Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenus

Q: Wait, why is the right angle at vertex $B$?

Great question! In a right-angled triangle, the hypotenuse is always the side opposite the $90^{\circ}$ angle. Since the problem explicitly tells us that $AC$ is the hypotenuse, the angle directly across from it—which is $\angle B$—must be the right angle.

Q: Why is side $AB$ vertical?

Think about the geometry: we established that $BC$ is a horizontal line because both points have a y-coordinate of $1$. Since $\angle B = 90^{\circ}$, the side $AB$ must be perpendicular to $BC$. A line perpendicular to a horizontal line is always vertical. This is a powerful constraint that simplifies our coordinates!

Q: Why do we get two different values for $k$?

That is the beauty of the absolute value! When we solve $\frac{1}{2} \times 1 \times |k - 1| = 1$, we get $|k - 1| = 2$. This equation means the distance between $k$ and $1$ is $2$ units. This can happen in two directions: $k$ could be $2$ units above $1$ (giving $k = 3$) or $2$ units below $1$ (giving $k = -1$). Both configurations form a valid triangle with the same area.

Question

Let $A(h, k), B(1, 1)$ and $C(2, 1)$ be the vertices of a right angled triangle with $AC$ as its hypotenuse. If the area of the triangle is 1 square unit, then the set of values which 'k' can take is given by

Options:

(A) $\{-1, 3\}$

(B) $\{-3, -2\}$

(C) $\{1, 3\}$

(D) $\{0, 2\}$

Sigma Percentile · Accepted Answer

### The Geometry of Possibility

Welcome, fellow traveler in the world of coordinate geometry! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful, symmetric truth.

We are given three vertices of a right-angled triangle: $A(h, k)$, $B(1, 1)$, and $C(2, 1)$. Our mission is to find the possible values of $k$ given that the area is $1$ square unit and $AC$ is the hypotenuse.

Let's embark on this journey step by step.

### Phase 1: The Horizontal Foundation

First, let's look at the points we know: $B(1, 1)$ and $C(2, 1)$. If you plot these on a Cartesian plane, you will notice something immediate and striking.

Both points share the same y-coordinate, $y = 1$. This means the line segment $BC$ is perfectly horizontal.

This is our anchor. It tells us that the base of our triangle, if we choose $BC$ as the base, is parallel to the x-axis. The length of this base is simply the difference in their x-coordinates: $|2 - 1| = 1$ unit.

### Phase 2: The Right-Angle Constraint

Now, let's address the hypotenuse. We are told $AC$ is the hypotenuse. By the fundamental properties of a right-angled triangle, the angle opposite the hypotenuse must be $90^{\circ}$.

Therefore, $\angle B = 90^{\circ}$. This is a massive piece of information!

It tells us that the side $AB$ is perpendicular to the side $BC$. Since we already know $BC$ is a horizontal line, $AB$ must be a vertical line.

This forces the x-coordinate of $A$ to be the same as the x-coordinate of $B$. Since $B$ is at $(1, 1)$, $A$ must have an x-coordinate of $1$. Thus, $h = 1$, and our vertex $A$ is simply $(1, k)$.

### Phase 3: The Algebra of Area

We are given that the area of this triangle is $1$ square unit. The area of a right-angled triangle is given by the formula:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

Here, our base is $BC$ and our height is $AB$. We know the length of $BC$ is $1$. The length of $AB$ is the vertical distance between $A(1, k)$ and $B(1, 1)$, which is $|k - 1|$.

Substituting these into our area formula, we get:

$$\frac{1}{2} 	imes 1 	imes |k - 1| = 1$$

This simplifies to:

$$|k - 1| = 2$$

### Phase 4: The Two Realities

This is where many students rush and miss a solution. The absolute value $|k - 1| = 2$ represents the distance between $k$ and $1$ on the number line.

This distance is $2$ units. This can happen in two ways: either $k - 1 = 2$ or $k - 1 = -2$.

Solving the first case, $k - 1 = 2$, we get $k = 3$. This corresponds to a triangle where vertex $A$ is at $(1, 3)$, sitting above the line $BC$.

Solving the second case, $k - 1 = -2$, we get $k = -1$. This corresponds to a triangle where vertex $A$ is at $(1, -1)$, sitting below the line $BC$.

Both triangles are perfectly valid, both have an area of $1$, and both satisfy the right-angle condition at $B$. The set of values for $k$ is therefore **$\{-1, 3\}$**.

### Conclusion

Isn't it elegant? The math didn't just give us a number; it gave us a geometric reality.

By respecting the absolute value, we uncovered the symmetry of the problem. Never fear the modulus—it is simply the math's way of telling you there is more than one way to build a triangle.

Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic!

Animated Solution for Mathematics - Straight Lines: Let $A (h, k), B (1, 1)$ and $C (2, 1)$ be the vertices of a right angled triangle with $A C$ as its hypotenuse. If the area of the triangle is 1 square unit, then the set of values which 'k' can take is given by

Select Answer:

Visualized Solution

Plotting the Given Vertices

Identifying the Right Angle

Direction of Side $A B$

Coordinates of Vertex $A$

Area of the Triangle

Calculating Base and Height

Substituting into Area Formula

Simplifying the Equation

Solving Case 1 (Positive)

Solving Case 2 (Negative)

Final Conclusion

The Sigma Insight: Area of Triangle

The Geometry of Possibility

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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