Quantitative Aptitude

Given three points A(1,2) B(3,-4) and C(5,-6).

To find the perpendicular bisectors of AB:-

• The slope of AB:
  = -4-2 / 3-1
  = -3
• The slope of line perpendicular to AB:
  = 1 / 3
• The mid-point of AB is:
  = (1 + 3 / 2, 2 - 4 / 2)
  = (2, -1)
• The equation of perpendicular line is:
  y = 1 / 3(x - 2) - 1

To find the perpendicular bisectors of AC:-

• The slope of AC:
  = -6 - 2 / 5 - 1
  = -2
• The slope of line perpendicular to AC:
  = 1 /2
• The mid-point of AC is:
  = (1 + 5 / 2, 2-6 / 2)
   (3, -2)
• The equation of perpendicular line is:
  y = 1 / 2(x - 3) - 2

Now, solve the above two equations:-

⇒ 1 / 3(x - 2) - 1 = 1 / 2(x - 3) - 2
⇒ 2(x - 2) - 6 = 3(x - 3) - 12
⇒×= 11
⇒ y = 1 / 2(x - 3) - 2 = 1 / 2(11 - 3) - 2 = 2

The coordinates of the points equidistant from the point A(1, 2), B(3, -4) and C(5, -6) are (11, 2)

3x - 8y =16 .....(i)
2x + 4y = 6 .....(ii)
By solving equation (i) and (ii), we get
x = 4 = a
y = -$$\frac{1}{2}$$ = b
∴
a² - 4b² = (4)² - 4(-$$\frac{1}{2}$$)²
= 16 - 4(1/4)
= 16 - 1
= 15
The value of (a² - 4b²) = 15.

OA = √3²+(-3)² = √9+9 = √18 = 3√2

Here, x₁ = 2, x₂ = 4, x₃ = 6 and y₁ = -5, y₂ = 9, y₃ = -1
= $$\frac{1}{2}$$ [2(9+1) + 4(-1+5) + 6(-5-9)] 
= $$\frac{1}{2}$$ [2(10) + 4(4) + 6(-14)] 
= $$\frac{1}{2}$$ [20 + 16 - 84] 
= $$\frac{1}{2}$$ [-48] 
= 24 sq.unit's

Let A (1, 7), B (4, 2), C(−1,−1) and D(−4,4) are the vertices of a quadrilateral ABCD.

The co-ordinates of A are A(0, -5)

Coordinates are given for  A(1 , 2) , B(1 , 0) and C(4 , 0)

Let coordinates of D be (x,y).
Since diagonals of a parallelogram bisect each other. at point O

Therefore O is the mid point of diagonal AC 

AB = √(b-0)²-(0-a)² 
= √b²+a² 
= √a²+b².