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A line has a slope of 3/4 and intersects the point (-12, -39). At whic [#permalink]
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bitenca wrote:
A line has a slope of 3/4 and intersects the point (-12, -39). At which point does this line intersect the x-axis?

A. (40,0)
B. (30,0)
C. (0,40)
D. (40,30)
E. (0,30)

My solutuion was:

y=3/4x+b
-39=3/4(-12)+b
-39=-9+b
b=-30

But he asks for x-axis, I´m confused here...

Please, send me your solutions.

thanks
Solution by the book: (40,0)


One last point to make:

Even if we aren't sure how to actually answer the question, we can still recognize that for any point on the x-axis, the y-coordinate will equal 0.
This one fact allows us to ELIMINATE answer choices C, D and E, leaving us with a 50% chance of guessing the correct answer.

Here's a free video about x- and y-intercepts on the GMAT - https://www.gmatprepnow.com/module/gmat- ... /video/996

Cheers,
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Re: A line has a slope of 3/4 and intersects the point (-12, -39). At whic [#permalink]
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Hi bitenca,

This question can be solved in a couple of different ways. If you're comfortable with the 'concept' of 'slope', then you can answer this question with just a bit of arithmetic.

We're given some information about a line:
1) The line has a slope of 3/4
2) The line intersects the point (-12,-39)

We're asked at which point the line will intersect the X-axis. Since THAT co-ordinate will be (X, 0), we can use the slope to our advantage and 'count up' to that point:

With a slope of 3/4, we know that the line will "go up 3 points and go forward 4 points", meaning that the first few co-ordinates would be...

(-12,-39)
(-8, -36)
(-4, -33)
(0, -30)
Etc.

We can continue to 'count up' from here OR do a bit of algebra:

From the original co-ordinate, we know that we have to increase the '-39' to a '0' in increments of 3. That will clearly take 13 increments (since 13x3 = 39). Those same 13 increments would 'add 4' each time to the original X co-ordinate, turning '-12' into.....-12 + (13x4) = 40. Thus, the answer is (40,0).

Final Answer:

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Re: A line has a slope of 3/4 and intersects the point (-12, -39). At whic [#permalink]
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The equation of the slope is (y2-y1)/(x2-x1). Here we have (y2+39)/(x2+12)=(3/4) ==> 4y +156=3x+36 . We need to find x-interscetion i.e the point where y=0 Hence we have 3x=120 and x=40 Point of x-intersection is (40,0) Answer A
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Re: A line has a slope of 3/4 and intersects the point (-12, -39). At whic [#permalink]
--> Equation of line with slope m and passing through (x1,y1) = (y-y1)=m(x-x1)
So here m=3/4 and (x1,y1)=(-12,-39). So equation of line: (y+39)=3/4(x+12) ==> (y+39)=(3/4x+9)
Now point where this line cuts x-axis is where y=0.
So equation becomes: 0+39=3/4x+9 Therefore, x=40 (when y=0)

Answer(x,y)=(40,0) >>>> correct answer A
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Re: A line has a slope of 3/4 and intersects the point (-12, -39). At whic [#permalink]
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Since we are looking for point on x axis, y coordinate should be zero,
Thus we can eliminate option C, D, E
Looking at option A
40,0
slope = \(\frac{change in y}{change in x}\)= \(\frac{y_2-y_1}{x_2-x_1}\)
ie. slope = \(\frac{-39-0}{-12-40}\)=\(\frac{3}{4}\)
thus option A is answer
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Re: A line has a slope of 3/4 and intersects the point (-12, -39). At whic [#permalink]
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One can solve these type of questions easily if he/she knows who to convert given points into to slope equation. For example, the author gives two points (x,y) (x1,y1), and requires to turn these into y=mx+k form.
Here is a great post bu Bunuel , check it out! https://gmatclub.com/forum/math-coordin ... ml#p659245
For this particular problem, the author gives us the slope of the equation, as well as one coordinate, such as slope= 3/4 and the coordinate, is (-12,-39).
\(\frac{y-(-39)}{x-(-12)}\) =\(\frac{3}{4}\)
4(y+39)=3(x+12), so y=3x/4-30. To the the x intercept, we should plug in 0 for the y, and find out the coordinate, (40,0).
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