Slope=-(Y intercept)/(X intercept)
=-3/2
gmat prep
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fskilnik@GMATH
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Hi there,
Let me give you a small summary on lines/slopes (just for lines in the plane xOy, for sure), because a folk asked me about this in private message...
Every line may be put on the form ax+by = c where a, b, c are real constants, the first two not simmultaneously zero.
This is the general form of a line (in a plane), and we have some important details on that:
> The line passes through the origin, that is, point (0,0) if and only if c is null.
> If a = 0 (then b is not), then y = c/b = constant, that means that the line is horizontal.
Important particular case: when c is zero, we have the "x axis", that may be written as (x,y) in the plane such that y = 0.
> If b = 0 (then a is not), then x = c/a = constant, that means that the line is vertical.
Important part. case: when c is zero, we have the "y axis", that may be written as (x,y) in the plane such that x = 0.
> If b is NOT null, that means the line may be put on the form (called reduced) y = mx +n, where m is the SLOPE of the line and n is the y-intersect ;
I repeat: slopes are defined if and only if the lines are non-vertical.
Slopes are the ratio of the variation on y over the variation on x ("raise over run", as some people say)...
When the slope is (say) 4, that means that going to the RIGHT (say) 1 unit of length, the line goes UP 4 units of length...
When the slope is (say) -0.3, that means that going to the RIGHT (say) 1 unit of length, the line goes DOWN 0.3 units of length...
As far as slopes are concerned, two important properties you should know:
1)
when the slope is positive, the line is "going upwards" (from left to right) ;
when the slope is null (zero), the line is horizontal
when the slope is negative, the line is "going downwards" (from left to right);
P.S.: put your LEFT hand up and down, to "see" slopes positives and negatives....
2)
when the MODULUS of the slope is between 0 and 1, the line is horizontalized, I mean, it´s more "near" horizontal than "near" vertical, considering that slopes 1 and -1 are the "equal" horizontal and vertical (lines with slopes 1 and -1 are parallel to the lines y = x and y = -x...)
when the MODULUS of the slope is greater than 1, the line is verticalized, I mean, it´s more "near" vertical than "near" horizontal, considering that slopes 1 and -1 are... (the same above)
I hope this helps!
As far as the problem itself is concerned, I will solve it below if someone hasn´t already done this while I was typing....
Regards,
Fabio.
Let me give you a small summary on lines/slopes (just for lines in the plane xOy, for sure), because a folk asked me about this in private message...
Every line may be put on the form ax+by = c where a, b, c are real constants, the first two not simmultaneously zero.
This is the general form of a line (in a plane), and we have some important details on that:
> The line passes through the origin, that is, point (0,0) if and only if c is null.
> If a = 0 (then b is not), then y = c/b = constant, that means that the line is horizontal.
Important particular case: when c is zero, we have the "x axis", that may be written as (x,y) in the plane such that y = 0.
> If b = 0 (then a is not), then x = c/a = constant, that means that the line is vertical.
Important part. case: when c is zero, we have the "y axis", that may be written as (x,y) in the plane such that x = 0.
> If b is NOT null, that means the line may be put on the form (called reduced) y = mx +n, where m is the SLOPE of the line and n is the y-intersect ;
I repeat: slopes are defined if and only if the lines are non-vertical.
Slopes are the ratio of the variation on y over the variation on x ("raise over run", as some people say)...
When the slope is (say) 4, that means that going to the RIGHT (say) 1 unit of length, the line goes UP 4 units of length...
When the slope is (say) -0.3, that means that going to the RIGHT (say) 1 unit of length, the line goes DOWN 0.3 units of length...
As far as slopes are concerned, two important properties you should know:
1)
when the slope is positive, the line is "going upwards" (from left to right) ;
when the slope is null (zero), the line is horizontal
when the slope is negative, the line is "going downwards" (from left to right);
P.S.: put your LEFT hand up and down, to "see" slopes positives and negatives....
2)
when the MODULUS of the slope is between 0 and 1, the line is horizontalized, I mean, it´s more "near" horizontal than "near" vertical, considering that slopes 1 and -1 are the "equal" horizontal and vertical (lines with slopes 1 and -1 are parallel to the lines y = x and y = -x...)
when the MODULUS of the slope is greater than 1, the line is verticalized, I mean, it´s more "near" vertical than "near" horizontal, considering that slopes 1 and -1 are... (the same above)
I hope this helps!
As far as the problem itself is concerned, I will solve it below if someone hasn´t already done this while I was typing....
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
As promised:
Obs.: the examiner asks for the slope, that means that we may admit (as hypothesis) that the line k is certainly NOT VERTICAL!
Answer: C
(1) BIFURCATES:
> If the line is (say) y = x-2, then (when y=0 we know that x = 2, so its x-intercept is really 2 and) the slope is 1
> If the line is (say) y = 2x-4, then (when y=0 we know that x = 2, so its x-intercept is really 2 and) the slope is 2
Obs.: you may consider the line given is different from the x-axis... otherwise the x-intercept would not be "only" 2... think about it...
(2) BIFURCATES:
> If the line is (say) y = x+3, then (when x=0 we know that y = 3, so its y-intercept is really 3 and) the slope is 1
> If the line is (say) y = 2x+3, then (when x=0 we know that y = 3, so its y-intercept is really 3 and) the slope is 2
(1+2) DECIDES!
Please note that points (0,3) and (2,0) are two points of the line k, therefore when x varies (runs) from 0 to 2 (+2) then y varies (raises) from 3 to 0 (IN THAT ORDER, so -3), hence the slope is really -3/2, as selango mentioned.
Regards,
Fabio.
Obs.: the examiner asks for the slope, that means that we may admit (as hypothesis) that the line k is certainly NOT VERTICAL!
Answer: C
(1) BIFURCATES:
> If the line is (say) y = x-2, then (when y=0 we know that x = 2, so its x-intercept is really 2 and) the slope is 1
> If the line is (say) y = 2x-4, then (when y=0 we know that x = 2, so its x-intercept is really 2 and) the slope is 2
Obs.: you may consider the line given is different from the x-axis... otherwise the x-intercept would not be "only" 2... think about it...
(2) BIFURCATES:
> If the line is (say) y = x+3, then (when x=0 we know that y = 3, so its y-intercept is really 3 and) the slope is 1
> If the line is (say) y = 2x+3, then (when x=0 we know that y = 3, so its y-intercept is really 3 and) the slope is 2
(1+2) DECIDES!
Please note that points (0,3) and (2,0) are two points of the line k, therefore when x varies (runs) from 0 to 2 (+2) then y varies (raises) from 3 to 0 (IN THAT ORDER, so -3), hence the slope is really -3/2, as selango mentioned.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
