Answer :
To find the equations of the lines that are perpendicular to the given lines and pass through the point [tex]\((3,0)\)[/tex], let's proceed step-by-step for each given line equation:
### Step 1: Identify the slopes of the given lines
First, rewrite the given line equations into slope-intercept form [tex]\(y = mx + c\)[/tex] to identify the slopes.
1. [tex]\(3x + 5y = -9\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ 5y = -3x - 9 \implies y = -\frac{3}{5}x - \frac{9}{5} \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(-\frac{3}{5}\)[/tex].
2. [tex]\(3x + 5y = 9\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ 5y = -3x + 9 \implies y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(-\frac{3}{5}\)[/tex].
3. [tex]\(5x - 3y = -15\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ -3y = -5x - 15 \implies y = \frac{5}{3}x + 5 \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(\frac{5}{3}\)[/tex].
4. [tex]\(5x - 3y = 15\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ -3y = -5x + 15 \implies y = \frac{5}{3}x - 5 \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(\frac{5}{3}\)[/tex].
### Step 2: Find the slopes of the perpendicular lines
The slope of a line perpendicular to another is the negative reciprocal of the original line's slope.
1. For [tex]\(y = -\frac{3}{5}x + \ldots\)[/tex]:
[tex]\[ \text{Perpendicular slope} = \frac{5}{3} \][/tex]
2. For [tex]\(y = \frac{5}{3}x + \ldots\)[/tex]:
[tex]\[ \text{Perpendicular slope} = -\frac{3}{5} \][/tex]
### Step 3: Write the equations of the perpendicular lines passing through [tex]\((3, 0)\)[/tex]
Using the point-slope form [tex]\(y - y_1 = m(x - x_1)\)[/tex] and the given point [tex]\((3,0)\)[/tex]:
1. Line with perpendicular slope [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \implies y = \frac{5}{3}x - 5 \][/tex]
In standard form:
[tex]\[ 3y = 5x - 15 \implies 5x - 3y = 15 \][/tex]
2. Line with perpendicular slope [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \implies y = \frac{5}{3}x - 5 \][/tex]
In standard form:
[tex]\[ 3y = 5x - 15 \implies 5x - 3y = 15 \][/tex]
3. Line with perpendicular slope [tex]\(-\frac{3}{5}\)[/tex]:
[tex]\[ y - 0 = -\frac{3}{5}(x - 3) \implies y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
In standard form:
[tex]\[ 5y = -3x + 9 \implies 3x + 5y = 9 \][/tex]
4. Line with perpendicular slope [tex]\(-\frac{3}{5}\)[/tex]:
[tex]\[ y - 0 = -\frac{3}{5}(x - 3) \implies y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
In standard form:
[tex]\[ 5y = -3x + 9 \implies 3x + 5y = 9 \][/tex]
### Final Step: Combine the equations
The perpendicular lines passing through [tex]\((3,0)\)[/tex] are:
1. [tex]\(5x - 3y = 15\)[/tex]
2. [tex]\(5x - 3y = 15\)[/tex]
3. [tex]\(3x + 5y = 9\)[/tex]
4. [tex]\(3x + 5y = 9\)[/tex]
In more standardized forms:
- [tex]\(Eq(3x + 5y, 9)\)[/tex]
- [tex]\(Eq(3x + 5y, 9)\)[/tex]
- [tex]\(Eq(5x - 3y, 15)\)[/tex]
- [tex]\(Eq(5x - 3y, 15)\)[/tex]
So, the resulting equations as per the requirement are:
[tex]\[ (Eq(5y, \frac{25}{3}x - 25), Eq(5y, \frac{25}{3}x - 25), Eq(3y, \frac{27}{5} - \frac{9}{5}x), Eq(3y, \frac{27}{5} - \frac{9}{5}x)) \][/tex]
Thus, the perpendicular lines passing through the point [tex]\((3,0)\)[/tex] are given as these standard form equations:
[tex]\[ 3x + 5y = 9, \, 3x + 5y = 9, \, 5x - 3y = 15, \, 5x - 3y = 15 \][/tex]
### Step 1: Identify the slopes of the given lines
First, rewrite the given line equations into slope-intercept form [tex]\(y = mx + c\)[/tex] to identify the slopes.
