Equation of rotated line.

Question

A line $L_1 \equiv 3y-2x-6=0 $ is rotated about its point of intersection with the y axis to in the clockwise direction to make it $L_2$ such that the area formed by $L_1,L_2, y=0, x=5$ is $49/3$ square units if $L_2's$ point of intersection with the line x= 5 lies below the X axis. Find the equation of $L_2$.

Diagram:

Attempt: We have the following points enclosing an area of $\dfrac{49}3$:

$(5,\dfrac {16} 3)$

$(5,0)$

$(x,y)$// Point on $L_2$

$(0,2)$

Using the Gauss's area formula, $5y + 2x = \dfrac{208}{3}$

But answer given is $x+y= 2$. Where have I gone wrong?

Answer 1

Let $F$ be a point on $BC$ such that $AF \perp BC$.

Then, $F=(5,2)$.

Now, let $E=(a,0)$.

The area of $\Delta ABF$ is $\dfrac12 (5) \left(\dfrac{10}3\right) = \dfrac{25}3$.

The area of $AECF$ is $\dfrac12 (5+(5-x)) (2) = 10-a$.

Therefore, $\dfrac{25}3 + 10 - a = \dfrac{49}3$, which means $a=2$.

Then, $L_2$ passes through $(0,2)$ and $(2,0)$, so its equation is $x+y=2$.

Answer 2

I figured out my mistake. The Gaussian area formula yields the modulus of area. SO it can be negative or positive. Also for $(x,y)$ I chose, $y=0$.

Thus,

$\dfrac{49}{3}= \dfrac{1}{2}|2x- \dfrac{110}{3}|$

which gives $ x= 2$ or $x= \dfrac{104}3$

Now, $\dfrac {104}{3}$ is rejected because $104/3$ exceeds $5$.

We already have a point of $L_2$ which is $(0,2)$

And the point just obtained is $(2,0)$.

The equation of $L_2$ can be found using intercept form. $x+y=2$

Alternatively,

I had chosen the points in clockwise order so the determinant of area is negative.

Cheers!