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2

$begingroup$

(just for context, this is from a study booklet for a military test. I haven't done algebra in about 10 years. I googled around but was having trouble finding specific information about the below.)

I was always taught that the equation to find the slop was $dfracy_2-y_1x_2-x_1$, so I was very surprised to see the answer was done via $dfracy_1-y_2x_1-x_2$ (never seen it done this way before). Is this typical/accepted way of finding the slope? Also was surprised that $dfrac-717 = dfrac7-17$, but I guess I can see why that would be since the slope is going to be negative either way when you divide the numbers?

algebra-precalculus slope

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edited Mar 14 at 19:48

Michael Rybkin

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asked Mar 14 at 8:10

Robert CRobert C

163

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2

$begingroup$

(just for context, this is from a study booklet for a military test. I haven't done algebra in about 10 years. I googled around but was having trouble finding specific information about the below.)

I was always taught that the equation to find the slop was $dfracy_2-y_1x_2-x_1$, so I was very surprised to see the answer was done via $dfracy_1-y_2x_1-x_2$ (never seen it done this way before). Is this typical/accepted way of finding the slope? Also was surprised that $dfrac-717 = dfrac7-17$, but I guess I can see why that would be since the slope is going to be negative either way when you divide the numbers?

algebra-precalculus slope

share|cite|improve this question

edited Mar 14 at 19:48

Michael Rybkin

3,939420

asked Mar 14 at 8:10

Robert CRobert C

163

New contributor

Robert C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

(just for context, this is from a study booklet for a military test. I haven't done algebra in about 10 years. I googled around but was having trouble finding specific information about the below.)

I was always taught that the equation to find the slop was $dfracy_2-y_1x_2-x_1$, so I was very surprised to see the answer was done via $dfracy_1-y_2x_1-x_2$ (never seen it done this way before). Is this typical/accepted way of finding the slope? Also was surprised that $dfrac-717 = dfrac7-17$, but I guess I can see why that would be since the slope is going to be negative either way when you divide the numbers?

algebra-precalculus slope

share|cite|improve this question

edited Mar 14 at 19:48

Michael Rybkin

3,939420

asked Mar 14 at 8:10

Robert CRobert C

163

New contributor

Robert C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 14 at 19:48

Michael Rybkin

3,939420

asked Mar 14 at 8:10

Robert CRobert C

163

New contributor

Robert C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 14 at 19:48

Michael Rybkin

3,939420

asked Mar 14 at 8:10

Robert CRobert C

163

New contributor

Robert C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 14 at 19:48

Michael Rybkin

3,939420

edited Mar 14 at 19:48

Michael Rybkin

3,939420

asked Mar 14 at 8:10

Robert CRobert C

163

New contributor

Robert C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 14 at 8:10

Robert CRobert C

163

2 Answers
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0

$begingroup$

Algebraically, those formulas are the same:
$$
fracy_1-y_2x_1-x_2=frac-(-y_1+y_2)-(-x_1+x_2)=
frac-y_1+y_2-x_1+x_2=
fracy_2-y_1x_2-x_1
$$

The minus signs just cancel each other out. And here's what your slope is going to be:

$$
frac1-84-(-13)=frac-74+13=-frac717
$$

share|cite|improve this answer

edited Mar 14 at 8:24

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

$endgroup$

add a comment | 

0

$begingroup$

It’s like asking $ -Deltay$ over $-Deltax$ which will obviously give the same result as the change over that same positive interval.

share|cite|improve this answer

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617

$endgroup$

add a comment | 

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0

$begingroup$

Algebraically, those formulas are the same:
$$
fracy_1-y_2x_1-x_2=frac-(-y_1+y_2)-(-x_1+x_2)=
frac-y_1+y_2-x_1+x_2=
fracy_2-y_1x_2-x_1
$$

The minus signs just cancel each other out. And here's what your slope is going to be:

$$
frac1-84-(-13)=frac-74+13=-frac717
$$

share|cite|improve this answer

edited Mar 14 at 8:24

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

$endgroup$

add a comment | 

0

$begingroup$

Algebraically, those formulas are the same:
$$
fracy_1-y_2x_1-x_2=frac-(-y_1+y_2)-(-x_1+x_2)=
frac-y_1+y_2-x_1+x_2=
fracy_2-y_1x_2-x_1
$$

The minus signs just cancel each other out. And here's what your slope is going to be:

$$
frac1-84-(-13)=frac-74+13=-frac717
$$

share|cite|improve this answer

edited Mar 14 at 8:24

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

$endgroup$

add a comment | 

$begingroup$

Algebraically, those formulas are the same:
$$
fracy_1-y_2x_1-x_2=frac-(-y_1+y_2)-(-x_1+x_2)=
frac-y_1+y_2-x_1+x_2=
fracy_2-y_1x_2-x_1
$$

The minus signs just cancel each other out. And here's what your slope is going to be:

$$
frac1-84-(-13)=frac-74+13=-frac717
$$

share|cite|improve this answer

edited Mar 14 at 8:24

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

$endgroup$

share|cite|improve this answer

edited Mar 14 at 8:24

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

answered Mar 14 at 8:18

Michael RybkinMichael Rybkin

3,939420

0

$begingroup$

It’s like asking $ -Deltay$ over $-Deltax$ which will obviously give the same result as the change over that same positive interval.

share|cite|improve this answer

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617

$endgroup$

add a comment | 

0

$begingroup$

It’s like asking $ -Deltay$ over $-Deltax$ which will obviously give the same result as the change over that same positive interval.

share|cite|improve this answer

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617

$endgroup$

add a comment | 

$begingroup$

It’s like asking $ -Deltay$ over $-Deltax$ which will obviously give the same result as the change over that same positive interval.

share|cite|improve this answer

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617

$endgroup$

share|cite|improve this answer

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617

answered Mar 16 at 15:18

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

617