Question

b) A parallelogram has 4 sides having equations as L1 →x
+ 2y – 2 = 0,L2 →y-1=0,L3 → x+2y + 2 = 0 and L4 →
y+2 = 0. The intersection point of the diagonals will be
(1) (1,-1/2)
(ii) (1/2,1/2)
(iii) (3/2,1)
(iv) (0,0)
​

Answer 1

Given :- A parallelogram has 4 sides having equations as L1 →x + 2y – 2 = 0,L2 →y-1=0,L3 → x+2y + 2 = 0 and L4 → y+2 = 0. The intersection point of the diagonals will be

(i) (1,-1/2)

(ii) (1/2,1/2)

(iii) (3/2,1)

(iv) (0,0)

Solution :-

Solving L1 :-

→ x + 2y - 2 = 0

→ x + 2y = 2

putting x = 0,

→ 0 + 2y = 2

→ 2y = 2

→ y = 1 .

so, coordinates of L1 are (0,1)

similarly,

Solving L3 :-

→ x + 2y + 2 = 0

→ x + 2y = -2

putting x = 0,

→ 0 + 2y = -2

→ 2y = -2

→ y = -1 .

so, coordinates of L1 are (0, -1)

now, we know that,

• Diagonals of a parallelogram bisect each other .

then,

→ Coordinates of diagonal = (x1 + x2)/2 and (y1 + y2)/2 = (0+0)/2 = 0, and (1 - 1)/2 = 0 => (0,0) . But no option is their .

checking L2 and L4 we get,

Solving L2 :-

→ y - 1 = 0

→ y = 1

since x is not given , it will be 0

so, coordinates of L2 are (0, 1) .

Solving L4 :-

→ y + 2 = 0

→ y = (-2)

so, coordinates of L4 are (0, -2) .

then,

→ Coordinates of diagonal = (0 + 0)/2 = 0 and (1 - 2)/2 = (-1/2) .

Since coordinates of diagonals will be same in both case we can conclude that, data is incorrect in the question .

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