Magical Math Explorer | Complete Geometry Adventure

Rectangle Midpoints Challenge

PQRS is a rectangle formed by joining the points P(-1,-1), Q(-1,4), R(5,4) and S(5,-1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus?

1

First, let's plot the rectangle PQRS with the given coordinates:

2

Find the midpoints A, B, C, D:

Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)

A (midpoint of PQ): ((-1 + -1)/2, (-1 + 4)/2) = (-1, 1.5)
B (midpoint of QR): ((-1 + 5)/2, (4 + 4)/2) = (2, 4)
C (midpoint of RS): ((5 + 5)/2, (4 + -1)/2) = (5, 1.5)
D (midpoint of SP): ((5 + -1)/2, (-1 + -1)/2) = (2, -1)

3

Plot the quadrilateral ABCD with these coordinates:

4

After analyzing the properties:

ABCD has all sides equal and diagonals perpendicular, making it a rhombus!

Triangle Area Mystery

The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3, -2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex.

1

We'll use the area formula for coordinates:

Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| = 5

2

Plug in the values with y = x + 3:

½ |2(-2 - (x+3)) + 3((x+3) - 1) + x(1 - (-2))| = 5

3

Simplify and solve:

|4x - 4| = 10 → x = 7/2 or -3/2
y = x + 3 → (7/2, 13/2) or (-3/2, 3/2)

4

Visual representation:

Possible third vertices: (7/2, 13/2) or (-3/2, 3/2)

Triangle Area from Lines

Find the area of a triangle formed by the lines 3x + y - 2 = 0, 5x + 2y - 3 = 0 and 2x - y - 3 = 0

1

Find intersection points of the lines:

2

Solve line pairs:

1. 3x + y = 2 and 5x + 2y = 3 → (1, -1)
2. 3x + y = 2 and 2x - y = 3 → (1, -1)
3. 5x + 2y = 3 and 2x - y = 3 → (9/7, -13/7)

3

All three lines meet at (1, -1)!

This means they don't form a triangle but intersect at a single point.

Therefore, the area is 0.

Quadrilateral Area Puzzle

Vertices of a quadrilateral are at A(-5,7), B(-4,k), C(-1,-6) and D(4,5) and its area is 72 sq. units. Find the value of k.

1

Using the shoelace formula:

Area = ½ |Σ(xᵢyᵢ₊₁) - Σ(yᵢxᵢ₊₁)| = 72

2

Calculate both sums:

Σ(xᵢyᵢ₊₁) = -5k + 24 - 5 + 28 = -5k + 47
Σ(yᵢxᵢ₊₁) = -28 - k - 24 - 25 = -k - 77

3

Solve the equation:

|-4k + 124| = 144 → k = -5 or 67

4

Visual verification:

Possible values: k = -5 or k = 67

Parallelogram Detective

Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.

1

We'll show that the midpoints of both diagonals are the same.

2

Find midpoints of diagonals AC and BD:

A(-2,-1), C(3,3) → midpoint = (0.5, 1)
B(4,0), D(-3,2) → midpoint = (0.5, 1)

3

Since both diagonals have the same midpoint, the quadrilateral is a parallelogram!

Line Equation Challenge

Find the equations of the lines, whose sum and product of intercepts are 1 and -6 respectively.

1

Let x-intercept = a, y-intercept = b

Given: a + b = 1
and a × b = -6

2

Solve the system:

b = 1 - a
a(1 - a) = -6 → a² - a - 6 = 0
Solutions: a = 3 or a = -2

3

Find corresponding b values:

When a = 3 → b = -2
When a = -2 → b = 3

4

Write line equations (intercept form):

1. x/3 + y/(-2) = 1 → 2x - 3y = 6
2. x/(-2) + y/3 = 1 → -3x + 2y = 6

Final equations: 2x - 3y - 6 = 0 and 3x - 2y + 6 = 0

Milk Store Economics

The owner of a milk store finds that he can sell 980 litres at ₹14/litre and 1220 litres at ₹16/litre. Assuming linear relationship, how many litres could he sell at ₹17/litre?

1

Establish linear relationship between price (p) and demand (q):

Points: (14, 980) and (16, 1220)
Slope (m) = (1220 - 980)/(16 - 14) = 120

2

Find the equation:

q - 980 = 120(p - 14)
q = 120p - 1680 + 980
q = 120p - 700

3

Find demand at p = 17:

q = 120(17) - 700 = 2040 - 700 = 1340

Projected sales at ₹17/litre: 1340 litres

Mirror Image Point

Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

1

To find the mirror image, we need to:

Find the equation of perpendicular line
Find intersection point (foot of perpendicular)
Extend the same distance to find image

2

Slope of given line: x + 3y = 7 → m₁ = -1/3

Slope of perpendicular line: m₂ = 3
Equation through (3,8): y - 8 = 3(x - 3) → y = 3x - 1

3

Find intersection point:

Solve x + 3y = 7 and y = 3x - 1
x + 3(3x - 1) = 7 → x = 1, y = 2
Foot of perpendicular: (1, 2)

4

Find image point:

Let image be (a,b). Midpoint of (3,8) and (a,b) is (1,2)
(3 + a)/2 = 1 → a = -1
(8 + b)/2 = 2 → b = -4

Image point: (-1, -4)

Equal Intercepts Line

Find the equation of a line passing through the intersection of 4x + 7y - 3 = 0 and 2x - 3y + 1 = 0 that has equal intercepts on the axes.

1

First find intersection point of given lines:

Solve 4x + 7y = 3 and 2x - 3y = -1
Solution: x = 1/13, y = 5/13

2

Line with equal intercepts can be written as:

x + y = a or x - y = b
Since it passes through (1/13, 5/13):
1/13 + 5/13 = a → a = 6/13
or 1/13 - 5/13 = b → b = -4/13

3

Possible equations:

1. x + y = 6/13 → 13x + 13y - 6 = 0
2. x - y = -4/13 → 13x - 13y + 4 = 0

Both lines satisfy the condition!

Shortest Path Problem

A person standing at the intersection of 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 wants to reach 6x - 7y + 8 = 0 in least time. Find the equation of the path he should follow.

1

Find the intersection point (starting position):

Solve 2x - 3y = -4 and 3x + 4y = 5
Solution: x = -1/17, y = 22/17

2

The shortest path is perpendicular to the target line 6x - 7y + 8 = 0

Slope of target line: 6/7
Slope of perpendicular path: -7/6

3

Equation of path line:

y - 22/17 = (-7/6)(x + 1/17)
Simplifies to: 119x + 102y - 125 = 0

Path equation: 119x + 102y - 125 = 0

✨ Magical Math Explorer