The figure shows ΔABC.

1. Find the gradient of line AC from the two given coordinates.

For $A(-2,4)$ and $C(6,1)$,
$m=\frac{1-4}{6-(-2)}=\frac{-3}{8}.$

2. Find the y-intercept of AC and check May's claim.

Using $y=mx+c$ and point $A(-2,4)$,
$4=-\frac{3}{8}(-2)+c$
$4=\frac{3}{4}+c$
$c=4-\frac{3}{4}=\frac{13}{4}=3.25.$
So the line intersects the $y$-axis at $3.25$.

1. Find the horizontal and vertical differences between A and B.

$A(-2,4)$ and $B(2,1)$

Horizontal difference: $2-(-2)=4$

Vertical difference: $1-4=-3$, so the vertical distance is $3$.

2. Apply the distance formula to find AB.

$AB=\sqrt{4^2+(-3)^2}$
$=\sqrt{16+9}$
$=\sqrt{25}=5$

1. Identify the adjacent side and hypotenuse for the angle at $B$, noting that the angle is obtuse.

From $B(2,1)$ to $A(-2,4)$, the horizontal change is $4$ units left and the vertical change is $3$ units up, so $AB=5$ units.

Since $BC$ is horizontal to the right, $\angle ABC$ is obtuse. Hence
$\cos \angle ABC = -\frac{4}{5}.$

1. Compare the areas using the bases on the same horizontal line.

Since $B$, $C$ and $P$ lie on $y=1$, triangles $ABC$ and $APB$ have the same perpendicular height from $A$.

Therefore,
$\frac{\text{area of }\triangle ABC}{\text{area of }\triangle APB}=\frac{BC}{BP}.$
Given
$\text{area of }\triangle ABC : \text{area of }\triangle APB = 4:3,$
so
$BC:BP=4:3.$

2. Find the possible positions of $P$ from the required length $BP$.

From the coordinates, $B=(2,1)$ and $C=(6,1)$, so
$BC=6-2=4.$
Since
$BC:BP=4:3,$
$BP=3.$

As $P$ lies on $y=1$, it is 3 units horizontally from $B$, whose $x$-coordinate is 2.

Hence
$x=2-3=-1$
or
$x=2+3=5.$