A) \[\frac{25}{4\sqrt{2}}\]
B) \[\frac{25}{3\sqrt{2}}\]
C) \[\frac{25}{2\sqrt{2}}\]
D) \[\frac{25}{\sqrt{2}}\]
Correct Answer: A
Solution :
[a] Two of the given lines \[y+3x-5=0\] and \[3y-x+10=0\] are \[\bot \] to each other, so the triangle is right angled and it circumradius is half the hypotenuse, \[y=x\] intersects other two lines in \[\left( \frac{5}{4},\frac{5}{4} \right)\] and \[(-5,-5),\] \[\therefore \,\,\,\,Circumradius=\frac{1}{2}\left[ \sqrt{{{\left( -5-\frac{5}{4} \right)}^{2}}+{{\left( -5-\frac{5}{4} \right)}^{2}}} \right]\] \[=\frac{1}{2}\left[ \sqrt{\frac{625}{16}+\frac{625}{16}} \right]=\frac{1}{2}\left[ \frac{25\sqrt{2}}{4} \right]=\frac{25}{4\sqrt{2}}\]
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