QuestionWhich coordinates for points $A^{\prime}$ and $B^{\prime}$ show that lines $AB$ and $A^{\prime}B^{\prime}$ are perpendicular?
1. $A^{\prime}:(p, m)$ and $B^{\prime}:(z, w)$
2. $A^{\prime}:(p, m)$ and $B^{\prime}:(z,-w)$
3. $A^{\prime}:(p,-m)$ and $B^{\prime}:(z, w)$
4. $A^{\prime}:(p,-m)$ and $B^{\prime}:(z,-w)$

Studdy Solution

STEP 1

Assumptions1. We are given two points A and B on a plane, but we do not know their coordinates. . We are given four options for the coordinates of points $A^{\prime}$ and $B^{\prime}$ .
3. We need to find which option would help prove that lines AB and $A^{\prime} B^{\prime}$ are perpendicular.
4. There are no additional constraints or conditions about the coordinates of points A, B, $A^{\prime}$ , and $B^{\prime}$ .

STEP 2

We know that two lines are perpendicular if the product of their slopes is -1.The slope of a line given two points $(x1, y1)$ and $(x2, y2)$ is calculated as $lope = \frac{y2 - y1}{x2 - x1}$

STEP 3

Let's denote the coordinates of points A and B as $(x1, y1)$ and $(x2, y2)$ respectively.The slope of line AB can be written as $lope_{AB} = \frac{y2 - y1}{x2 - x1}$

STEP 4

Now, let's calculate the slopes for the lines $A^{\prime} B^{\prime}$ for each of the given options and check if their product with $lope_{AB}$ is -1.
Option1 $A^{\prime}:(p, m)$ and $B^{\prime}:(z, w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{w - m}{z - p}$

STEP 5

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -1.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y2 - y1}{x2 - x1} \times \frac{w - m}{z - p}$ We cannot determine if this product is -1 without knowing the coordinates of points A and B.

STEP 6

Option2 $A^{\prime}:(p, m)$ and $B^{\prime}:(z,-w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{-w - m}{z - p}$

STEP 7

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -1.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y2 - y1}{x2 - x1} \times \frac{-w - m}{z - p}$ Again, we cannot determine if this product is -1 without knowing the coordinates of points A and B.

STEP 8

Option3 $A^{\prime}:(p,-m)$ and $B^{\prime}:(z, w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{w - (-m)}{z - p} = \frac{w + m}{z - p}$

STEP 9

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y2 - y}{x2 - x} \times \frac{w + m}{z - p}$ Again, we cannot determine if this product is - without knowing the coordinates of points A and B.

STEP 10

Option4 $A^{\prime}:(p,-m)$ and $B^{\prime}:(z,-w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{-w - (-m)}{z - p} = \frac{-w + m}{z - p}$

STEP 11

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y - y}{x - x} \times \frac{-w + m}{z - p}$ Again, we cannot determine if this product is - without knowing the coordinates of points A and B.

STEP 12

From the above steps, we can see that we cannot determine which option would help prove that lines AB and $A^{\prime} B^{\prime}$ are perpendicular without knowing the coordinates of points A and B. Therefore, we cannot solve this problem with the given information.

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Studdy Solution

STEP 1

STEP 2

We know that two lines are perpendicular if the product of their slopes is -1.The slope of a line given two points $(x1, y1)$ and $(x2, y2)$ is calculated as $lope = \frac{y2 - y1}{x2 - x1}$

STEP 3

Let's denote the coordinates of points A and B as $(x1, y1)$ and $(x2, y2)$ respectively.The slope of line AB can be written as $lope_{AB} = \frac{y2 - y1}{x2 - x1}$

STEP 4

Now, let's calculate the slopes for the lines $A^{\prime} B^{\prime}$ for each of the given options and check if their product with $lope_{AB}$ is -1.
Option1 $A^{\prime}:(p, m)$ and $B^{\prime}:(z, w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{w - m}{z - p}$

STEP 5

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -1.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y2 - y1}{x2 - x1} \times \frac{w - m}{z - p}$ We cannot determine if this product is -1 without knowing the coordinates of points A and B.

