A triangle with vertices (4, 0), (-1, -1), (3, 5) is (a) isosceles and right angled (b) isosceles bu

Q: Wait, why do we use the distance formula instead of just calculating the slopes?

Great question! While slopes are excellent for checking if lines are perpendicular (which would tell us if the triangle is right-angled), they don't tell us anything about the lengths of the sides. To classify a triangle as isosceles, we absolutely need the side lengths. The distance formula gives us both the lengths for the isosceles check and the values needed for the Pythagorean theorem check, making it the most efficient path.

Q: Do I really need to calculate all three sides?

Yes, you do! In coordinate geometry, you cannot assume the properties of a triangle just by looking at a sketch. Calculating all three sides ($AB$, $BC$, and $CA$) is the only rigorous way to prove whether the triangle is isosceles (two sides equal) or equilateral (three sides equal). It also provides the necessary values to test for the right-angled property.

Q: What if the sum of the squares of the two sides didn't equal the square of the third side?

That is a fantastic hypothetical! If $AB^2 + CA^2 \neq BC^2$, it would simply mean the triangle is not right-angled. In that case, you would look at your side lengths and conclude it is 'isosceles but not right-angled.' Always trust the math over your visual intuition!

Question

A triangle with vertices $(4, 0), (-1, -1), (3, 5)$ is

Options:

(A) isosceles and right angled

(B) isosceles but not right angled

(C) right angled but not isosceles

(D) neither right angled nor isosceles

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a blank coordinate plane. You have three points: $A(4, 0)$, $B(-1, -1)$, and $C(3, 5)$.

These are the anchors of a geometric shape waiting to be understood. In the world of JEE Advanced, classification is about uncovering the hidden relationships between coordinates.

### The Visual Intuition

Before we touch a single algebraic expression, let us visualize. Point $A(4, 0)$ sits proudly on the x-axis.

Point $B(-1, -1)$ pulls us into the third quadrant, while $C(3, 5)$ reaches up into the first. By connecting these, we create a triangle. We need the language of algebra to determine its specific properties.

### The Distance Formula as Our Compass

To know the nature of this triangle, we must measure its sides. The distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, is our most reliable tool.

It is the algebraic manifestation of the Pythagorean theorem, allowing us to bridge the gap between two points in space. Let us calculate the lengths of the sides.

For side $AB$ using $A(4, 0)$ and $B(-1, -1)$:

$$AB = \sqrt{(-1 - 4)^2 + (-1 - 0)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$$

For side $BC$ using $B(-1, -1)$ and $C(3, 5)$:

$$BC = \sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$$

For side $CA$ using $C(3, 5)$ and $A(4, 0)$:

$$CA = \sqrt{(4 - 3)^2 + (0 - 5)^2} = \sqrt{1^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}$$

### The Synthesis

Look at what we have uncovered: $AB = \sqrt{26}$, $BC = \sqrt{52}$, and $CA = \sqrt{26}$. The symmetry is beautiful!

Since $AB = CA$, we have confirmed that the triangle is **isosceles**. Now, we must test for the right-angled property using the converse of the Pythagorean theorem: does $AB^2 + CA^2 = BC^2$?

Substituting our values:

$$(\sqrt{26})^2 + (\sqrt{26})^2 = 26 + 26 = 52$$

Since $BC^2 = (\sqrt{52})^2 = 52$, the condition $52 = 52$ holds perfectly. The triangle is **right-angled at vertex $A$**.

### Conclusion: The Elegance of Proof

We have moved from raw coordinates to a definitive classification. Our triangle is both **isosceles and right-angled**.

This problem reminds us that behind every set of coordinates lies a story of symmetry and logic. Keep this rigor in your toolkit, and no geometry problem will ever be able to hide its true nature from you.

Animated Solution for Mathematics - Straight Lines: A triangle with vertices $(4, 0), (- 1, - 1), (3, 5)$ is

Select Answer:

Visualized Solution

Visualizing the Vertices

The Distance Formula

Setting up Side $A B$

Calculating Side $A B$

Setting up Side $B C$

Calculating Side $B C$

Setting up Side $C A$

Calculating Side $C A$

Checking the Isosceles Property

Checking the Right-Angled Property

Final Conclusion

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Visual Intuition

The Distance Formula as Our Compass

The Synthesis

Conclusion

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A triangle with vertices (4,0),(−1,−1),(3,5) is