Question

Lines $x+y=1$ and $3y=x+3$ intersect the ellipse $x^2+9y^2=9$ at the points P, Q, R. The area of the triangle PQR is ?

• A. $\frac{36}{5}$
• B. $\frac{18}{5}$
• C. $\frac{9}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{18}{5}$

R.E.F image

The given diagram is of ellipse whose equation is $x^2+9y^2=9$

And the two lines with eq. (1) $x+y=1$ (2) $3y=x+3$

It is clear from figure where these lines are intersecting with

ellipse, two points are (0,1) and (-3, 0). For the third point we

have to solve the eq. of ellipse and line(1).

From eq. of line (1), we get, $x=1-y$ ...(i)

Putting this in eq. of ellipse,

$(1-y{)}^2+9y^2=9$

$1+y^2-2y+9y^2=9$

$y^2-2y-8=0$

By solving, we get

$y=1$ and $y=\frac{-4}{5}$

by putting $y=\frac{-4}{5}$ in eq. (i), we get, $x=\frac{9}{5}$

Now, the three points are $A(0,1),B(\frac{9}{5},\frac{-4}{5})$ and $C(-3,0)$.

So, area of triangle $ABC$ is:

$A=\frac{1}{2}\left|\begin{array}{ccc}0& 1& 1\\ -3& 0& 1\\ \frac{9}{5}& \frac{-4}{5}& 1\end{array}\right|$

By solving along $R1$, we get

$A=\frac{18}{5}sq.unit$