Straight Line Question 266
Question: The circumradius of the triangle formed by the three lines $ y+3x-5=0;y=x $ and $ 3y-x+10=0 $ is
Options:
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 $ \therefore Circumradius=\frac{1}{2}[ \sqrt{{{( -5-\frac{5}{4} )}^{2}}+{{( -5-\frac{5}{4} )}^{2}}} ] $ $ =\frac{1}{2}[ \sqrt{\frac{625}{16}+\frac{625}{16}} ]=\frac{1}{2}[ \frac{25\sqrt{2}}{4} ]=\frac{25}{4\sqrt{2}} $
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Answer:
Solution:
BETA
