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Sajjad1994
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In the figure to the right, if the equation of the line is y = px + a, 03 Jun 2017, 10:17
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C
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74% (01:00) correct 26% (00:57) wrong based on 99 sessions

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In the figure below, if the equation of the line is y = px + a, then p =

(A) b/a
(B) −b/a
(C) −a/b
(D) a/b
(E) there is not enough information to decide

Source: Nova GMAT
Difficulty Level: 550
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C

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In the figure to the right, if the equation of the line is y = px + a, 03 Jun 2017, 10:58
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When we write a line like this: y = px + a .. in that case, 'p' is the slope and 'a' is the y-intercept of this line.
Slope of any line can be calculated by taking the difference of y coordinates and dividing it by the difference of x coordinates.

We already have two points on this line given as: (0,a) and (b,0). Hence slope p = (0-a)/(b-0) = -a/b

Hence C answer
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In the figure to the right, if the equation of the line is y = px + a, 04 Jun 2017, 00:50
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y=px+a
Slope of line(m) =p

m= y2-y1/x2-x1
p=0-a/(b-0)=-a/b

Answer c

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In the figure to the right, if the equation of the line is y = px + a, 08 Jun 2017, 05:26
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Curious to know why cant we take (b,0) as (x1,y1) and (0,a) as (x2,y2). So the formula will become
m = (y2-y1)/(x2-x1) = a/b
Is it becoz the equation is given as y=px+a we consider (x1,y1) as (0,a) and (x2,y2) as (b,0).? Just trying to clarify the basics since the wrong answer will always be present in the choices given.
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In the figure to the right, if the equation of the line is y = px + a, 08 Jun 2017, 10:58
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sarugiri
Curious to know why cant we take (b,0) as (x1,y1) and (0,a) as (x2,y2). So the formula will become
m = (y2-y1)/(x2-x1) = a/b
Is it becoz the equation is given as y=px+a we consider (x1,y1) as (0,a) and (x2,y2) as (b,0).? Just trying to clarify the basics since the wrong answer will always be present in the choices given.
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Hi

It doesn't matter whether you take (x1,y1) as (0,a) or (b,0). Slope is going to be the same in either case.
Its upto you which point you take as (x1,y1) and which point you take as (x2,y2):
If you apply the formula of slope (y2-y1)/(x2-x1) correctly, you will get the same answer.

Let me show you:

Case 1. Lets take (0,a) as (x1,y1) and (b,0) as (x2,y2).
Then Slope = (y2-y1)/(x2-x1) = (0-a)/(b-0) = -a/b. Here the slope is -a/b

Case 2. Lets take (b,0) as (x1,y1) and (0,a) as (x2,y2).
Then Slope = (y2-y1)/(x2-x1) = (a-0)/(0-b) = -a/b. Here ALSO the slope is -a/b

Hope this helps
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In the figure to the right, if the equation of the line is y = px + a, 10 Jun 2017, 06:27
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In the figure to the right, if the equation of the line is y = px + a, 10 Jun 2017, 08:04
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SajjadAhmad
In the figure below, if the equation of the line is y = px + a, then p =

(A) b/a
(B) −b/a
(C) −a/b
(D) a/b
(E) there is not enough information to decide
Show more

Recall that the slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept of the line. We are given that the equation of a line is y = px + a, so we see that the slope = p. Since points (0,a) and (b,0) are on the line, we can determine the slope, which is p.

slope = change in y/change in x

slope = (0 - a)/(b - 0) = -a/b

Answer: C
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In the figure to the right, if the equation of the line is y = px + a, 13 Jun 2017, 08:47
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SajjadAhmad
In the figure below, if the equation of the line is y = px + a, then p =

(A) b/a
(B) −b/a
(C) −a/b
(D) a/b
(E) there is not enough information to decide
Show more

Recall that the slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept of the line. We are given that the equation of a line is y = px + a, so we see that the slope = p. Since points (0,a) and (b,0) are on the line, we can determine the slope, which is p.

slope = change in y/change in x

slope = (0 - a)/(b - 0) = -a/b

Answer: C
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