In the xy-plane, if line k has negative slope and passes through the

SOLUTION

In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

This question can be done with graphic approach (just by drawing the lines) or with algebraic approach.

Algebraic approach:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)), and \(x\) is the independent variable of the function \(y\).

We are told that slope of line \(k\) is negative (\(m<0\)) and it passes through the point (-5,r): \(y=mx+b\) --> \(r=-5m+b\).

Question: is x-intercept of line \(k\) positive? x-intercep is the value of \(x\) for \(y=0\) --> \(0=mx+b\) --> is \(x=-\frac{b}{m}>0\)? As we know that \(m<0\), then the question basically becomes: is \(b>0\)?.

(1) The slope of line \(k\) is -5 --> \(m=-5<0\). We've already known that slope was negative and there is no info about \(b\), hence this statement is insufficient.

(2) \(r>0\) --> \(r=-5m+b>0\) --> \(b>5m=some \ negative \ number\), as \(m<0\) we have that \(b\) is more than some negative number (\(5m\)), hence insufficient, to say whether \(b>0\).

(1)+(2) From (1) \(m=-5\) and from (2) \(r=-5m+b>0\) --> \(r=-5m+b=25+b>0\) --> \(b>-25\). Not sufficient to say whether \(b>0\).

Answer: E.

Graphic approach:

If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

When we take both statement together all we know is that slope is negative and that it crosses some point in II quadrant (-5, r>0) (this info is redundant as we know that if the slope of the line is negative, the line WILL intersect quadrants II). Basically we just know that the slope is negative - that's all. We can not say whether x-intercept is positive or negative from this info.

Below are two graphs with positive and negative x-intercepts. Statements that the slope=-5 and that the line crosses (-5, r>0) are satisfied.

\(y=-5x+5\):


\(y=-5x-20\):


For more on Coordinate Geometry check: https://gmatclub.com/forum/math-coordina ... 87652.html


Answer: E.


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I will not repeat the solutions above because I did the same way.

I will just repeat the most important takeaway: When slope is negative then both intercepts will have equal sign but when the slope is positive the intercepts will have opposite signs. Remember this and bully your way through these type of questions.

TIPS ON SLOPE AND QUADRANTS:

1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.

3. Every line (but the one crosses origin OR parallel to X or Y axis OR X and Y axis themselves) crosses three quadrants. Only the line which crosses origin \((0,0)\) OR is parallel to either of axis crosses only two quadrants.

4. If a line is horizontal it has a slope of \(0\), is parallel to X-axis and crosses quadrant I and II if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative. Equation of such line is y=b, where b is y intersect.

5. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses quadrant I and IV, if the X intersect is positive and quadrant II and III, if the X intersect is negative. Equation of such line is \(x=a\), where a is x-intercept.

6. For a line that crosses two points \((x_1,y_1)\) and \((x_2,y_2)\), slope \(m=\frac{y_2-y_1}{x_2-x_1}\)

7. If the slope is 1 the angle formed by the line is \(45\) degrees.

8. Given a point and slope, equation of a line can be found. The equation of a straight line that passes through a point \((x_1, y_1)\) with a slope \(m\) is: \(y - y_1 = m(x - x_1)\)

Check for more here: math-coordinate-geometry-87652.html

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i solved like this-

line passes thru (-5,r). Eqn r= -5m + c. since slope is -ve, if c is +ve , x intercept = +ve. otherwise -ve.

statement 1. m=-5. r= 25 +c . c= r-25. c could be +ve or -ve depending on r.

statement 2. r> 0. c = r+5m. c could be +ve or -ve depending on value of m.

together- c= r-25; r>0. c could still be +ve or - ve . Hence x intercept could be +ve or -ve.

Hence ans = E

Is it ok?

Responding to a pm:

Check out this post first: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/09 ... -vertices/

Slope of a line = - y intercept/x intercept

If slope is negative, x intercept will be positive when y intercept is positive too. So we need to know the sign of the y intercept.

(1) The slope of line k is –5.
Does not matter what the actual slope is. We already know it is negative. We need the sign of y intercept.

(2) r > 0
The line passes through (-5, r) where r is positive. This just gives one point through which the line passes. The y intercept could still be positive or negative as shown by the two diagrams in Bunuel's first post above.

Using both also we don't know the sign of the y intercept.

Answer (E)
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Given: (x, y) = (-5, r)

St1: Slope of line k = -5
To find the x intercept let the other co-ordinate be (x, 0)
(r - 0)/(-5 - x) = -5
r = 25 + 5x
x = (r - 25)/5
x intercept is positive if r > 25 and negative if r < 25
Not Sufficient

St2: r > 0 --> Clearly insufficient

Combining St1 and St2: r may still be < 25 or > 25
Not Sufficient

Answer: E

We need to determine whether the x-intercept of line k is positive, given that line k has a negative slope and passes through the point (-5, r).

We can let the slope of line k be m and the x-intercept of line k be a; that is, line k passes through the point (a, 0). Using the slope formula m = (y_2 - y_1)/(x_2 - x_1), we have:

m = (0 - r)/(a -(-5))= -r/(a + 5)

Statement One Alone:

The slope of line k is -5.

Using the information in statement one, we can say the following:

-r/(a + 5) = -5

a + 5 = r/5

a = r/5 - 5

Since we don't know the value of r, we cannot determine the value of a.

