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Finding the slope of a line, slightly confused by the answer
Calculating gradient of a line: how do you know which way to order the points?Using distance formula to find slope, any reason to use the concluding equation?Simplest way to calculate the width of a segment of a convex shapeThe sum of the abscissae of the intersections of a cubic and a lineFinding a point d distance away from another point only given a slopeHow to shift a line in a graph regardless of slope?Hint on writing a proof for slopePositive slope with negative valuesA line in the xy-plane contains the points (5, 4) and (2, –1)Understanding calculating the intercept C between two points
$begingroup$
(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
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.
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$endgroup$
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
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
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.
$endgroup$
(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
algebra-precalculus slope
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
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
edited Mar 14 at 19:48
Michael Rybkin
3,939420
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
asked Mar 14 at 8:10
Robert CRobert C
163
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.
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.
Robert C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$
frac1-84-(-13)=frac-74+13=-frac717
$$
answered Mar 14 at 8:18
Michael RybkinMichael Rybkin
3,939420
$endgroup$
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.
answered Mar 16 at 15:18
Roshan Klein-SeetharamanRoshan Klein-Seetharaman
617
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
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:
$$
frac1-84-(-13)=frac-74+13=-frac717
$$
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:
$$
frac1-84-(-13)=frac-74+13=-frac717
$$
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:
$$
frac1-84-(-13)=frac-74+13=-frac717
$$
answered Mar 14 at 8:18
Michael RybkinMichael Rybkin
3,939420
$endgroup$
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:
$$
frac1-84-(-13)=frac-74+13=-frac717
$$
answered Mar 14 at 8:18
Michael RybkinMichael Rybkin
3,939420
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
3,939420
add a comment |
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.
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.
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.
answered Mar 16 at 15:18
Roshan Klein-SeetharamanRoshan Klein-Seetharaman
617
$endgroup$
It’s like asking $ -Deltay$ over $-Deltax$ which will obviously give the same result as the change over that same positive interval.
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
answered Mar 16 at 15:18
Roshan Klein-SeetharamanRoshan Klein-Seetharaman
617
617
add a comment |
add a comment |
Robert C is a new contributor. Be nice, and check out our Code of Conduct.
Robert C is a new contributor. Be nice, and check out our Code of Conduct.
Robert C is a new contributor. Be nice, and check out our Code of Conduct.
Robert C is a new contributor. Be nice, and check out our Code of Conduct.
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