IDNLearn.com: Where questions are met with accurate and insightful answers. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.
Sagot :
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]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.