Animated Solution for Mathematics - Straight Lines: Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$, for which the points $(a^2,a+1)$ lie in $R$, is :

Coordinate geometry is not just about plotting points; it is about understanding the architecture of the plane. Imagine you are standing on a vast, infinite sheet of paper.

Two lines, $L_1:3x-y+1=0$ and $L_2:x+2y-5=0$, are drawn across this paper. These lines are not merely equations; they are boundaries that partition the plane into four distinct regions.

Our goal is to find the 'interior' region $R$ that contains the origin $(0,0)$ and then identify the values of $a$ such that the point $P(a^2,a+1)$ resides within this sanctuary.

To solve this, we must embrace the 'Same Side' condition. Think of a line as a fence; if you and your friend are on the same side of the fence, you are in the same region.

Mathematically, this means that if you plug your coordinates into the line's equation, the result will have the same sign as your friend's coordinates. If the origin $(0,0)$ is our reference, we simply need to ensure that our point $P(a^2,a+1)$ yields the same sign as the origin when substituted into both $L_1$ and $L_2$.

Let us test our anchor, the origin, against the first line, $L_1:3x-y+1=0$. Substituting $x=0$ and $y=0$, we get $3(0)-0+1=1$.

Since $1>0$, the origin lies on the positive side of $L_1$. Therefore, for our point $P(a^2,a+1)$ to be in the same region, it must also yield a positive result:

Simplifying this, we get $3a^2-a-1+1>0$, which reduces to $3a^2-a>0$. Factoring this, we have $a(3a-1)>0$.

Using the wavy curve method, we find the roots at $a=0$ and $a=\frac{1}{3}$. The inequality holds for:

Now, we turn our attention to $L_2:x+2y-5=0$. Testing the origin again, we get $0+2(0)-5=-5$.

Since $-5<0$, the origin lies on the negative side of $L_2$. Thus, our point $P$ must also yield a negative result:

Expanding and simplifying, we get $a^2+2a+2-5<0$, or $a^2+2a-3<0$. This quadratic factorizes beautifully into $(a+3)(a-1)<0$.

For this product to be negative, $a$ must lie strictly between the roots, giving us:

We have two conditions that must be satisfied simultaneously. We need the intersection of $a\in (-\infty ,0)\cup (\frac{1}{3},\infty )$ and $a\in (-3,1)$.

Visualizing this on a number line, we see the overlap occurs between $-3$ and $0$, and between $\frac{1}{3}$ and $1$.

Thus, the set of all values of $a$ is:

This is the exact region where our point $P$ is guaranteed to be safe within the boundaries of $R$. You have successfully navigated the logic of coordinate geometry!