Find the equation of straight line which passes through the intersection of the lines $x-2y-5=0$ and $7x+y=50$.... 
Find the equation of straight line which passes through the intersection of the lines $x-2y-5=0$ and $7x+y=50$ and divides the circumference of the circle $x^2+y^2=100$ into arcs of ratio 2:1

The point of intersection of the two lines was found to be $(7,1)$
Since the lines divides the arc in the ratio 2:1, the angles subtended by the line at the center is $2\pi/3$
If we consider $\Delta ABC$, where A is the origin, line BC is the required straight line and $D(h,k)$ is the foot of the perpendicular drawn from A. 
Length AD is 5 units. 
From here I can find the locus of D, but that isnt helpful. I just need the point D, and from there I can easily find the equation of the line. 
 A: Let $D(a,b)$. Then, the distance from the Origin to $D$ and the distance from $(7,1)$ to $D$ will give us 2 equations:
$$
\begin{cases}
a^2+b^2=25 \\ 
(a-7)^2+(b-1)^2=25
\end{cases}
$$ By solving these, we get $(a,b)=(3,4)$ or $(4,-3)$
(Reason for the second equation: Let $M(7,1)$. Both $D$ and $M$ are located on the line $BC$. So, the triangle $ADM$ is a right triangle. $AM=5\sqrt{2}$ and the leg $AD=5$ implies that the leg $DM=5$)
A: You didn’t really need to compute the intersection point explicitly. Every line that passes through that point has an equation that’s a linear combination of the equations of the two lines: $$\lambda(x-2y-5)+\mu(7x+y-50) = (\lambda+7\mu)x+(\mu-2\lambda)y-5(\lambda+10\mu) = 0\tag1$$ with $\lambda$ and $\mu$ not both zero. As you’ve already found, since the sought-after line divides the circumference in a $2:1$ ratio, the chord subtends an angle of $\frac23\pi$, so the line must be at a distance of $10\cos{\frac\pi3}$ from the circle’s center. Plugging these values into the standard point-line distance formula and squaring, we have $${25(\lambda+10\mu)^2 \over (\lambda+7\mu)^2+(\mu-2\lambda)^2} = 25,$$ which simplifies to $$2\lambda^2-5\lambda\mu-25\mu^2 = (\lambda-5\mu)(2\lambda+5\mu) = 0.\tag2$$ If this didn’t happen to factor nicely, we could observe that neither of the original lines is a solution, so we can set either $\lambda$ or $\mu$ to $1$ to obtain a straightforward quadratic equation in one variable. This amounts to forming either the combination $L_1+\mu L_2$ or $\lambda L_1+L_2$, which omits one of the two lines, but we’ve determined that this is safe to do. (You could always start with one of these combinations, but you must remember to check the line that is omitted from that family separately.)  
Alternatively, you can “split” the conic represented by equation (2) by finding a value of $\alpha$ which makes the matrix $$\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}2&-\frac52\\-\frac52&-25\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}+\alpha\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ singular, i.e., solve $\alpha^2-\frac{225}4=0.$The resulting matrix has rank one, so any nonzero row and nonzero column give you the coefficients $\lambda$ and $\mu$ of the two solutions. Here, using the positive square root yields $\tiny{\begin{bmatrix}-25&-5\\10&2\end{bmatrix}}$, so two independent solutions are $\lambda=5$, $\mu=1$ and $\lambda=-5$, $\mu=2$, which agrees with the factorization in (2).
