Answer:
(i) (-3/2, 3/2) and (1/2, - 3/2)
6x + 4y = - 3
(ii) Points are col linear
Step-by-step explanation:
Let the coordinates of the point(lower) be (a, b) and that of upper be (x, y).
Using mid point formula,
Mid point of diagonal(using the given points) is:
= ( (-2+1)/2 , (-1+1)/2) = (-½ , 0)
Using (a, b) and (x, y) :
= ( (a+x)/2 , (b+y)/2 ) = (-1/2, 0)
Which means, a+x= -1 & b + y = 0
Therefore, x = - 1 - a & b = - y
Using distance formula, diagonal = √(-2-1)² + (-1-1)² = √13
Knowing the relation in diagonal and side, side = diagonal/√2 = √13/√2 = √(13/2)
Again using distance formula, diagonal = √13
=> √(x - a)² + (y - b)² = √13
=> (-1-a -a)² + (-b - b)² = 13
=> a² + b² + a = 3 ... (1) [solved directly]
Length of side = √(13/2)
=> √(a-(-2))² + (b-(-1))² = √(13/2)
=> a² + b² + 4a + 2b = 3/2 ...(2)
On solving (1) and (2):
a = 1/2 or -3/2, but a lies in 4th quadrant so a > 0, thus, a = 1/2
b = -3/2 or 3/2, but b lies in 4th quadrant so b < 0, b = -3/2
Therefore,
x = - 1 - a = -1 - 1/2 = -3/2
y = - b = - (-3/2) = 3/2
Vertices are (x, y) = (-3/2, 3/2) and (a, b) = (1/2, -3/2)
Equation is just a relation in y and x, for a relation:
Subtract (1) from (2):
3a + 2b = -3/2 => 6a + 4b = - 3
By merely replacing a, b by x, y
6x + 4y = -3 is the required equation
(ii):
Ar. of ∆ = ½ | 1(3 - 5) + 2(5 - 2) + 4(2 - 3) |
Ar. of ∆ = ½ | -2 + 6 - 4 | = 0
As the area of the ∆ is 0, the given points don't form ∆, they are col linear.
