IDNLearn.com makes it easy to find precise answers to your specific questions. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.
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].
### 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].
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.