Other Lines - GMAT Quantitative

Consider linear functions h(t) and g(t).

I) at the point

II)

Is the point on the line h(t)?

We can use II) and I) to find the slope of h(t)

Recall that perpendicular lines have opposite reciprocal slope. Thus, the slope of h(t) must be

Next, we know that h(t) must pass through (6,4), so lets us that to find the y-intercept:

Next, check if (10,4) is on h(t) by plugging it in.

So, the point is not on the line, and we needed both statements to know.

To find the equation of a linear function, we need some combination of slope and a point.

Statement I gives us a clue to find the slope of the desired function. It must be the opposite reciprocal of the slope of . This makes the slope of equal to

Statement II gives us a point on our desired function, .

Using slope-intercept form, we get the following:

So our equation is as follows

Find the equation for linear function p(x)

I) and

II)

To begin:

I) Tells us that p(x) must have a slope of 16

II) Tells us a point on p(x). Plug it in and solve for b:

Given that the square of a negative is still positive, it is possible for a to have an x-intercept that is negative, while still having a positive slope. The example above shows how the square of the x-intercept for line a could be greater, while having still giving line a a slope that is less than that of b.

Either statement is sufficient.

Line j, as a line, has an equation of the form

Statement 1 gives the equation of a perpindicular line, so the slopes of the two lines are negative reciprocals of each other:

Statement 2 allows the slope to be found using rise over run:

Then, since the x-intercept is known:

Assume both statements to be true. Then the line shares its - and -intercepts with the graph of , which is a parabola. The common -intercept can be found by setting and solving for :

,

making the -intercept of the parabola, and that of the line, .

The common -intercept can be found by setting and solving for :

, in which case , or

, in which case ,

The parabola therefore has two -intercepts, and , so it is not clear which one is the -intercept of the line. Therefore, the equation of the line is also unclear.

, in slope-intercept form, is

Therefore, the sign of is the sign of the slope.

The first statement means that is positive - all that means is that both and are nonzero and of like sign. can be either positive or negative, and consequently, so can slope .

The second statement - that is positive - makes , the sign of the slope, negative.

The equation is in standard form, with shapes replacing coefficients; that is, the equation is

.

Rewrite the equation in slope-intercept form:

The slope of the line of the equation is - and it depends on the number in the square, , and the triangle . Each statement alone gives the number in only one of the shapes; the two statements together give the numbers in both shapes and allow the slope to be calculated, thereby answering the question of Quinn's success.

The slope of a line through and is

From Statement 1 alone, we can tell that

,

so we know the sign of the slope.

From Statement 2 alone, we can tell that

But this can be positive or negative - for example:

but

The slope of a line through and is

If and have the same sign, then , making the slope negative; if and have the same sign, then , making the slope positive.

Statement 1 is not enough to determine the sign of .

Case 1:

Case 2:

So if we only know Statement 1, we do not know whether and have the same sign, and, subsequently, we do not know the sign of slope . A similar argument can be made that Statement 2 provides insufficient information.

If we know both statements, we can solve the system of equations as follows:

Therefore, we know and have unlike sign and .

The slope of a line through and is

If and have the same sign, then , making the slope negative; if and have the same sign, then , making the slope positive.

If we know both statements, we try to solve the system of equations as follows:

This means that the system is dependent, and that the statements are essentially the same.

Case 1:

Then

Case 2:

Then

Thus from Statement 1 alone, it cannot be determined whether and have the same sign, and the sign of the slope cannot be determined. Since Statement 2 is equivalent to Statement 1, the same holds of this statement, as well as both statements together.

Statement 2 alone tells us that the lines are parallel, not perpendicular. Statement 1 alone is neither necessary nor helpful, as the sum of the slopes is irrelevant.

The slope of the line that includes points and is .

For the question of the sign of the slope to be answered, it must be known whether and are of the same sign or of different signs, or whether one of them is equal to zero.

Statement 1 alone does not answer this question, as it only states that the denominator is greater; it is possible for this to happen whether both are of like sign or unlike sign. Statement 2 only proves that - that is, that the denominator is positive.

If the two statements together are assumed, we know that . Since both the numerator and the denominator are positive, the slope of the line must be positive.

If Statement 1 alone holds - that is, if it is known only that the line is parallel to the line of - then this equation can be rewritten in slope-intercept form. The slope can be deduced from this, and the two lines, being parallel, will have the same slope.

If Statement 2 alone holds - that is, if it is known only that the line is perpendicular to the line of the equation - then this equation can be rewritten in slope-intercept form. The slope can be deduced from this, and the first line, which is perpendicular to this one, will have the slope that is the opposite of the reciprocal of that.

Either statement alone will yield an answer.

Examine the diagram below.

It can be seen from the red lines that no conclusions about the sign of the slope of a line can be drawn from Statement 1, since lines of positive, negative, and zero slope can contain points in both Quadrant I and Quadrant II.

If a line contains a point in Quadrant I and a point in Quadrant III, then it contains a point with positive coordinates and a point with negative coordinates ; its slope is

which is a positive slope.

Therefore, Statement 2 alone, but not Statement 1 alone, provides a definitive answer.

The -axis is horizontal, so any line perpendicular to it is vertical and has undefined slope. Statement 1 is sufficient.

A line on the coordinate plane with no -intercept does not intersect the -axis and therefore must be parallel to it - subsequently, it must be vertical and have undefined slope. This makes Statement 2 sufficient.

A line with undefined slope is a vertical line.

Infinitely many lines, some vertical and some not, pass through , and infinitely many lines, some vertical and some not, pass through each quadrant, so neither statement alone is sufficient to answer the question.

Now assume both statements are true. Then, since the line passes through Quadrant II, it passes through at least one point with negative -coordinate and positive -coordinate, which we call . Its slope will be

,

a negative slope. Therefore, the slope is not undefined.

Infinitely many lines pass through the origin, and infinitely many lines pass through each quadrant, so neither statement alone is sufficient to answer the question.

Suppose that both statements are known to be true. Since the line passes through quadrant II, it passes through a point , where are positive. It also passes through so its slope will be

which is a negative slope.

Examine the diagram below.

Both the red line and the green line fit the descriptions in both statements. The red line has negative slope and the green line has positive slope.

The two statements together give insufficient infomation.