Without using the Pythagoras theorem, show that the points $(4, 4)$, $(3, 5)$ and $(-1, -1)$ are the vertices of a right angled triangle.

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> [!TLDR]
> Use the slope criterion: two lines are perpendicular if and only if the product of their slopes is $-1$. Computing slopes of the three sides shows that two are perpendicular, confirming a right angle without invoking Pythagoras.

The Pythagorean theorem tells us *that* a triangle is right-angled by checking side lengths. But the deeper geometric fact is that a right angle means **perpendicular sides**. In coordinate geometry, two non-vertical lines are perpendicular precisely when the product of their slopes equals $-1$. This gives us a direct, algebraic way to verify a right angle.

Let's label the points: $A = (4, 4)$, $B = (3, 5)$, and $C = (-1, -1)$.

A triangle has a right angle at one of its vertices if the two sides meeting there are perpendicular. We'll compute the slopes of all three sides and check whether any pair multiplies to $-1$.

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**1. Slope of side $AB$**

The slope between $A(4, 4)$ and $B(3, 5)$ is

$$m_{AB} = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1.$$

**2. Slope of side $BC$**

The slope between $B(3, 5)$ and $C(-1, -1)$ is

$$m_{BC} = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}.$$

**3. Slope of side $CA$**

The slope between $C(-1, -1)$ and $A(4, 4)$ is

$$m_{CA} = \frac{4 - (-1)}{4 - (-1)} = \frac{5}{5} = 1.$$

**4. Check for perpendicularity**

Now we test the products:

| Pair of sides | Product of slopes | Perpendicular? |
|---------------|-------------------|----------------|
| $AB$ and $BC$ | $(-1) \cdot \frac{3}{2} = -\frac{3}{2}$ | No |
| $BC$ and $CA$ | $\frac{3}{2} \cdot 1 = \frac{3}{2}$ | No | …

Q.Without using the Pythagoras theorem, show that the points $(4, 4)$ , $(3, 5)$ and $(-1, -1)$ are the vertices of a right angled triangle.

Concept understanding — Perpendicular Slopes Condition

Perpendicular Slopes Condition

The Intuition

The Precise Statement

What About Vertical and Horizontal Lines?

Quick Check

Why It Works (A Short Proof)

Pair of sides	Product of slopes	Perpendicular?
$AB$ and $BC$	$(-1) \cdot \frac{3}{2} = -\frac{3}{2}$	No
$BC$ and $CA$	$\frac{3}{2} \cdot 1 = \frac{3}{2}$	No

Q.Without using the Pythagoras theorem, show that the points (4,4)(4, 4)(4,4), (3,5)(3, 5)(3,5) and (−1,−1)(-1, -1)(−1,−1) are the vertices of a right angled triangle.

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Q.Without using the Pythagoras theorem, show that the points $(4, 4)$ , $(3, 5)$ and $(-1, -1)$ are the vertices of a right angled triangle.

Unlock everything free for 14 days