In right triangle ABC the angle at C is 90 degrees, AC = 6 and BC = 8. Point P moves along BC without reaching B or C, and PC = x. Express the area y of triangle APB as a function of x and give the range of x.

Step-by-step solution

1. 1

   Choose the base so that the height is already known. Both $P$ and $B$ lie on segment $BC$, so take $PB$ as the base of triangle $APB$. Since $\angle C=90^\circ$, the leg $AC$ is perpendicular to line $BC$ — and therefore perpendicular to $PB$ no matter where $P$ sits. The height is the constant

   $h=AC=6.$

   This is the whole idea of the problem: the moving point changes the base but never the height.

2. 2

   Write the base in terms of x. $P$ lies between $C$ and $B$ with $PC=x$ and $CB=8$, so

   $PB=BC-PC=8-x.$

3. 3

   Apply the triangle area formula.

   $y=\frac{1}{2}\cdot PB\cdot AC=\frac{1}{2}(8-x)(6)=3(8-x).$

4. 4

   Expand to the requested linear form.

   $y=24-3x.$

   The area falls at a constant rate of $3$ square units per unit that $P$ slides toward $B$, which is what a linear relationship should look like.

5. 5

   Determine the range of x from the geometry, not from the formula. $P$ is on segment $BC$ and coincides with neither endpoint, so $PC$ is strictly between $0$ (at $C$) and $8$ (at $B$):

   $0<x<8.$

   Both endpoints are excluded, which is also why $y$ never reaches its extreme values $24$ or $0$ — at $x=8$ the three points would be collinear and there would be no triangle at all.