Rearranging the equations,
$$L_1:y=\frac12x+2$$
$$L_2:y=-2x+5$$
As product of slope of $L_1$ and $L_2$ $=(-2)\cdot(\frac12)=-1$,
$L_1$ is perpendicular to $L_2$.
Let length of one of the side of triangle be $r$.
$$\frac12r^2=10$$ $$r=\sqrt{20}$$
Let a point on $L_1$ be A: $(t,\frac{t+4}2)$, by considering the distance between the intersection point $(\frac65,\frac{13}5)$ and A,we have
$$\sqrt{20}=\sqrt{(t-\frac65)^2+(\frac{t+4}2-\frac{13}5)^2}$$
$$t=5.2\text{ or }-2.8$$
$$A_1= (5.2,4.6)\text{ or }A_2=(-2.8,0.6)$$
Similarly, for a point on $L_2$, B: $(p,-2p+5)$, by considering the distance between the intersection point $(\frac65,\frac{13}5)$ and B,we have
$$\sqrt{20}=\sqrt{(p-\frac65)^2+(-2p+5-\frac{13}5)^2}$$
$$p=3.2\text{ or }-0.8$$
$$B_1= (3.2,-1.4)\text{ or }B_2=(-0.8,6.6)$$
So we have line $L_{11}=A_1B_1, L_{21}=A_2B_1,L_{22}=A_2B_2,L_{12}=A_1B_2$
$$L_{11}:\frac{y-4.6}{x-5.2}=\frac{4.6-(-1.4)}{5.-3.2}$$
$$L_{11}:y=3x-11$$
Similar for $L_{21},L_{22}\text{ and }L_{12}$.
Repeat your reasoning starting with this form of triangular inequalities:
$$
|a-b|<c,\quad |b-c|<a,\quad |c-a|<b,
$$
that is:
$$
(a-b)^2<c^2,\quad (b-c)^2<a^2,\quad (c-a)^2<b^2.
$$
My solution:-
Best Answer
The problem doesn't say that Q, R, and S are the vertices of a (non-degenerate) triangle, only that they are points on the plane. They may all be on the same line. Therefore, you want $7\le x\le 29$, and (D) is the correct answer.