STEP 6

Option2 $A^{\prime}:(p, m)$ and $B^{\prime}:(z,-w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{-w - m}{z - p}$

STEP 7

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -1.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y2 - y1}{x2 - x1} \times \frac{-w - m}{z - p}$ Again, we cannot determine if this product is -1 without knowing the coordinates of points A and B.

STEP 8

Option3 $A^{\prime}:(p,-m)$ and $B^{\prime}:(z, w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{w - (-m)}{z - p} = \frac{w + m}{z - p}$

STEP 9

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y2 - y}{x2 - x} \times \frac{w + m}{z - p}$ Again, we cannot determine if this product is - without knowing the coordinates of points A and B.

STEP 10

Option4 $A^{\prime}:(p,-m)$ and $B^{\prime}:(z,-w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{-w - (-m)}{z - p} = \frac{-w + m}{z - p}$

STEP 11

Check if the product of $lope_{AB}$ and $lope_{A^{\prime} B^{\prime}}$ is -.
$lope_{AB} \times Slope_{A^{\prime} B^{\prime}} = \frac{y - y}{x - x} \times \frac{-w + m}{z - p}$ Again, we cannot determine if this product is - without knowing the coordinates of points A and B.

STEP 12

From the above steps, we can see that we cannot determine which option would help prove that lines AB and $A^{\prime} B^{\prime}$ are perpendicular without knowing the coordinates of points A and B. Therefore, we cannot solve this problem with the given information.

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Studdy Solution

STEP 1

STEP 2

We know that two lines are perpendicular if the product of their slopes is -1.The slope of a line given two points (x1,y1)(x1, y1)(x1,y1) and (x2,y2)(x2, y2)(x2,y2) is calculated aslope=y2−y1x2−x1lope = \frac{y2 - y1}{x2 - x1}lope=x2−x1y2−y1​

STEP 3

Let's denote the coordinates of points A and B as (x1,y1)(x1, y1)(x1,y1) and (x2,y2)(x2, y2)(x2,y2) respectively.The slope of line AB can be written aslopeAB=y2−y1x2−x1lope_{AB} = \frac{y2 - y1}{x2 - x1}lopeAB​=x2−x1y2−y1​

STEP 4

STEP 5

STEP 6

Option2 A′:(p,m)A^{\prime}:(p, m)A′:(p,m) and B′:(z,−w)B^{\prime}:(z,-w)B′:(z,−w)The slope of line A′B′A^{\prime} B^{\prime}A′B′ islopeA′B′=−w−mz−plope_{A^{\prime} B^{\prime}} = \frac{-w - m}{z - p}lopeA′B′​=z−p−w−m​

STEP 7

STEP 8

STEP 9

STEP 10

STEP 11

STEP 12

From the above steps, we can see that we cannot determine which option would help prove that lines AB and A′B′A^{\prime} B^{\prime}A′B′ are perpendicular without knowing the coordinates of points A and B. Therefore, we cannot solve this problem with the given information.

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

We know that two lines are perpendicular if the product of their slopes is -1.The slope of a line given two points $(x1, y1)$ and $(x2, y2)$ is calculated as $lope = \frac{y2 - y1}{x2 - x1}$

Let's denote the coordinates of points A and B as $(x1, y1)$ and $(x2, y2)$ respectively.The slope of line AB can be written as $lope_{AB} = \frac{y2 - y1}{x2 - x1}$

Option2 $A^{\prime}:(p, m)$ and $B^{\prime}:(z,-w)$ The slope of line $A^{\prime} B^{\prime}$ is $lope_{A^{\prime} B^{\prime}} = \frac{-w - m}{z - p}$

From the above steps, we can see that we cannot determine which option would help prove that lines AB and $A^{\prime} B^{\prime}$ are perpendicular without knowing the coordinates of points A and B. Therefore, we cannot solve this problem with the given information.