4.4 Q-9

Question Statement

Find the coordinates of the triangle formed by the lines:

$x - 2y - 6 = 0$
$3x - y + 3 = 0$
$2x + y - 4 = 0$

Also, find the measures of the angles of the triangle formed by these lines.

Background and Explanation

To solve this, we need to find the points of intersection of the given lines. The points of intersection will form the vertices of the triangle. Once we have the coordinates of the vertices, we can proceed to calculate the angles of the triangle using the slopes of the sides.

Solution

We begin by solving the system of equations to find the points where the lines intersect.

Step 1: Find the coordinates of the first vertex (Intersection of lines 1 and 2)

From the equations of lines 1 and 2:

$x - 2y - 6 = 0$
$3x - y + 3 = 0$

Solve these equations simultaneously.

\frac{x}{-6-6} = \frac{-y}{3+18} = \frac{1}{-1+6}

This simplifies to:

\frac{x}{-12} = \frac{y}{-21} = \frac{1}{5}

Thus,

x = \frac{-12}{5}, \quad y = \frac{-21}{5}

So, the first vertex is $\left[ \frac{-12}{5}, \frac{-21}{5} \right]$ .

Step 2: Find the coordinates of the second vertex (Intersection of lines 2 and 3)

From the equations of lines 2 and 3:

$3x - y + 3 = 0$
$2x + y - 4 = 0$

Solve these equations simultaneously.

\frac{x}{4-3} = \frac{-y}{-12-6} = \frac{1}{3+2}

This simplifies to:

\frac{x}{1} = \frac{y}{18} = \frac{1}{5}

Thus,

x = \frac{1}{5}, \quad y = \frac{18}{5}

So, the second vertex is $\left[ \frac{1}{5}, \frac{18}{5} \right]$ .

Step 3: Find the coordinates of the third vertex (Intersection of lines 1 and 3)

From the equations of lines 1 and 3:

$x - 2y - 6 = 0$
$2x + y - 4 = 0$

Solve these equations simultaneously.

\frac{x}{8+6} = \frac{-y}{-4+12} = \frac{1}{1+4}

This simplifies to:

\frac{x}{14} = \frac{y}{-8} = \frac{1}{5}

Thus,

x = \frac{14}{5}, \quad y = \frac{-8}{5}

So, the third vertex is $\left[ \frac{14}{5}, \frac{-8}{5} \right]$ .

Step 4: Find the measures of the angles of the triangle

Now that we have the coordinates of the vertices:

$A\left[\frac{-12}{5}, \frac{-21}{5}\right]$
$B\left[\frac{1}{5}, \frac{18}{5}\right]$
$C\left[\frac{14}{5}, \frac{-8}{5}\right]$

We calculate the slopes of the sides of the triangle.

Slope of side AB:

m_1 = \frac{\frac{18}{5} - \frac{-21}{5}}{\frac{1}{5} - \frac{-12}{5}} = \frac{39}{13} = 3

Slope of side BC:

m_2 = \frac{\frac{-8}{5} - \frac{18}{5}}{\frac{14}{5} - \frac{1}{5}} = \frac{-26}{13} = -2

Slope of side AC:

m_3 = \frac{\frac{-8}{5} - \frac{-21}{5}}{\frac{14}{5} - \frac{-12}{5}} = \frac{13}{26} = \frac{1}{2}

Step 5: Calculate the angles

Now we can use the formula for the angle between two lines with slopes $m_1$ and $m_2$ :

\theta = \tan^{-1}\left( \frac{m_1 - m_2}{1 + m_1 m_2} \right)

Angle at A:

\angle A = \tan^{-1}\left( \frac{3 - \frac{1}{2}}{1 + 3 \times \frac{1}{2}} \right) = \tan^{-1}\left( \frac{6 - 1}{2 + 3} \right) = \tan^{-1}(1) = 45^\circ

Angle at B:

\angle B = \tan^{-1}\left( \frac{-2 - 3}{1 + (-2) \times 3} \right) = \tan^{-1}\left( \frac{-5}{1 - 6} \right) = \tan^{-1}(1) = 45^\circ

Angle at C:

\angle C = \tan^{-1}\left( \frac{\frac{1}{2} - (-2)}{1 + \frac{1}{2} \times (-2)} \right) = \tan^{-1}\left( \frac{\frac{1}{2} + 2}{1 - 1} \right)

This gives us:

\angle C = 90^\circ

Key Formulas or Methods Used

Point of intersection: Solving two linear equations simultaneously.
Slope of a line: $m = \frac{y_2 - y_1}{x_2 - x_1}$
Angle between two lines:

\theta = \tan^{-1}\left( \frac{m_1 - m_2}{1 + m_1 m_2} \right)

Summary of Steps

Solve the system of equations to find the points of intersection of the lines.
Use the intersection points to determine the coordinates of the vertices.
Calculate the slopes of the sides of the triangle.
Use the angle formula to find the measures of the angles.
Conclude that the triangle is an isosceles right-angled triangle with angles $45^\circ, 45^\circ, 90^\circ$ .