## University Calculus: Early Transcendentals (3rd Edition)

$a.\:\:\:y\:=\:-x+1$ $b.\:\:\:\mathrm{Area}\:=\:-2x^2+2x$
$a.\:\:$ Since the given triangle is an isosceles triangle, its base angles measure $\:45^{\circ}.\:$ We will also notice that the right and left halves of the given triangle are also isosceles triangles, so that the point $\:B\:$ is at $\:(0,1).$ We have two points $\:A=(1,0)\:\:\mathrm{and}\:\:B(0,1).$ We can write the equation of line $\:\overline{AB}\:$ as: $y-0=\frac{1-0}{0-1}(x-1)$ $y=-x+1$ As the point $\:P\:$ is on the line $\:\overline{AB}\:$, coordinates of it are $\:(x,-x+1).$ $b.\:\:$ Area of the triangle is the product of base and height. We have the base of the triangle as $\:2x\:$ and the height as $\:-x+1.$ So the area is represented as: $\mathrm{Area}\:=\:2x(-x+1)$ $=-2x^2+2x$