For example, if r = 5, a = -4, which is negative. However, if r = 50, a = 5, which is positive. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

r > 0

Knowing r is positive does not give us enough information to determine whether a, the x-intercept of line k, is positive. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Looking at our work from statement one, and keeping in mind that r > 0 from statement two, we see that a can be positive or negative.

For example, if r = 5 then a = -4, which is negative. However, if r = 50 then a = 5, which is positive. The two statements together are still not sufficient to answer the question.

Answer: E
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Target question: Is the x-intercept of line k positive?

When I scan the two statements, I can see that, if r is POSITIVE, then the point (-5, r) can be very close to the x-axis or the point (-5, r) can be very far from the x-axis.
Given this, let's go straight to the statements COMBINED, and see if we can find two cases that yield DIFFERENT answers to the target question.

CASE A: If r =1, then the point (-5, r) is very close to the x-axis, and we get something like this:

In this case, the answer to the target question is NO, the x-intercept of line k is NOT positive


CASE B: If r =50, then the point (-5, r) is far from the x-axis, and we get something like this:

In this case, the answer to the target question is YES, the x-intercept of line k IS positive

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

Hi,

Just to add here we are told that, slope is negative and slope is (-5) This means as that x increases by 1 y decreases by 5. So Points on this line would be (-4, r-5), (-3, r-10), (-2, r-15),(-1, r-20),(0, r-25)

So we see that at x=0 y is r-25. Now since we know that if slope is negative both the x intercept and y intercept have same sign. So if r-25<0 then x intercept is negative or if r-25>0 then x intercept is positive

So what we rally have to know is that if r> 25 or r<25.

So we see that Both the statements are insufficient to conclude if x intercept is positive/ negative
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Given: In the xy-plane, if line k has negative slope and passes through the point (-5,r).

Asked: Is the x-intercept of line k positive?

Let the slope of line k be m where m<0
In the xy-plane, if line k has negative slope and passes through the point (-5,r) , is the x-intercept of line k positive?

(y-r)=m(x+5)
For x-intercept, y =0
-r/m = x+5 => x=-5 -r/m
Asked: Whether -5-r/m>0?

(1) The slope of line k is -5.
m=-5
x-intercept = -5+r/5
Since r is unknown
NOT SUFFICIENT

(2) r > 0
x-intercept = -5-r/m
Since r & m are unknown
NOT SUFFICIENT

Combining (1) & (2)
(1) The slope of line k is -5.
m=-5
x-intercept = -5+r/5
(2) r > 0
If r>25 => x-intercept is positive
If r<=25 => x-intercept is NOT positive
NOT SUFFICIENT

IMO E

Hi JeffTargetTestPrep ScottTargetTestPrep

can you pls explain why did you assume that line k passes through the point (a, 0) "0" is not mentioned . Doesn`t "0" change the target question and why

BrentGMATPrepNow hey Brent perhaps you can explain why most people set y = 0 to find x intercept or vice versa and doest it change target question
as far as i know i can apply this formula \((y-y_1) = m (x-x_1)\) when i know slope and one point, but i know that one point is ( -5, r) where R is unknown hence i dont have complete information about this point

Once we know the slope is -5, the equation of the line becomes y = -5x + b
We also know that the line passes through the point (-5,r) (and statement 2 tells us r is POSITIVE)

The x-intercept (what the target question is asking about) will be the x-value that satisfies the line's equation when y = 0

At this point we can test various values of r.

If r = 1, then the line passes through the point (-5, 1)
Plug these values into our equation, y = -5x + b, to get: 1 = -5(-5) + b, which means b = -24
So, in this case, the equation of a line is: y = -5x - 24
To find the x-intercept, plug in y = 0 to get: 0 = -5x - 24
Solve to get: x = -24/5, which means the x-intercept is LESS THAN zero.

If r = 50, then the line passes through the point (-5, 50)
Plug these values into our equation, y = -5x + b, to get: 50 = -5(-5) + b, which means b = 25
So, in this case, the equation of a line is: y = -5x + 25
To find the x-intercept, plug in y = 0 to get: 0 = -5x + 25
Solve to get: x = 5, which means the x-intercept is GREATER THAN zero.

So the combined statements are not sufficient.

Answer: E

BrentGMATPrepNow many thanks great explanation !

Response:
The definition of x-intercept is the point where some line crosses the x-axis. Since the y-coordinates of every point on the x-axis is 0, the y-coordinate of the x-intercept of a line must also be 0. Thus, whenever we are told that the x-intercept of some line is a, we should understand that the line is passing through (a, 0). Similarly for the y-intercept; if we are told that the y-intercept of some line is b, then that line is passing through (0, b). Therefore, the question does not need to mention “0;” the y-coordinate of every x-intercept is 0.
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Bunuel the images aren't showing up on my screen. Please could you check if there is any issue with the upload?

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Fixed. Thank you for notifying.
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One can quickly do these questions by drawing graphs and considering scenarios. You could also consider taking imaginary values that are missing. You would quickly realize that there are multiple scenarios that may be possible giving a yes/no answer. Therefore the correct answer is option E
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could someone explain why B is not correct?

X intercept is 0=kx+b;kx=-b; x=-(b/k)>0?
Since we know that k is negative we either need to know the sign of b or the sign of b/k
From the (2) statement we have
r=-5k+b; since r>0; -5k+b>0; -5k>-b;b>5k;b/k>5
Since b/k is bigger than 5 we know that it is positive. therefore the x intercept is positive