Finding the equation of a line that splits the area in half I came across a question which I couldn't find the answer to and need some help with.
So there is a parabola of $~y=-3x^2 + 18x -15~$ which is bound by the $~x-~$axis.
Find the equation of the line $~y=mx+c~$ that goes through $~(1,0)~$ so that it splits the area of the parabola into $~2~$ equal parts.
Thanks in advance.
Edit:
Using the help below I still don't seem to be able to get it as I got h= to about 1.1094 which doesn't seem like its splitting the area of the parabola into 2 equal parts
 A: The total area enclosed by the parabola and the $x$-axis is
$$\int_1^5 (-3x^2+18x-15)dx = 32$$
Assume the line intersects the parabola at (a,b). Then, the area above the line and under the parabola should be half of 32, i.e.
$$\int_1^a (-3x^2+18x-15)dx - \frac{1}{2}(a-1)b = 16\tag{1}$$
Carry out the integral
$$\int_1^a (-3x^2+18x-15)dx = (7-a)(a-1)^2\tag{2}$$
Plugging (2) and the replacement $b=3(5-a)(a-1)$ into (1) leads to a clean equation for $a$
$$ (a-1)^3=32$$
and the solution
$$a = 1+2^{5/3}$$
Correspondingly, $b=24(2^{2/3}-2^{1/3})$. Then, it is straightforward to find the equation of the line
$$y=6(2-2^{2/3})(x-1)$$
A: Let the other point of intersection be $(h,k)$
So the equation of the line is $y-0 = \frac{k-0}{h-1}(x-1) \implies y = \frac{k(x-1)}{h-1} $
As we have equal areas,

$$\int^h_1\big[(-3x^2+18x-15) -\frac{k(x-1)}{h-1}\big]dx   = \int^h_1\big[\frac{k(x-1)}{h-1} - 0\big]dx + \int^5_h\big[(-3x^2+18x-15) - 0\big]dx$$
From this you'll get a relation between $h$ and $k$. Also $(h,k)$ lies on the parabola.
So, $k = -3h^2+18h-15$ 
On solving these equations you'll get $h$ and $k$
A: Start off by finding where the parabola intersects the $x$-axis and the resulting area. I assume that you know how to do this, so I’ll just present the answers: $x=1$ and $x=5$, with area equal to $32$.  
An equation of a line through $(1,0)$ with slope $m$ is $y=m(x-1)$, so we find that $c=-m$ after virtually no work. Solving the system of equations, which again I assume you know how to do, gives the other intersection point of this line with the parabola: $(x_1,y_1) = \left(\frac{15-m}3,-\frac13m(m-12)\right)$.  
Here I’ll depart from other answers. Instead of setting up an integral to find the area bounded by the parabola and this chord, I’ll use a handy property of parabolas: The area enclosed by a parabola and its chord is equal to two-thirds of the area of the bounding paralellogram (see here for details). Differentiating and solving for $x$ gives $x=3-\frac m6$ as the point at which the tangent to the parabola has slope $m$ and the equation of this tangent works out to be $$y=mx+\left(\frac{m^2}{12}-3m+12\right).$$ Subtracting the two $y$-intercepts produces $\frac1{12}(m-12)^2$, which is the length of the vertical sides of the bounding paralellogram, so the area bounded by the chord is $$\frac23 \left(\frac{15-m}3-1\right)\frac1{12}(m-12)^2 = \frac1{54}(12-m)^3.$$ Setting this equal to half the area computed previously results in the equation $(12-m)^3 = 864$. I’ll leave solving that to you.