1. [tex]\(3x + 5y = -9\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ 5y = -3x - 9 \implies y = -\frac{3}{5}x - \frac{9}{5} \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(-\frac{3}{5}\)[/tex].
2. [tex]\(3x + 5y = 9\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ 5y = -3x + 9 \implies y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(-\frac{3}{5}\)[/tex].
3. [tex]\(5x - 3y = -15\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ -3y = -5x - 15 \implies y = \frac{5}{3}x + 5 \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(\frac{5}{3}\)[/tex].
4. [tex]\(5x - 3y = 15\)[/tex]
Rewrite in slope-intercept form:
[tex]\[ -3y = -5x + 15 \implies y = \frac{5}{3}x - 5 \][/tex]
The slope ([tex]\(m\)[/tex]) is [tex]\(\frac{5}{3}\)[/tex].
### Step 2: Find the slopes of the perpendicular lines
The slope of a line perpendicular to another is the negative reciprocal of the original line's slope.
1. For [tex]\(y = -\frac{3}{5}x + \ldots\)[/tex]:
[tex]\[ \text{Perpendicular slope} = \frac{5}{3} \][/tex]
2. For [tex]\(y = \frac{5}{3}x + \ldots\)[/tex]:
[tex]\[ \text{Perpendicular slope} = -\frac{3}{5} \][/tex]
### Step 3: Write the equations of the perpendicular lines passing through [tex]\((3, 0)\)[/tex]
Using the point-slope form [tex]\(y - y_1 = m(x - x_1)\)[/tex] and the given point [tex]\((3,0)\)[/tex]:
1. Line with perpendicular slope [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \implies y = \frac{5}{3}x - 5 \][/tex]
In standard form:
[tex]\[ 3y = 5x - 15 \implies 5x - 3y = 15 \][/tex]
2. Line with perpendicular slope [tex]\(\frac{5}{3}\)[/tex]:
[tex]\[ y - 0 = \frac{5}{3}(x - 3) \implies y = \frac{5}{3}x - 5 \][/tex]
In standard form:
[tex]\[ 3y = 5x - 15 \implies 5x - 3y = 15 \][/tex]
3. Line with perpendicular slope [tex]\(-\frac{3}{5}\)[/tex]:
[tex]\[ y - 0 = -\frac{3}{5}(x - 3) \implies y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
In standard form:
[tex]\[ 5y = -3x + 9 \implies 3x + 5y = 9 \][/tex]
4. Line with perpendicular slope [tex]\(-\frac{3}{5}\)[/tex]:
[tex]\[ y - 0 = -\frac{3}{5}(x - 3) \implies y = -\frac{3}{5}x + \frac{9}{5} \][/tex]
In standard form:
[tex]\[ 5y = -3x + 9 \implies 3x + 5y = 9 \][/tex]
### Final Step: Combine the equations
The perpendicular lines passing through [tex]\((3,0)\)[/tex] are:
1. [tex]\(5x - 3y = 15\)[/tex]
2. [tex]\(5x - 3y = 15\)[/tex]
3. [tex]\(3x + 5y = 9\)[/tex]
4. [tex]\(3x + 5y = 9\)[/tex]
In more standardized forms:
- [tex]\(Eq(3x + 5y, 9)\)[/tex]
- [tex]\(Eq(3x + 5y, 9)\)[/tex]
- [tex]\(Eq(5x - 3y, 15)\)[/tex]
- [tex]\(Eq(5x - 3y, 15)\)[/tex]
So, the resulting equations as per the requirement are:
[tex]\[ (Eq(5y, \frac{25}{3}x - 25), Eq(5y, \frac{25}{3}x - 25), Eq(3y, \frac{27}{5} - \frac{9}{5}x), Eq(3y, \frac{27}{5} - \frac{9}{5}x)) \][/tex]
Thus, the perpendicular lines passing through the point [tex]\((3,0)\)[/tex] are given as these standard form equations:
[tex]\[ 3x + 5y = 9, \, 3x + 5y = 9, \, 5x - 3y = 15, \, 5x - 3y = 15 \][/tex]
