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### Problem 5: Find the equation of the line parallel to the line [tex]\(3x + 4y = 8\)[/tex] and at a distance of 5 units from the origin.
To solve this problem, follow these steps:
1. Identify the coefficients of the given line:
The given line [tex]\(3x + 4y = 8\)[/tex] can be written in the general form [tex]\(Ax + By = C\)[/tex], where [tex]\(A = 3\)[/tex], [tex]\(B = 4\)[/tex], and [tex]\(C = 8\)[/tex].
2. Form of the parallel line:
Any line parallel to [tex]\(3x + 4y = 8\)[/tex] will have the same coefficients for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Therefore, the general form of the lines parallel to [tex]\(3x + 4y = 8\)[/tex] will be [tex]\(3x + 4y = C\)[/tex], where [tex]\(C\)[/tex] is some constant.
3. Distance from the origin:
The formula for the distance [tex]\(p\)[/tex] from a point (in this case, the origin [tex]\((0, 0)\)[/tex]) to a line [tex]\(Ax + By + C = 0\)[/tex] is given by:
[tex]\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \][/tex]
Here, [tex]\(A = 3\)[/tex], [tex]\(B = 4\)[/tex], and the distance [tex]\(p = 5\)[/tex].
4. Calculate the constant [tex]\(C\)[/tex]:
Using the distance formula:
[tex]\[ 5 = \frac{|C|}{\sqrt{3^2 + 4^2}} = \frac{|C|}{\sqrt{9 + 16}} = \frac{|C|}{\sqrt{25}} = \frac{|C|}{5} \][/tex]
Solving for [tex]\(C\)[/tex]:
[tex]\[ |C| = 5 \times 5 = 25 \][/tex]
Thus, [tex]\(C\)[/tex] can be either [tex]\(25\)[/tex] or [tex]\(-25\)[/tex].
5. Form the equations:
Therefore, the two possible equations of the lines parallel to [tex]\(3x + 4y = 8\)[/tex] and at a distance of 5 units from the origin are:
[tex]\[ \boxed{3x + 4y = 25 \quad \text{and} \quad 3x + 4y = -25} \][/tex]
---
### Problem 6: If [tex]\(p\)[/tex] is the length of the perpendicular dropped from the origin on the line [tex]\(\frac{x}{a} + \frac{y}{b} = 1\)[/tex], prove that [tex]\(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}\)[/tex].
To prove this, follow these steps:
1. Standard form of the line equation:
Convert [tex]\(\frac{x}{a} + \frac{y}{b} = 1\)[/tex] into the standard form [tex]\(Ax + By = C\)[/tex].
Multiply through by [tex]\(ab\)[/tex]:
[tex]\[ bx + ay = ab \][/tex]
So, the line equation in standard form is:
[tex]\[ bx + ay = ab \][/tex]
Here, [tex]\(A = b\)[/tex], [tex]\(B = a\)[/tex], and [tex]\(C = ab\)[/tex].
2. Formula for perpendicular distance:
The perpendicular distance [tex]\(p\)[/tex] from the origin [tex]\((0, 0)\)[/tex] to the line [tex]\(Ax + By = C\)[/tex] is given by:
[tex]\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \][/tex]
Plug in [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
[tex]\[ p = \frac{|ab|}{\sqrt{b^2 + a^2}} \][/tex]
Since [tex]\(ab\)[/tex] is positive:
[tex]\[ p = \frac{ab}{\sqrt{b^2 + a^2}} \][/tex]
3. Square both sides:
[tex]\[ p^2 = \left(\frac{ab}{\sqrt{b^2 + a^2}}\right)^2 = \frac{(ab)^2}{b^2 + a^2} \][/tex]
Simplify:
[tex]\[ p^2 = \frac{a^2b^2}{a^2 + b^2} \][/tex]
4. Rearrange to prove the desired identity:
[tex]\[ \frac{1}{p^2} = \frac{a^2 + b^2}{a^2b^2} \][/tex]
[tex]\[ \frac{1}{p^2} = \frac{a^2}{a^2b^2} + \frac{b^2}{a^2b^2} \][/tex]
[tex]\[ \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2} \][/tex]
Rewriting, we obtain:
[tex]\[ \boxed{\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}} \][/tex]
To solve this problem, follow these steps:
1. Identify the coefficients of the given line:
The given line [tex]\(3x + 4y = 8\)[/tex] can be written in the general form [tex]\(Ax + By = C\)[/tex], where [tex]\(A = 3\)[/tex], [tex]\(B = 4\)[/tex], and [tex]\(C = 8\)[/tex].
