QuestionDetermine the relationship (parallel, perpendicular, or neither) for the lines: $y=4x+5$ , $4x-16y=32$ , $y=-\frac{1}{4}x-2$ .

Studdy Solution

STEP 1

Assumptions1. The equations of the lines are given in the form $y = mx + c$ , where $m$ is the slope of the line and $c$ is the y-intercept. . Two lines are parallel if their slopes are equal.
3. Two lines are perpendicular if the product of their slopes is -1.
4. If neither of the above conditions are met, the lines are neither parallel nor perpendicular.

STEP 2

First, we need to find the slope of each line. For Line1, the slope is directly given in the equation as4.
$m1 =4$

STEP 3

For Line2, we need to rearrange the equation into the form $y = mx + c$ to find the slope. $x -16y =32 \Rightarrow y = \frac{1}{}x -2$ So, the slope of Line2 is $\frac{1}{}$ .
$m2 = \frac{1}{}$

STEP 4

For Line3, the slope is directly given in the equation as $-\frac{1}{4}$ .
$m3 = -\frac{1}{4}$

STEP 5

Now that we have the slopes of all three lines, we can compare them to determine if the lines are parallel, perpendicular, or neither.First, let's compare Line1 and Line2. Since $m1 \neq m2$ , Line1 and Line2 are not parallel.Since $m1 \times m2 =4 \times \frac{1}{4} =1 \neq -1$ , Line1 and Line2 are not perpendicular.Therefore, Line1 and Line2 are neither parallel nor perpendicular.

STEP 6

Next, let's compare Line1 and Line3. Since $m1 \neq m3$ , Line1 and Line3 are not parallel.Since $m1 \times m3 =4 \times -\frac{1}{4} = -1$ , Line1 and Line3 are perpendicular.

STEP 7

Finally, let's compare Line2 and Line3. Since $m2 = m3$ , Line2 and Line3 are parallel.Therefore, Line1 and Line2 are neither parallel nor perpendicular, Line1 and Line3 are perpendicular, and Line2 and Line3 are parallel.

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QuestionDetermine the relationship (parallel, perpendicular, or neither) for the lines: $y=4x+5$ , $4x-16y=32$ , $y=-\frac{1}{4}x-2$ .

Studdy Solution

STEP 1

STEP 2

First, we need to find the slope of each line. For Line1, the slope is directly given in the equation as4.
$m1 =4$

STEP 3

For Line2, we need to rearrange the equation into the form $y = mx + c$ to find the slope. $x -16y =32 \Rightarrow y = \frac{1}{}x -2$ So, the slope of Line2 is $\frac{1}{}$ .
$m2 = \frac{1}{}$

STEP 4

For Line3, the slope is directly given in the equation as $-\frac{1}{4}$ .
$m3 = -\frac{1}{4}$

STEP 5

STEP 6

Next, let's compare Line1 and Line3. Since $m1 \neq m3$ , Line1 and Line3 are not parallel.Since $m1 \times m3 =4 \times -\frac{1}{4} = -1$ , Line1 and Line3 are perpendicular.

STEP 7

Finally, let's compare Line2 and Line3. Since $m2 = m3$ , Line2 and Line3 are parallel.Therefore, Line1 and Line2 are neither parallel nor perpendicular, Line1 and Line3 are perpendicular, and Line2 and Line3 are parallel.

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QuestionDetermine the relationship (parallel, perpendicular, or neither) for the lines: y=4x+5y=4x+5y=4x+5, 4x−16y=324x-16y=324x−16y=32, y=−14x−2y=-\frac{1}{4}x-2y=−41​x−2.

Studdy Solution

STEP 1

STEP 2

First, we need to find the slope of each line. For Line1, the slope is directly given in the equation as4.m1=4m1 =4m1=4

STEP 3

For Line2, we need to rearrange the equation into the form y=mx+cy = mx + cy=mx+c to find the slope.x−16y=32⇒y=1x−2x -16y =32 \Rightarrow y = \frac{1}{}x -2x−16y=32⇒y=1​x−2So, the slope of Line2 is 1\frac{1}{}1​.m2=1m2 = \frac{1}{}m2=1​

STEP 4

For Line3, the slope is directly given in the equation as −14-\frac{1}{4}−41​.m3=−14m3 = -\frac{1}{4}m3=−41​

STEP 5

STEP 6

Next, let's compare Line1 and Line3. Since m1≠m3m1 \neq m3m1=m3, Line1 and Line3 are not parallel.Since m1×m3=4×−14=−1m1 \times m3 =4 \times -\frac{1}{4} = -1m1×m3=4×−41​=−1, Line1 and Line3 are perpendicular.

STEP 7

Finally, let's compare Line2 and Line3. Since m2=m3m2 = m3m2=m3, Line2 and Line3 are parallel.Therefore, Line1 and Line2 are neither parallel nor perpendicular, Line1 and Line3 are perpendicular, and Line2 and Line3 are parallel.

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QuestionDetermine the relationship (parallel, perpendicular, or neither) for the lines: $y=4x+5$ , $4x-16y=32$ , $y=-\frac{1}{4}x-2$ .

First, we need to find the slope of each line. For Line1, the slope is directly given in the equation as4.
$m1 =4$

For Line2, we need to rearrange the equation into the form $y = mx + c$ to find the slope. $x -16y =32 \Rightarrow y = \frac{1}{}x -2$ So, the slope of Line2 is $\frac{1}{}$ .
$m2 = \frac{1}{}$

For Line3, the slope is directly given in the equation as $-\frac{1}{4}$ .
$m3 = -\frac{1}{4}$

Next, let's compare Line1 and Line3. Since $m1 \neq m3$ , Line1 and Line3 are not parallel.Since $m1 \times m3 =4 \times -\frac{1}{4} = -1$ , Line1 and Line3 are perpendicular.

Finally, let's compare Line2 and Line3. Since $m2 = m3$ , Line2 and Line3 are parallel.Therefore, Line1 and Line2 are neither parallel nor perpendicular, Line1 and Line3 are perpendicular, and Line2 and Line3 are parallel.