The image shows the calculation of the equation of a line that passes through the point $P(-1, 2)$ and is perpendicular to the line $L_1: 2x - y - 3 = 0$. Also, the image presents four pairs of linear equations and asks us to find the intersection point of each pair. We will solve the four systems of linear equations.

1. Problem Description

The image shows the calculation of the equation of a line that passes through the point $P(-1, 2)$ and is perpendicular to the line $L_1: 2x - y - 3 = 0$. Also, the image presents four pairs of linear equations and asks us to find the intersection point of each pair.

We will solve the four systems of linear equations.

2. Solution Steps

Let's analyze the solutions for each system:

1) $x - 2y - 4 = 0$ and $2x + 3y - 1 = 0$.

We can rewrite the first equation as $x = 2y + 4$. Substituting into the second equation:

$2(2y + 4) + 3y - 1 = 0$

$4y + 8 + 3y - 1 = 0$

$7y + 7 = 0$

$7y = -7$

$y = -1$

Then $x = 2(-1) + 4 = -2 + 4 = 2$.

The intersection point is $(2, -1)$.

2) $x + 5y - 15 = 0$ and $2x - y + 3 = 0$.

We can rewrite the first equation as $x = 15 - 5y$. Substituting into the second equation:

$2(15 - 5y) - y + 3 = 0$

$30 - 10y - y + 3 = 0$

$33 - 11y = 0$

$11y = 33$

$y = 3$

Then $x = 15 - 5(3) = 15 - 15 = 0$.

The intersection point is $(0, 3)$.

3) $x - 2y - 1 = 0$ and $3x + y - 3 = 0$.

From the first equation, $x = 2y + 1$. Substituting into the second equation:

$3(2y + 1) + y - 3 = 0$

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

$7y = 0$

$y = 0$

Then $x = 2(0) + 1 = 1$.

The intersection point is $(1, 0)$.

4) $2x + y - 6 = 0$ and $3x - y - 4 = 0$.

Adding the two equations:

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

$5x - 10 = 0$

$5x = 10$

$x = 2$

Substituting into the first equation:

$2(2) + y - 6 = 0$

$4 + y - 6 = 0$

$y - 2 = 0$

$y = 2$

The intersection point is $(2, 2)$.

3. Final Answer

1) The intersection point of $x - 2y - 4 = 0$ and $2x + 3y - 1 = 0$ is $(2, -1)$.

2) The intersection point of $x + 5y - 15 = 0$ and $2x - y + 3 = 0$ is $(0, 3)$.

3) The intersection point of $x - 2y - 1 = 0$ and $3x + y - 3 = 0$ is $(1, 0)$.

4) The intersection point of $2x + y - 6 = 0$ and $3x - y - 4 = 0$ is $(2, 2)$.