4.4 Q-1

Question Statement

Find the points of intersection of the following pairs of lines:

• a. $x - 2y + 1 = 0$ and $2x - y + 12 = 0$
• b. $3x + y + 12 = 0$ and $x + 2y - 1 = 0$
• c. $x + 4y - 12 = 0$ and $x - 3y + 3 = 0$

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Background and Explanation

To find the point of intersection of two lines, we need to:

1. Solve the two equations simultaneously (usually by substitution or elimination).
2. Check the slopes of the lines to ensure they are not parallel. Parallel lines will never intersect, while non-parallel lines will.

The slope of a line in the form $Ax + By + C = 0$ is given by:

$$m = -\frac{A}{B}$$

If the slopes of the two lines are different, the lines are not parallel, and they will intersect at a unique point.

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Solution

Part a: Intersection of $x - 2y + 1 = 0$ and $2x - y + 12 = 0$

Step 1: Write down the equations

$$x - 2y + 1 = 0 \quad \text{(1)}$$ $$2x - y + 12 = 0 \quad \text{(2)}$$

Step 2: Calculate the slopes of the lines

• Slope of Line (1):

$$m_1 = -\left( \frac{1}{2} \right) = \frac{1}{2}$$

• Slope of Line (2):

$$m_2 = -\left( \frac{-2}{1} \right) = 2$$

Since $m_1 \neq m_2$, the lines are not parallel and will intersect.

Step 3: Solve the system of equations

To solve, we can use substitution or elimination. Solving these equations gives:

$$x = -1, \quad y = 0$$

Thus, the point of intersection is $(-1, 0)$.

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Part b: Intersection of $3x + y + 12 = 0$ and $x + 2y - 1 = 0$

Step 1: Write down the equations

$$3x + y + 12 = 0 \quad \text{(1)}$$ $$x + 2y - 1 = 0 \quad \text{(2)}$$

Step 2: Calculate the slopes of the lines

• Slope of Line (1):

$$m_1 = -\left( \frac{-3}{1} \right) = -3$$

• Slope of Line (2):

$$m_2 = -\left( \frac{-1}{2} \right) = \frac{1}{2}$$

Since $m_1 \neq m_2$, the lines are not parallel and will intersect.

Step 3: Solve the system of equations

Solving the equations simultaneously gives:

$$x = -5, \quad y = 3$$

Thus, the point of intersection is $(-5, 3)$.

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Part c: Intersection of $x + 4y - 12 = 0$ and $x - 3y + 3 = 0$

Step 1: Write down the equations

$$x + 4y - 12 = 0 \quad \text{(1)}$$ $$x - 3y + 3 = 0 \quad \text{(2)}$$

Step 2: Calculate the slopes of the lines

• Slope of Line (1):

$$m_1 = -\left( \frac{1}{4} \right) = \frac{1}{4}$$

• Slope of Line (2):

$$m_2 = -\left( \frac{-1}{3} \right) = \frac{1}{3}$$

Since $m_1 \neq m_2$, the lines are not parallel and will intersect.

Step 3: Solve the system of equations

Solving these equations simultaneously gives:

$$x = \frac{24}{7}, \quad y = \frac{15}{7}$$

Thus, the point of intersection is $\left( \frac{24}{7}, \frac{15}{7} \right)$.

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Key Formulas or Methods Used

• Slope formula for a line in the form $Ax + By + C = 0$:

$$m = -\frac{A}{B}$$

• Solving systems of linear equations using substitution or elimination to find the intersection point.

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Summary of Steps

1. Write the equations of the two lines.
2. Calculate the slopes of the lines. If the slopes are different, the lines will intersect.
3. Solve the system of equations using substitution or elimination to find the coordinates of the point of intersection.
4. Repeat for each pair of lines to find their points of intersection.

For each part:

• a. The point of intersection is $(-1, 0)$.
• b. The point of intersection is $(-5, 3)$.
• c. The point of intersection is $\left( \frac{24}{7}, \frac{15}{7} \right)$.