2. Form of the parallel line:
Any line parallel to [tex]\(3x + 4y = 8\)[/tex] will have the same coefficients for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Therefore, the general form of the lines parallel to [tex]\(3x + 4y = 8\)[/tex] will be [tex]\(3x + 4y = C\)[/tex], where [tex]\(C\)[/tex] is some constant.
3. Distance from the origin:
The formula for the distance [tex]\(p\)[/tex] from a point (in this case, the origin [tex]\((0, 0)\)[/tex]) to a line [tex]\(Ax + By + C = 0\)[/tex] is given by:
[tex]\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \][/tex]
Here, [tex]\(A = 3\)[/tex], [tex]\(B = 4\)[/tex], and the distance [tex]\(p = 5\)[/tex].
4. Calculate the constant [tex]\(C\)[/tex]:
Using the distance formula:
[tex]\[ 5 = \frac{|C|}{\sqrt{3^2 + 4^2}} = \frac{|C|}{\sqrt{9 + 16}} = \frac{|C|}{\sqrt{25}} = \frac{|C|}{5} \][/tex]
Solving for [tex]\(C\)[/tex]:
[tex]\[ |C| = 5 \times 5 = 25 \][/tex]
Thus, [tex]\(C\)[/tex] can be either [tex]\(25\)[/tex] or [tex]\(-25\)[/tex].
5. Form the equations:
Therefore, the two possible equations of the lines parallel to [tex]\(3x + 4y = 8\)[/tex] and at a distance of 5 units from the origin are:
[tex]\[ \boxed{3x + 4y = 25 \quad \text{and} \quad 3x + 4y = -25} \][/tex]
---
### Problem 6: If [tex]\(p\)[/tex] is the length of the perpendicular dropped from the origin on the line [tex]\(\frac{x}{a} + \frac{y}{b} = 1\)[/tex], prove that [tex]\(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}\)[/tex].
To prove this, follow these steps:
1. Standard form of the line equation:
Convert [tex]\(\frac{x}{a} + \frac{y}{b} = 1\)[/tex] into the standard form [tex]\(Ax + By = C\)[/tex].
Multiply through by [tex]\(ab\)[/tex]:
[tex]\[ bx + ay = ab \][/tex]
So, the line equation in standard form is:
[tex]\[ bx + ay = ab \][/tex]
Here, [tex]\(A = b\)[/tex], [tex]\(B = a\)[/tex], and [tex]\(C = ab\)[/tex].
2. Formula for perpendicular distance:
The perpendicular distance [tex]\(p\)[/tex] from the origin [tex]\((0, 0)\)[/tex] to the line [tex]\(Ax + By = C\)[/tex] is given by:
[tex]\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \][/tex]
Plug in [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
[tex]\[ p = \frac{|ab|}{\sqrt{b^2 + a^2}} \][/tex]
Since [tex]\(ab\)[/tex] is positive:
[tex]\[ p = \frac{ab}{\sqrt{b^2 + a^2}} \][/tex]
3. Square both sides:
[tex]\[ p^2 = \left(\frac{ab}{\sqrt{b^2 + a^2}}\right)^2 = \frac{(ab)^2}{b^2 + a^2} \][/tex]
Simplify:
[tex]\[ p^2 = \frac{a^2b^2}{a^2 + b^2} \][/tex]
4. Rearrange to prove the desired identity:
[tex]\[ \frac{1}{p^2} = \frac{a^2 + b^2}{a^2b^2} \][/tex]
[tex]\[ \frac{1}{p^2} = \frac{a^2}{a^2b^2} + \frac{b^2}{a^2b^2} \][/tex]
[tex]\[ \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2} \][/tex]
Rewriting, we obtain:
[tex]\[ \boxed{\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}} \][/tex]
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