A) \[(a_{2}^{2}+b_{2}^{2}){{({{c}_{1}}-{{c}_{2}})}^{2}}=(a_{1}^{2}+b_{1}^{2}){{({{d}_{1}}-{{d}_{2}})}^{2}}\]
B) \[(a_{1}^{2}+b_{1}^{2})|{{d}_{1}}-{{d}_{2}}|\,\,=(a_{2}^{2}+b_{2}^{2})|{{c}_{1}}-{{c}_{2}}|\]
C) \[(a_{2}^{2}+b_{2}^{2}){{({{d}_{1}}-{{d}_{2}})}^{2}}=(a_{1}^{2}+b_{1}^{2}){{({{c}_{1}}-{{c}_{2}})}^{2}}\]
D) \[(a_{1}^{2}+b_{1}^{2})|{{c}_{1}}-{{c}_{2}}|\,\,=(a_{2}^{2}+b_{2}^{2})|{{d}_{1}}-{{d}_{2}}\]
Correct Answer: A
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 \[\therefore \] \[\frac{|{{c}_{1}}-{{c}_{2}}|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{|{{d}_{1}}-{{d}_{2}}|}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\] \[\Rightarrow \]\[(a_{2}^{2}+b_{2}^{2}){{({{c}_{1}}-{{c}_{2}})}^{2}}=(a_{1}^{2}+a_{1}^{2}){{({{d}_{1}}-{{d}_{2}})}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
