Question:

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

Updated On: Oct 26, 2023
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Solution and Explanation

The vertices of the given triangle are A (4, 4), B (3, 5), and C (-1, -1).

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by.

\(m = \frac {y_2-y_1}{x_2-x_1}, \ x_2≠x_1\)

∴ Slope of AB \((m_1) = \frac {5-4}{3-4} = -1\)

Slope of BC \((m_2) = \frac {-1-5}{-1-3}= \frac {-6}{-4} = \frac 32\)

Slope of CA \((m_3) =\frac { 4+1 }{ 4+1 }= \frac 55 = 1\)

It is observed that \(m_1m_3 = -1\)

This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (-1, -1) are the vertices of a right-angled triangle.

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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c