Question

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

Answer 1

Step-by-step explanation:

first let the three points be

A=(4,4)

B=(3,5)

C=(-1-1)

let's calculate slope of AB ,BC,andAC..

you can check the images for further steps

Answer 2

Given :-

Point A = (4,4)

Point B = (3,5)

Point C = (–1, –1)

To Prove :-

The points are vertices of a right angled triangle.

Solution :-

Given that,

The vertices of the given triangle are (4, 4), (3, 5) and (–1, –1).

The slope (m) of the line non-vertical line passing through the point $\sf (x_1, \: y_1)$ and,

$\sf (x_2,\: y_2)$ is given by m = $\sf \dfrac{(y_2-y_1)}{(x_2-x_1)}$ where, $\sf x \neq x_1$

So, the slope of the line AB

$\sf (m_1) = \dfrac{(5-4)}{(3-4)} -\dfrac{1}{-1} =-1$

The slope of the line BC $\sf (m_2),$

$\sf \dfrac{(-1-5)}{(-1-3)} =\dfrac{-6}{-4} =\dfrac{3}{2}$

The slope of the line CA $\sf (m_3),$

$\sf (m_3) =\dfrac{(4+1)}{(4+1)} =\dfrac{5}{5} =1$

It is observed that, $\sf m_1.m_3 = -1.1 = -1$

Hence, the lines AB and CA are perpendicular to each other.

∴ Given triangle is right-angled at A (4, 4)

And the vertices of the right-angled ∆ are (4, 4), (3, 5) and (-1, -1)