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Find the equations of the lines which pass through the point of intersection of the lines [tex]x + 2y = 12[/tex] and [tex]x - y = 0[/tex] and are 5 units from the point [tex](-2, 4)[/tex].

Sagot :

To find the equations of the lines that pass through the point of intersection of the given lines [tex]\(x + 2y = 12\)[/tex] and [tex]\(x - y = 0\)[/tex], and are 5 units away from the point [tex]\((-2, 4)\)[/tex], let's solve it step by step:

### Step 1: Find the Point of Intersection

The given equations of the lines are:
1. [tex]\(x + 2y = 12 \)[/tex]
2. [tex]\(x - y = 0 \)[/tex]

To find the point of intersection, solve the system of linear equations.

From equation (2), we can express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = y \][/tex]

Substitute [tex]\(x = y\)[/tex] into equation (1):
[tex]\[ y + 2y = 12 \][/tex]
[tex]\[ 3y = 12 \][/tex]
[tex]\[ y = 4 \][/tex]

Now substitute [tex]\(y = 4\)[/tex] back into [tex]\(x = y\)[/tex]:
[tex]\[ x = 4 \][/tex]

So, the point of intersection is [tex]\((4, 4)\)[/tex].

### Step 2: Determine the Distance from a Point to a Line

The distance [tex]\(D\)[/tex] from a point [tex]\((x_1, y_1)\)[/tex] to a line represented by [tex]\(ax + by + c = 0\)[/tex] is given by:
[tex]\[ D = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \][/tex]

### Step 3: Apply the Distance Formula

We need the lines that pass through the point [tex]\((4, 4)\)[/tex] and are 5 units away from the point [tex]\((-2, 4)\)[/tex].

Let's assume the equation of one such line is:
[tex]\[ x - y + c = 0 \][/tex]

(Here we use the form [tex]\(x - y + c = 0\)[/tex] as it will make the calculations simpler.)

Using the distance formula, set up the equation with the given distance:
[tex]\[ D = \frac{|(-2) - 4 + c|}{\sqrt{1^2 + (-1)^2}} = 5 \][/tex]

Simplify:
[tex]\[ 5 = \frac{|-6 + c|}{\sqrt{2}} \][/tex]
[tex]\[ 5\sqrt{2} = |-6 + c| \][/tex]

This gives two cases:
[tex]\[ -6 + c = 5\sqrt{2} \implies c = 5\sqrt{2} + 6 \][/tex]
or
[tex]\[ -6 + c = -5\sqrt{2} \implies c = -5\sqrt{2} + 6 \][/tex]

### Step 4: Form the Equations of the Lines

Substitute the values of [tex]\(c\)[/tex] back into the line equation.

For [tex]\(c = 5\sqrt{2} + 6\)[/tex]:
[tex]\[ x - y + 5\sqrt{2} + 6 = 0 \][/tex]

For [tex]\(c = -5\sqrt{2} + 6\)[/tex]:
[tex]\[ x - y - 5\sqrt{2} + 6 = 0 \][/tex]

### Final Equations of the Lines

The required equations of the lines are:
[tex]\[ x - y + 5\sqrt{2} + 6 = 0 \][/tex]
and
[tex]\[ x - y - 5\sqrt{2} + 6 = 0 \][/tex]

These lines pass through the intersection point [tex]\((4, 4)\)[/tex] and are 5 units away from the point [tex]\((-2, 4)\)[/tex].