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

Asked by Abhisek | 11 months ago |  87

##### Solution :-

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 (x1, y1) and

(x2, y2) is given by m =

$$\dfrac{ (y_2 – y_1)}{(x_2 – x_1)}$$ where, x ≠ x1

So, the slope of the line AB (m1

$$\dfrac{(5-4)}{(3-4)}$$

$$\dfrac{1}{-1}$$ = -1

the slope of the line BC (m2) = $$\dfrac{(1-5)}{(1-3)}$$

= $$\dfrac{-6}{-4}$$$$\dfrac{3}{2}$$

the slope of the line CA (m3) = $$\dfrac{(4+1)}{(4+1)}$$

$$\dfrac{5}{5}$$ = 1

It is observed that, m1.m3 = -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)

Answered by Pragya Singh | 11 months ago

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