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Two sides of a parallelogram are along the lines, $x + y =3$ and $x - y + 3 = 0$ . If its diagonals intersect at $(2,4)$ , then one of its vertex is:
Option 1)
(3,6)
Option 2)
(2,6)
Option 3)
(3,5)
Option 4)
(2,1)

Answers (1)

best_answer

Mid-point formula -

$x= \frac{x_{1}+x_{2}}{2}$

$y= \frac{y_{1}+y_{2}}{2}$

- wherein

If the point P(x,y) is the mid point of line joining A(x₁,y₁) and B(x₂,y₂) .

Two lines

$x+y=3 \: \: and\: \: x-y=-3$ intersects at A (0,3)

Point C is $(x_{1},y_{1})$

So,

$\frac{x_{1}+0}{2}=2$ , $\frac{y_{1}+4}{2}=4$

$=>C(x_{1},y_{1})=C(4,5)$

So, Equation of BC is $x-y=-1$

and equation of CD is $x+y=9$

Solve $x+y=9$ and $x-y=-3$

$D(3,6)$

Option 1)
(3,6)
Option 2)

(2,6)

Option 3)

(3,5)

Option 4)

(2,1)

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