CAT — Coordinate Geometry

Given: The vertices of the triangle are
A(1, 2), B(7, 2), and C(1, 10).
Find the radius of the incircle.

Step 1: Find the lengths of the sides
AB = 7 − 1 = 6
AC = 10 − 2 = 8
Using the distance formula,
BC = √[(7 − 1)² + (2 − 10)²]
= √(36 + 64)
= 10
Thus, the triangle is a right-angled triangle with sides 6, 8 and 10.

Step 2: Find the area
Area
= (1/2) × 6 × 8
= 24 sq. units

Step 3: Find the semiperimeter
Semiperimeter
= (6 + 8 + 10)/2
= 12

Step 4: Find the inradius
For any triangle,
Inradius = Area / Semiperimeter
= 24/12
= 2

Answer: 2

Step 1: Rewrite the second inequality
Complete the squares:
x² + y² + 4x − 4y = (x+2)² − 4 + (y−2)² − 4 = (x+2)² + (y−2)² − 8
Hence: −4 ≤ (x+2)² + (y−2)² − 8 ≤ 0
Adding 8: 4 ≤ (x+2)² + (y−2)² ≤ 8
This is the annular region between two concentric circles with centre (−2, 2), inner radius 2, outer radius 2√2.

Step 2: Interpret the line
The line y = x + 4. Substituting the centre (−2, 2): 2 = −2 + 4 ✓
The line passes through the centre, dividing the annular region into two equal halves.
The required region satisfies y ≥ x + 4, so it is half the annulus.

Step 3: Find the required area
Area of annulus = π[(2√2)² − 2²] = π(8 − 4) = 4π
Required area = (1/2) × 4π = 2π

Step 1: Assign coordinates to the hexagon. Let the side length of the regular hexagon be 2. Coordinates: A = (2, 0), B = (1, √3), C = (−1, √3), D = (−2, 0), E = (−1, −√3), F = (1, −√3). Since P and Q are midpoints: P = (3/2, √3/2), Q = (−3/2, √3/2).

Step 2: Find the lengths of the parallel sides of trapezium PBCQ. PQ = distance between P and Q = 3. BC = 2.

Step 3: Find the height of the trapezium. The y-coordinate of BC is √3. The y-coordinate of PQ is √3/2. Hence, height = √3 − √3/2 = √3/2.

Step 4: Find the area of trapezium PBCQ. Area = (1/2) × (sum of parallel sides) × height = (1/2) × (3 + 2) × (√3/2) = 5√3/4.

Step 5: Find the area of the hexagon. A regular hexagon consists of six equilateral triangles of side 2. Area of one equilateral triangle = (√3/4) × 2² = √3. Therefore, area of the hexagon = 6√3.

Step 6: Find the required ratio. Area of trapezium : Area of hexagon = (5√3/4) : 6√3 = 5 : 24.

In parallelogram PQRS, S = P + R − Q = (−3+9−1, −2+1+5) = (5, 4).

Slope of SQ = (−5−4)/(1−5) = −9/−4 = 9/4. Line SQ through S(5,4): y − 4 = (9/4)(x−5).

At y = 0: −4 = (9/4)(x−5) → −16 = 9(x−5) → 9x = 29 → a = 29/9.

Step 1: Simplify for 0 ≤ x ≤ 5 Since |x − 5| = 5 − x in this region: |x − y| = 2 + (5 − x) = 7 − x Step 2: Split into two cases Case 1: x − y = 7 − x → y = 2x − 7 Case 2: x − y = −(7 − x) → y = 7 Bounding lines: y = 7 and y = 2x − 7 Step 3: Find the vertices At x = 0: y = 7 and y = −7 → (0, 7) and (0, −7) At x = 5: y = 7 and y = 3 → (5, 7) and (5, 3) The figure is a trapezium with parallel sides 14 and 4, distance = 5. Step 4: Find the area Area = (1/2) × (14 + 4) × 5 = (1/2) × 18 × 5 = 45

Given:
In △ABC,
● M, N and P are the midpoints of AB, BC and AC, respectively.
● The medians from A, B and C intersect MP, MN and NP at X, Y and Z, respectively.
● Area of △ABC = 1440 sq cm.
Find the area of △XYZ.

Step 1: Choose convenient coordinates
Since area ratios are preserved under affine transformations, take
A = (0, 0)
B = (2, 0)
C = (0, 2)
Then,
Area of △ABC
= (1/2) × 2 × 2
= 2 sq units.

Step 2: Find the midpoints
M = (1, 0)
N = (1, 1)
P = (0, 1)

Step 3: Find X
Median from A passes through N.
Equation:
y = x
Line MP joins (1, 0) and (0, 1).
Equation:
x + y = 1
Solving,
x = y
2x = 1
x = y = 1/2
Hence,
X = (1/2, 1/2)

Step 4: Find Y
Median from B joins B(2, 0) and P(0, 1).
Equation:
y = 1 − x/2
Line MN is
x = 1
Hence,
Y = (1, 1/2)

Step 5: Find Z
Median from C joins C(0, 2) and M(1, 0).
Equation:
y = 2 − 2x
Line NP is
y = 1
Hence,
Z = (1/2, 1)

Step 6: Find the area of △XYZ
XY = 1/2
YZ = 1/2
Thus,
Area of △XYZ
= (1/2) × (1/2) × (1/2)
= 1/8 sq units.
Since
Area of △ABC = 2 sq units,
Area ratio
= (1/8)/2
= 1/16

Step 7: Find the required area
Area of △XYZ
= (1/16) × 1440
= 90 sq cm

Final Answer
90