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Sagot :
To determine if each line is parallel, perpendicular, or neither parallel nor perpendicular to the line [tex]\( -3x + 5y = -15 \)[/tex], we need to analyze the slopes of these lines.
1. The first line to compare is:
[tex]\( -3x + 5y = 15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 5y = 3x + 15 \Rightarrow y = \frac{3}{5}x + 3 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( \frac{3}{5} \)[/tex].
Comparing this with the reference line [tex]\( -3x + 5y = -15 \)[/tex]:
- The reference line can be written as:
[tex]\[ 5y = 3x - 15 \Rightarrow y = \frac{3}{5}x - 3 \][/tex]
- The slope is also [tex]\( \frac{3}{5} \)[/tex].
Since the slopes are equal, the lines are parallel.
2. The second line to compare is:
[tex]\( 5x + 3y = 15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 3y = -5x + 15 \Rightarrow y = -\frac{5}{3}x + 5 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{5}{3} \)[/tex].
Comparing this with the reference line:
- The reference line has slope [tex]\( \frac{3}{5} \)[/tex].
If we multiply these slopes ([tex]\( \frac{3}{5} \times -\frac{5}{3} \)[/tex]), we get:
[tex]\[ \frac{3}{5} \times -\frac{5}{3} = -1 \][/tex]
Since the product of the slopes is [tex]\(-1\)[/tex], the lines are perpendicular.
3. The third line to compare is:
[tex]\( 3x + 5y = 15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 5y = -3x + 15 \Rightarrow y = -\frac{3}{5}x + 3 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{3}{5} \)[/tex].
Comparing this with the reference line:
- The reference line has slope [tex]\( \frac{3}{5} \)[/tex].
The slopes [tex]\(\frac{3}{5} \neq -\frac{3}{5}\)[/tex] and their product is not [tex]\(-1\)[/tex], indicating the lines are neither parallel nor perpendicular.
4. The fourth line to compare is:
[tex]\( 3x + 5y = -15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 5y = -3x - 15 \Rightarrow y = -\frac{3}{5}x - 3 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{3}{5} \)[/tex].
Comparing this with the reference line:
- The reference line slope is [tex]\( \frac{3}{5} \)[/tex].
The slopes [tex]\(\frac{3}{5} \neq -\frac{3}{5}\)[/tex], and their product is not [tex]\(-1\)[/tex], indicating the lines are neither parallel nor perpendicular.
So, the table should be completed as follows:
- Parallel: [tex]\( -3x + 5y = 15 \)[/tex]
- Perpendicular: [tex]\( 5x + 3y = 15 \)[/tex]
- Neither: [tex]\( 3x + 5y = 15 \)[/tex]
- Neither: [tex]\( 3x + 5y = -15 \)[/tex]
1. The first line to compare is:
[tex]\( -3x + 5y = 15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 5y = 3x + 15 \Rightarrow y = \frac{3}{5}x + 3 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( \frac{3}{5} \)[/tex].
Comparing this with the reference line [tex]\( -3x + 5y = -15 \)[/tex]:
- The reference line can be written as:
[tex]\[ 5y = 3x - 15 \Rightarrow y = \frac{3}{5}x - 3 \][/tex]
- The slope is also [tex]\( \frac{3}{5} \)[/tex].
Since the slopes are equal, the lines are parallel.
2. The second line to compare is:
[tex]\( 5x + 3y = 15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 3y = -5x + 15 \Rightarrow y = -\frac{5}{3}x + 5 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{5}{3} \)[/tex].
Comparing this with the reference line:
- The reference line has slope [tex]\( \frac{3}{5} \)[/tex].
If we multiply these slopes ([tex]\( \frac{3}{5} \times -\frac{5}{3} \)[/tex]), we get:
[tex]\[ \frac{3}{5} \times -\frac{5}{3} = -1 \][/tex]
Since the product of the slopes is [tex]\(-1\)[/tex], the lines are perpendicular.
3. The third line to compare is:
[tex]\( 3x + 5y = 15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 5y = -3x + 15 \Rightarrow y = -\frac{3}{5}x + 3 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{3}{5} \)[/tex].
Comparing this with the reference line:
- The reference line has slope [tex]\( \frac{3}{5} \)[/tex].
The slopes [tex]\(\frac{3}{5} \neq -\frac{3}{5}\)[/tex] and their product is not [tex]\(-1\)[/tex], indicating the lines are neither parallel nor perpendicular.
4. The fourth line to compare is:
[tex]\( 3x + 5y = -15 \)[/tex]
- To convert this to slope-intercept form [tex]\( y = mx + c \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ 5y = -3x - 15 \Rightarrow y = -\frac{3}{5}x - 3 \][/tex]
- The slope ([tex]\( m \)[/tex]) is [tex]\( -\frac{3}{5} \)[/tex].
Comparing this with the reference line:
- The reference line slope is [tex]\( \frac{3}{5} \)[/tex].
The slopes [tex]\(\frac{3}{5} \neq -\frac{3}{5}\)[/tex], and their product is not [tex]\(-1\)[/tex], indicating the lines are neither parallel nor perpendicular.
So, the table should be completed as follows:
- Parallel: [tex]\( -3x + 5y = 15 \)[/tex]
- Perpendicular: [tex]\( 5x + 3y = 15 \)[/tex]
- Neither: [tex]\( 3x + 5y = 15 \)[/tex]
- Neither: [tex]\( 3x + 5y = -15 \)[/tex]
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