Q106 / 200 · Manipal (MET) 2010 · Mathematics
If the straight line $a_1 x+b_1 y+c_1=0$, $a_1 x+b_1 y+c_2=0, \quad a_2 x+b_2 y+d_1=0$ and $a_2 x+b_2 y+d_2=0$ are the sides of rhombus, then
- Correct answerA$\left(a_2^2+b_2^2\right)\left(c_1-c_2\right)^2=\left(a_1^2+b_1^2\right)\left(d_1-d_2\right)^2$
- B$\left(a_1^2+b_1^2\right)\left|d_1-d_2\right|=\left(a_2^2+b_2^2\right)\left|c_1-c_2\right|$
- C$\left(a_2^2+b_2^2\right)\left(d_1-d_2\right)^2=\left(a_1^2+b_1^2\right)\left(c_1-c_2\right)^2$
- D$\left(a_1^2+b_1^2\right)\left|c_1-c_2\right|=\left(a_2^2+b_2^2\right)\left|d_1-d_2\right|$
Solution
Since, given straight lines are the sides of a rhombus. Therefore, distance between the parallel lines
$a_1 x+b_1 y+c_1=0, a_1 x+b_1 y+c_2=0$ and
$a_2 x+b_2 y+d_1=0, a_2 x+b_2 y+d_2=0$ must be equal
$\therefore \frac{\left|c_1-c_2\right|}{\sqrt{a_1^2+b_1^2}}=\frac{\left|d_1-d_2\right|}{\sqrt{a_2^2+b_2^2}}$
$\Rightarrow\left(a_2^2+b_2^2\right)\left(c_1-c_2\right)^2=\left(a_1^2+a_1^2\right)\left(d_1-d_2\right)^2$
Hence, the correct answer is option: 'A'.