QuestionDetermine if the lines $-3y=5x+2$ , $10x-6y=-2$ , and $y=-\frac{3}{5}x+1$ are parallel, perpendicular, or neither.

Studdy Solution

STEP 1

Assumptions1. The equations of the three lines are given as follows Line1 $-3y =5x +$ Line $10x -6y = -$ Line3 $y = -\frac{3}{5}x +1$ . We need to determine the relationship between each pair of lines Line1 and Line, Line1 and Line3, Line and Line3.
3. Two lines are parallel if their slopes are equal.
4. Two lines are perpendicular if the product of their slopes is -1.
5. If neither of these conditions are met, the lines are neither parallel nor perpendicular.

STEP 2

First, we need to rewrite each line in slope-intercept form, $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.
For Line1, we can rewrite the equation as follows $-y =5x +2 \Rightarrow y = -\frac{5}{}x - \frac{2}{}$

STEP 3

For Line2, we can rewrite the equation as follows $10x -6y = -2 \Rightarrow y = \frac{5}{3}x + \frac{1}{3}$

STEP 4

Line3 is already in slope-intercept form $y = -\frac{3}{}x +1$

STEP 5

Now, we can compare the slopes of each pair of lines to determine whether they are parallel, perpendicular, or neither.
For Line1 and Line2, the slopes are $-\frac{5}{3}$ and $\frac{5}{3}$ , respectively. The product of these slopes is $-\frac{5}{3} \times \frac{5}{3} = -1$ , so these lines are perpendicular.

STEP 6

For Line1 and Line3, the slopes are $-\frac{5}{3}$ and $-\frac{3}{5}$ , respectively. The product of these slopes is $-\frac{5}{3} \times -\frac{3}{5} =1$ , so these lines are neither parallel nor perpendicular.

STEP 7

For Line2 and Line3, the slopes are $\frac{5}{3}$ and $-\frac{3}{5}$ , respectively. The product of these slopes is $\frac{5}{3} \times -\frac{3}{5} = -1$ , so these lines are perpendicular.
In conclusion, Line1 and Line2 are perpendicular, Line1 and Line3 are neither parallel nor perpendicular, and Line2 and Line3 are perpendicular.

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QuestionDetermine if the lines $-3y=5x+2$ , $10x-6y=-2$ , and $y=-\frac{3}{5}x+1$ are parallel, perpendicular, or neither.

Studdy Solution

STEP 1

STEP 2

First, we need to rewrite each line in slope-intercept form, $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.
For Line1, we can rewrite the equation as follows $-y =5x +2 \Rightarrow y = -\frac{5}{}x - \frac{2}{}$

STEP 3

For Line2, we can rewrite the equation as follows $10x -6y = -2 \Rightarrow y = \frac{5}{3}x + \frac{1}{3}$

STEP 4

Line3 is already in slope-intercept form $y = -\frac{3}{}x +1$

STEP 5

STEP 6

For Line1 and Line3, the slopes are $-\frac{5}{3}$ and $-\frac{3}{5}$ , respectively. The product of these slopes is $-\frac{5}{3} \times -\frac{3}{5} =1$ , so these lines are neither parallel nor perpendicular.

STEP 7

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QuestionDetermine if the lines −3y=5x+2-3y=5x+2−3y=5x+2, 10x−6y=−210x-6y=-210x−6y=−2, and y=−35x+1y=-\frac{3}{5}x+1y=−53​x+1 are parallel, perpendicular, or neither.

Studdy Solution

STEP 1

STEP 2

First, we need to rewrite each line in slope-intercept form, y=mx+by = mx + by=mx+b, where mmm is the slope and bbb is the y-intercept.For Line1, we can rewrite the equation as follows−y=5x+2⇒y=−5x−2-y =5x +2 \Rightarrow y = -\frac{5}{}x - \frac{2}{}−y=5x+2⇒y=−5​x−2​

STEP 3

For Line2, we can rewrite the equation as follows10x−6y=−2⇒y=53x+1310x -6y = -2 \Rightarrow y = \frac{5}{3}x + \frac{1}{3}10x−6y=−2⇒y=35​x+31​

STEP 4

Line3 is already in slope-intercept formy=−3x+1y = -\frac{3}{}x +1y=−3​x+1

STEP 5

STEP 6

For Line1 and Line3, the slopes are −53-\frac{5}{3}−35​ and −35-\frac{3}{5}−53​, respectively. The product of these slopes is −53×−35=1-\frac{5}{3} \times -\frac{3}{5} =1−35​×−53​=1, so these lines are neither parallel nor perpendicular.

STEP 7

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QuestionDetermine if the lines $-3y=5x+2$ , $10x-6y=-2$ , and $y=-\frac{3}{5}x+1$ are parallel, perpendicular, or neither.

First, we need to rewrite each line in slope-intercept form, $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.
For Line1, we can rewrite the equation as follows $-y =5x +2 \Rightarrow y = -\frac{5}{}x - \frac{2}{}$

For Line2, we can rewrite the equation as follows $10x -6y = -2 \Rightarrow y = \frac{5}{3}x + \frac{1}{3}$

Line3 is already in slope-intercept form $y = -\frac{3}{}x +1$

For Line1 and Line3, the slopes are $-\frac{5}{3}$ and $-\frac{3}{5}$ , respectively. The product of these slopes is $-\frac{5}{3} \times -\frac{3}{5} =1$ , so these lines are neither parallel nor perpendicular.