Triangle in Parabola

$$ The area of a triangle whose vertices are $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ is given by: $$ $A=\frac{1}{2}\left|\begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{matrix}\right|.$ $$ Therefore, the area of a triangle whose vertices are $(1,0)$ , $(4,3)$ , and $(x,y)$ where $(x,y)$ lies on the parabola $\,y=-x^2+6x-5$ is $$ $\begin{aligned} A&=\frac{1}{2}\left|\begin{matrix} x & y & 1 \\ 1 & 0 & 1 \\ 4 & 3 & 1\end{matrix}\right|\\ &=\frac{1}{2}(3y-3x+3)\\ &=\frac{3}{2}(y-x+1)\\ &=\frac{3}{2}(-x^2+6x-5-x+1)\\ &=-\frac{3}{2}x^2+\frac{15}{2}x-6. \end{aligned}$ $$ In general, quadratic equation $\,f(x)=ax^2+bx+c$ , for $a<0$ , will have a maximum value at $x=-\dfrac{b}{2a}$ . Thus $$ $x=-\frac{\frac{15}{2}}{2\left(-\frac{3}{2}\right)}=\frac{5}{2}$ and $\begin{aligned} A&=-\frac{3}{2}x^2+\frac{15}{2}x-6\\ &=-\frac{3}{2}\left(\frac{5}{2}\right)^2+\frac{15}{2}\left(\frac{5}{2}\right)-6\\ &=\boxed{\dfrac{27}{8}} \end{aligned}$ $$ $\text{\# }\mathbb{Q.E.D.}\text{ \#}$

find the area of the triangle with co-ordinates (1,0),(4,3),(a,b)...then (a,b) also lies on the parabola...so we find a relation between a and b...using this relation we reduce the area of triangle to one variable either in terms of a or b...then differentiate the area to use maxima and minima condition...hence we find the value of either a or b for which the area will be maximum....hence the result...

eqn of parabola is (y-4)=-(x-3)^{2} so, for simplicity shift the coordinates A(1,0) and B(4,3) to new C(-2,-4) and D(1,-1) respectively and parabola to simple eqn :y=-(x^{2}) so b=-a^{2} and distance of (a,-a^{2}) from the line CD=>Y-X+2=0 is |-a^2-a+2|/sqrt{2} is to be maximum for maxm area hence \frac{d-a^2-a+2}{da}=0; => a=-0.5 hence find area