Mathematics Asked on December 10, 2021
The line $6x+8y=48$ intersects the $x$-axis in point $A$ and the $y$-axis in point $B$. A line $L$ bisects the area and the perimeter of the triangle $OAB$,where $O$ is origin.
Find possible equation(s) of $L$.
My Attempt
I feel there can be three lines but how to proceed.
Let the line $L$ be $y= k x +b$ and it intersects with $6x+8y=48$ at $(p,q)$, where
$$p= frac{24-4b}{3+4k },>>>>> q= frac{24k+3b}{3+4k } $$
Given the equal area and perimeter, establish the equations below
$$(6-b)p=bp+8q$$ $$(6-b)+sqrt{p^2+(6-q)^2} = b+8+sqrt{(8-p)^2+q^2}$$
Solve the system of equations above to obtain $b=sqrt6$ and $k=1-sqrt{frac32}$. Thus, the bisecting line $L$ is
$$y= left(1-sqrt{frac32}right)x+sqrt6$$
(Note that above equations corresponds to $L$ intersecting the vertical leg and the hypotenuse; other configurations do not yield valid solutions.)
Answered by Quanto on December 10, 2021
There are three conditions possible :
Area of $displaystyletriangle ACD=frac{1}{2}(pq)sin37^circ=12implies pq=40$. Also, $p+q=12$ for perimeter bisection.
Solve for $p,q$.
Do the same for other two cases
As corrected by @Moko19, also check the other three cases, too which I didn't mentioned in the diagram.
Answered by SarGe on December 10, 2021
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