MHT CET · Maths · Pair of Lines
Area of the triangle formed by the line \(y^2-9 x y+18 x^2=0\) and \(y=9\) is
- A \(\frac{27}{3}\) sq. units
- B \(\frac{27}{2}\) sq. units
- C \(\frac{27}{4}\) sq. units
- D 27 sq. units
Answer & Solution
Correct Answer
(C) \(\frac{27}{4}\) sq. units
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& y^2-9 x y+18 x^2=0 \\
& \therefore(y-3 x)(y-6 x)=0
\end{aligned}
\)
Thus three lines forming triangle are \(y=3 x, y=6 x, y=9\)
Their point of intersections are \((0,0),(3,9),\left(\frac{3}{2}, 9\right)\)
\(\therefore\) Area of triangle
\(=\frac{1}{2} \times \frac{3}{2} \times 9\)
\(=\frac{27}{4}\) sq. units
\begin{aligned}
& y^2-9 x y+18 x^2=0 \\
& \therefore(y-3 x)(y-6 x)=0
\end{aligned}
\)
Thus three lines forming triangle are \(y=3 x, y=6 x, y=9\)
Their point of intersections are \((0,0),(3,9),\left(\frac{3}{2}, 9\right)\)
\(\therefore\) Area of triangle
\(=\frac{1}{2} \times \frac{3}{2} \times 9\)
\(=\frac{27}{4}\) sq. units
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