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Line k lies in the xy-plane. The x-intercept of line k is a

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Line k lies in the xy-plane. The x-intercept of line k is a [#permalink] New post 14 Dec 2015, 07:15
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Line k lies in the xy-plane. The x-intercept of line k is -4, and line k passes through the midpoint of the line segment whose endpoints are and (2, 9), and (2, 0). What is the slope of line k ?

Give your answer as a fraction.


[Reveal] Spoiler:
\(\frac{3}{4}\)



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Re: Line k lies in the xy-plane. The x-intercept of line k is a [#permalink] New post 14 Dec 2015, 07:41
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Solution


The coordinates of the first point are since the x-coordinate is and (−4, 0), −4 the y-coordinate of every point on the x-axis is 0. For the second point, the mid-point of the line segment is halfway between the endpoints and (2, 9) (2, 0). Thus, the midpoint has x-coordinate 2 and y-coordinate
\(\frac{9}{2}\) the number halfway between 9 and 0. Based on the coordinates (−4, 0) and \((2, \frac{9}{2})\), the slope of line k is


\(\frac{9}{2}- 0 / 2 - (- 4 )\)

\(=\frac{9}{2}/6\)\(=\frac{3}{4}\)
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Re: Line k lies in the xy-plane. The x-intercept of line k is a [#permalink] New post 13 Sep 2018, 09:36
Can someone explain this in simpler terms? How can one solve this using slope equation?
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Re: Line k lies in the xy-plane. The x-intercept of line k is a [#permalink] New post 14 Sep 2018, 15:53
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rajlal wrote:
Can someone explain this in simpler terms? How can one solve this using slope equation?


It is better not to solve this by using a standard line equation. The reason being a line equation is represented as:

\(y=mx+c\) where m is the slope and c is the y intercept. Here we have the x intercept.

Now we can rewrite the equation \(y=mx+c\) as \(x=\frac{1}{m}y+k\) where \(k=\frac{-c}{m}\). This value is given n the question as -4.
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Re: Line k lies in the xy-plane. The x-intercept of line k is a [#permalink] New post 15 Sep 2018, 01:40
sandy wrote:
rajlal wrote:
Can someone explain this in simpler terms? How can one solve this using slope equation?


It is better not to solve this by using a standard line equation. The reason being a line equation is represented as:

\(y=mx+c\) where m is the slope and c is the y intercept. Here we have the x intercept.

Now we can rewrite the equation \(y=mx+c\) as \(x=\frac{1}{m}y+k\) where \(k=\frac{-c}{m}\). This value is given n the question as -4.


I am sorry, but I don't understand your approach! How does this match the answer?
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Re: Line k lies in the xy-plane. The x-intercept of line k is a [#permalink] New post 15 Sep 2018, 12:06
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rajlal wrote:
I am sorry, but I don't understand your approach! How does this match the answer?


For you to solve the line using line equation \(y=mx+c\) you need to find the values of both m and c.

You can solve them by putting points (-4, 0) and the mid point of (2, 9), and (2, 0) into \(y=mx+c\).

However that would make it unecessarily long. You can directly solve for the slope using the metod presented in the solution post by Carcass.
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Re: Line k lies in the xy-plane. The x-intercept of line k is a   [#permalink] 15 Sep 2018, 12:06
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