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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].
### 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].
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