Find the maximum $x$ coordinate of a point so that the area of a quadrilateral is $48$

Question

In the $Oxy$ rectangular coordinate system we're given points $O(0,0), A(0,6)$ and $B(8,0)$. The point P is chosen so that $OAPB$ is a convex quadrilateral with area of $48$. Find such P with maximum $x \in \mathbb{Z}$ value.

Here's what I did: first off, we can draw a line from $A$ to $B$, we get a right triangle with area $24$. Therefore, the area of triangle $PAB = 48 - 24 = 24$. We also know, from Pythagoras, that $AB = 10$, so the height from point $P$ to side $AB $ will be $\frac{2\cdot24}{10} = 4.8$. I'm not sure how to go from here.

Edit: I've added a picture, so it's easier to see

Answer 1

You're almost there. What is the locus of points that are at a fixed perpendicular distance from another line? It would be a parallel line, at the given distance

Now, the line AB is $6x +8y = 48$. Hence the line you need is of the form $6x+8y = c$, where $c$ is found using the perpendicular distance

To find the distance between the two lines, we use the following

$$d = \frac{|c-48|}{\sqrt{6^2 + 8^2}} \implies \frac{c-48}{10} = 4.8$$

$$\implies c = 96$$

If you notice, even $c=0$ would give us the same distance, but that would not give us the maximum x coordinate

Now, To maximize x-coordinate, we need to find the solution with biggest $x$ coordinate for $6x+8y = 96$

Now, $x = 15$ is the clearly largest allowable solution, as $x=16$ would result in a triangle, and $x > 16$ would make a concave quadilateral

Answer 2

Apply the formula for distance of a point from the line $\frac{24}5=\frac{\left|\frac x8+\frac y6-1\right|}{\sqrt{\frac 1{8^{^2}}+\frac 1{6^{^2}}}}$and take the positive distance since you want a convex quadrilateral. You should get $\frac x8+\frac y6-1=1$ and you get the largest integral $x$ by choosing $y=\frac34$.