Animated Solution for Mathematics - Straight Lines: All points lying inside the triangle formed by the points $(1,3),(5,0)$ and $(-1,2)$ satisfy

Navigating the Triangle

Imagine you are standing on a vast, flat Cartesian plane. Before you lies a triangle, defined by three anchors: $A(1,3)$, $B(5,0)$, and $C(-1,2)$.

You are asked to determine which mathematical 'laws'—represented by linear inequalities—govern every single point within this triangular territory. It feels like a daunting task, doesn't it? How can we possibly check an infinite number of points?

Here is the secret that separates the novice from the master: the triangle is a convex set. In the world of geometry, convexity is a superpower.

It means that if you take any two points inside the triangle and draw a line segment between them, that entire segment stays trapped inside the triangle.

Because the boundaries of our triangle are straight lines, and our candidate inequalities are also linear, we are dealing with half-planes. A linear inequality $ax+by+c\ge 0$ essentially divides the entire universe into two sides.

If all three vertices of our triangle fall on the 'correct' side of this line, then the entire triangle—by the grace of convexity—must also reside on that same side. We don't need to check the infinite points inside; we only need to check the three sentinels at the corners: $A$, $B$, and $C$.

Let us test our first candidate: $3x+2y\ge 0$. We approach vertex $A(1,3)$ first.

Substituting these coordinates, we get: Since $9\ge 0$, the law holds at $A$.

Now, we move to $B(5,0)$. Here: This is also $\ge 0$.

Finally, we check $C(-1,2)$, yielding: Again, $1\ge 0$. Because all three vertices satisfy the condition, we have found our first truth: the entire triangle obeys $3x+2y\ge 0$.

Not every law is universal. Consider $2x+y-13\ge 0$.

When we test vertex $A(1,3)$, we calculate: Since $-8$ is clearly not $\ge 0$, this law fails at the very first gate. We can immediately discard this option.

It is a beautiful moment of efficiency—one calculation, and we have saved ourselves from wasting time on the rest of the triangle.

Finally, let us examine $2x-3y-12\le 0$. We test $A(1,3)$: This is $\le 0$, so $A$ is satisfied.

Moving to $B(5,0)$, we get: This is also $\le 0$.

Finally, at $C(-1,2)$, we find: Since $-20\le 0$, the condition holds for all three vertices.

The Harmony of Mathematics

Through the lens of convexity, we have transformed a seemingly impossible problem into a simple verification task.

We have discovered that both $3x+2y\ge 0$ and $2x-3y-12\le 0$ act as the boundaries of truth for our triangle.

Remember, in JEE Advanced, the most complex problems often yield to the most fundamental principles. Keep your eyes on the geometry, trust the properties of the shapes you study, and the math will always reveal its secrets to you.