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 (x_{1}, y_{1}) and

(x_{2}, y_{2}) is given by m =

\(
\dfrac{ (y_2 – y_1)}{(x_2 – x_1)}\) where, x ≠ x_{1}

So, the slope of the line AB (m_{1})

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

= \(\dfrac{1}{-1}\) = -1

the slope of the line BC (m_{2}) = \(
\dfrac{(1-5)}{(1-3)}\)

= \( \dfrac{-6}{-4}\)= \( \dfrac{3}{2}\)

the slope of the line CA (m_{3}) = \(
\dfrac{(4+1)}{(4+1)}\)

= \( \dfrac{5}{5}\) = 1

It is observed that, 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)

Answered by Pragya Singh | 11 months agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the angles between pairs of straight lines 3x – y + 5 = 0 and x – 3y + 1 = 0

Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0