Pair Of Straight Lines Question 107
Question: Area of the triangle formed by the lines $ y^{2}-9xy+18x^{2}=0 $ and $ y=9 $ is
Options:
A) $ \frac{27}{4}sq $ . units
B) $ 27sq. $ units
C) $ \frac{27}{2}sq. $ units
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The lines represented by $ y^{2}-9xy+18x^{2}=0 $ are $ 6x-y=0 $ and $ 3x-y=0 $ and third line is $ y=9 $ .
Therefore, coordinates of vertices of triangle are given by $ A(0,0) $ ; $ B(3,9) $ and $ C( \frac{3}{2},9 ) $ .
Hence, area of DABC $ =\frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 3 & 9 & 1 \\ 3/2 & 9 & 1 \\ \end{vmatrix} =\frac{27}{4}sq.units. $
A liter: Applying the formula given in the theory part, the required area is $ \frac{{{(-9)}^{2}}\sqrt{{{(9/2)}^{2}}-18}}{18\times 1+9\times 0\times 1+1\times 0}=\frac{81}{18}\sqrt{\frac{81}{4}-18} $
$ =\frac{81}{18}\times \frac{3}{2}=\frac{27}{4}sq.\ units. $
BETA
