: Solve for p
|12 + 6p| = 36
12 + 6p = 36 or 12 + 6p = -36
6p = 24 or 6p = -48
p = 4 or p = -8
∴ The values of p are 4 and -8.
7
Find the points of trisection of the line segment joining the points (-2, 1) and (7, 4).
(M'18)
Step 1: First point (dividing in ratio 1:2)
Using section formula:
x = (1×7 + 2×(-2))/(1+2) = (7 - 4)/3 = 3/3 = 1
y = (1×4 + 2×1)/(1+2) = (4 + 2)/3 = 6/3 = 2
First point: (1, 2)
Step 2: Second point (dividing in ratio 2:1)
x = (2×7 + 1×(-2))/(2+1) = (14 - 2)/3 = 12/3 = 4
y = (2×4 + 1×1)/(2+1) = (8 + 1)/3 = 9/3 = 3
Second point: (4, 3)
∴ The points of trisection are (1, 2) and (4, 3).
8
Show that the points A(-1, -2), B(4, 3), C(2, 5) and D(-3, 0) in that order form a rectangle.
(J'18)
Step 1: Check opposite sides are equal
AB = √[(4-(-1))² + (3-(-2))²] = √[5² + 5²] = √50
BC = √[(2-4)² + (5-3)²] = √[(-2)² + 2²] = √8
CD = √[(-3-2)² + (0-5)²] = √[(-5)² + (-5)²] = √50
DA = √[(-1-(-3))² + (-2-0)²] = √[2² + (-2)²] = √8
AB = CD and BC = DA
Step 2: Check diagonals are equal
AC = √[(2-(-1))² + (5-(-2))²] = √[3² + 7²] = √58
BD = √[(-3-4)² + (0-3)²] = √[(-7)² + (-3)²] = √58
AC = BD
Step 3: Check if adjacent sides are perpendicular
Slope of AB = (3-(-2))/(4-(-1)) = 5/5 = 1
Slope of BC = (5-3)/(2-4) = 2/(-2) = -1
Product of slopes = 1 × (-1) = -1
So AB ⟂ BC
∴ Since opposite sides are equal, diagonals are equal, and adjacent sides are perpendicular, ABCD is a rectangle.
9
Find the ratio in which X-axis divides the line segment joining the points (2, -3) and (5, 6). Then find the intersecting point on X-axis.
(M'19)
Step 1: Let the ratio be k:1
Point on X-axis has y-coordinate = 0
Using section formula: y = (my₂+ny₁)/(m+n)
0 = (k×6 + 1×(-3))/(k+1)
Step 2: Solve for k
0 = (6k - 3)/(k+1)
6k - 3 = 0
6k = 3
k = 1/2
Step 3: Find the point
x = (1/2×5 + 1×2)/(1/2+1) = (2.5+2)/(1.5) = 4.5/1.5 = 3
Point = (3, 0)
∴ The X-axis divides the segment in ratio 1:2 and the point of intersection is (3, 0).
10
Find the area of the Rhombus ABCD, whose vertices taken in order are A(-1, 1), B(1, -2), C(3, 1), D(1, 4).
(J'19)
Step 1: Calculate diagonals
AC = √[(3-(-1))² + (1-1)²] = √[4² + 0] = 4
BD = √[(1-1)² + (4-(-2))²] = √[0 + 6²] = 6
Step 2: Area of rhombus
Area = ½ × d₁ × d₂ = ½ × 4 × 6 = 12
∴ The area of rhombus ABCD is 12 square units.
11
Show that the distances of the points (5, 12), (7, 24) and (35, 12) from the origin are arranged in ascending order forms an arithmetic progression. Find the common difference of the progression.
(May 2022)
Step 1: Calculate distances from origin
Distance from origin = √(x² + y²)
d₁ = √(5² + 12²) = √(25 + 144) = √169 = 13
d₂ = √(7² + 24²) = √(49 + 576) = √625 = 25
d₃ = √(35² + 12²) = √(1225 + 144) = √1369 = 37
Step 2: Check if they form AP
Ascending order: 13, 25, 37
Common difference = 25 - 13 = 12
37 - 25 = 12
∴ The distances form an AP with common difference 12.
12
If A(-2, 2), B(a, 6), C(4, b) and D(2, -2) are the vertices of a parallelogram ABCD, then find the values of a and b. Also find the lengths of its sides.
(Apr'23)
Step 1: Use midpoint property of parallelogram
In a parallelogram, diagonals bisect each other
Midpoint of AC = Midpoint of BD
Step 2: Equate midpoints
Midpoint of AC = [(-2+4)/2, (2+b)/2] = (1, (2+b)/2)
Midpoint of BD = [(a+2)/2, (6+(-2))/2] = ((a+2)/2, 2)
Step 3: Solve for a and b
(a+2)/2 = 1 ⇒ a+2 = 2 ⇒ a = 0
(2+b)/2 = 2 ⇒ 2+b = 4 ⇒ b = 2
Step 4: Find side lengths
AB = √[(0-(-2))² + (6-2)²] = √[2² + 4²] = √20 = 2√5
BC = √[(4-0)² + (2-6)²] = √[4² + (-4)²] = √32 = 4√2
CD = √[(2-4)² + (-2-2)²] = √[(-2)² + (-4)²] = √20 = 2√5
DA = √[(-2-2)² + (2-(-2))²] = √[(-4)² + 4²] = √32 = 4√2
∴ a = 0, b = 2, and side lengths are AB = CD = 2√5, BC = DA = 4√2.
13
Show that the quadrilateral formed by joining the points (-4, 2), (4, 4), (2, 12) and (-6, 10) taken in order is a square.
(Jun'23)
Step 1: Check all sides are equal
AB = √[(4-(-4))² + (4-2)²] = √[8² + 2²] = √68
BC = √[(2-4)² + (12-4)²] = √[(-2)² + 8²] = √68
CD = √[(-6-2)² + (10-12)²] = √[(-8)² + (-2)²] = √68
DA = √[(-4-(-6))² + (2-10)²] = √[2² + (-8)²] = √68
All sides equal
Step 2: Check diagonals are equal
AC = √[(2-(-4))² + (12-2)²] = √[6² + 10²] = √136
BD = √[(-6-4)² + (10-4)²] = √[(-10)² + 6²] = √136
Diagonals equal
Step 3: Check if adjacent sides are perpendicular
Slope of AB = (4-2)/(4-(-4)) = 2/8 = 1/4
Slope of BC = (12-4)/(2-4) = 8/(-2) = -4
Product = (1/4) × (-4) = -1
So AB ⟂ BC
∴ Since all sides are equal, diagonals are equal, and adjacent sides are perpendicular, the quadrilateral is a square.
