# How NOT to Write the Equation of a Line on the GMAT

A question brought an interesting situation to our notice. Let’s start by asking a question: How do we write the equation of a line? There are two formulas:

y = mx + c (where m is the slope and c is the y-intercept)
and
yy1 = m * (xx1) [where m is the slope and (x1,y1) is a point on the line]

We also know that m = (y2y1)/(x2x1) – this is how we find the slope given two points that lie on a line. The variables are x1, y1 and x2, y2, and they represent specific values.

But think about it, is m = (y2y)/(xx1) really the equation of a line? Let’s further clarify this idea using a GMAT practice question:

In the coordinate plane, line k passes through the origin and has slope 2. If points (3,y) and (x,4) are on line k, then x + y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

We have been given that the line passes through (0, 0) and has a slope of 2. We can find the equation of the line from this information.

y = mx + c
y = 2x + 0 (Since the line passes through (0, 0), its y-intercept is 0 – when x is 0, y is also 0.)
y = 2x

Since we are given two other points, (3, y) and (x, 4), on the line and we have a slope of 2, many test-takers will be tempted to make another equation for the line using this information.

(4 – y)/(x – 3) = 2
(4 – y) = 2*(x – 3)
Thus, 2+ y = 10

Here, test-takers will use the two equations to solve for x and y and get x = 5/2 and y = 5.

After adding x and y together, they then wonder why 7.5 is not one of the answer choices. If this were an actual GMAT question, it is quite likely that 7.5 would have been one of the options. So all in all, the test-taker would not even have realized that he or she made a mistake, and would choose 7.5 as the (incorrect) answer.

The error is conceptual here. Note that the equation of the line, 2x + y = 10, is not the same as the equation we obtained above, y = 2x. They represent two different lines, but we have only a single line in the question. So which is the actual equation of that line?

To get the second equation, we have used m = (y2y)/(xx1). But is this really the equation of a line? No. This formula doesn’t have y and x, the generic variables for the x– and y-coordinates in the equation of a line.

To further clarify, instead of x and y, try using the variables a and b in the question stem and see if it makes sense:

“In the coordinate plane, line k passes through the origin and has slope 2. If points (3, a) and (b, 4) are on line k, then a + b =”

You can write (4 – a)/(b – 3) = 2 and this would be correct. But can we solve for both a and b here? No – we can write one of them in terms of the other, but we can’t get their exact values.

We know a and b must have specific values. (3, a) is a point on the line y = 2x. For x = 3, the value of of the y-coordinate, a, will be y = 2*3 = 6. Therefore, a = 6.

(b, 4) is also on the line y = 2x. So if the y-coordinate is 4, the x-coordinate, b, will be 4 = 2b, i.e. b = 2. Thus, a + b = 6 + 2 = 8, and our answer is C.

This logic remains the same even if the variables used are x and y, although test-takers often get confused because of it. Let’s solve the question in another way using the variables as given in the original question.

Recall what we have learned about slope in the past. If the slope of the line is 2 and the point (0, 0) lies on the line, the value of y – if point (3, y) also lies on the line – will be 6 (a slope of 2 means a 1-unit increase in x will lead to a 2-unit increase in y).

Again, if point (x, 4) lies on the line too, an increase of 4 in the y-coordinate implies an increase of 2 in the x-coordinate. So x will be 2, and again, x + y = 2 + 6 = 8.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Quarter Wit, Quarter Wisdom: An Innovative Use of the Slope of a Line on the GMAT

Let’s continue our discussion on coordinate geometry today.

The concept of slope is extremely important on the GMAT – it is not sufficient to just know how to calculate it using (y2 – y1)/(x2 – x1).

In simple terms, the slope of a line specifies the units by which the y-coordinate changes and the direction in which it changes with each 1 unit increase in the x-coordinate. If the slope (m) is positive, the y-coordinate changes in the same direction as the x-coordinate. If m is negative, however, the y-coordinate changes in the opposite direction.

For example, if the slope of a line is 2, it means that every time the x-coordinate increases by 1 unit, the y-coordinate increases by 2 units. So if the point (3, 5) lies on a line with a slope of 2, the point (4, 7) will also lie on it. Here, when the x-coordinate increases from 3 to 4, the y-coordinate increases from 5 to 7 (by an increase of 2 units). Similarly,  the point (2, 3) will also lie on this same line – if the x-coordinate decreases by 1 unit (from 3 to 2), the y-coordinate will decrease by 2 units (from 5 to 3). Since the slope is positive, the direction of change of the x-coordinate will be the same as the direction of change of the y-coordinate.

Now, if we have a line where the slope is -2 and the point (3, 5) lies on it, when the x-coordinate increases by 1 unit, the y-coordinate DECREASES by 2 units – the point (4, 3) will also lie on this line. Similarly, if the x-coordinate decreases by 1 unit, the y-coordinate will increase by 2 units. So, for example, the point (2, 7) will also lie on this line.

This understanding of the concept of slope can be very helpful, as we will see in this GMAT question:

Line L and line K have slopes -2 and 1/2 respectively. If line L and line K intersect at (6,8), what is the distance between the x-intercept of line L and the y-intercept of line K?

(A) 5
(B) 10
(C) 5√(5)
(D) 15
(E) 10√(5)

Traditionally, one would solve this question like this:

The equation of a line with slope m and constant c is given as y = mx + c. Therefore, the equations of lines L and K would be:

Line L: y = (-2)x + a
and
Line K: y = (1/2)x + b

As both these lines pass through (6,8), we would substitute x=6 and y=8 to get the values of a and b.

Line L: 8 = (-2)*6 + a
a = 20

Line K: 8 = (1/2)*6 + b
b = 5

Thus, the equations of the 2 lines become:

Line L: y = (-2)x + 20
and
Line K: y = (1/2)x + 5

The x-intercept of a line is given by the point where y = 0. So, the x-intercept of line L is given by:

0 = (-2)x + 20
x = 10

This means line L intersects the x-axis at the point (10, 0).

Similarly, the y-intercept of a line is given by the point where x = 0. So, y-intercept of line K is given by:

y = (1/2)*0 + 5
y = 5

This means that line K intersects the y-axis at the point (0, 5).

Looking back at our original question, the distance between these two points is given by √((10 – 0)^2 + (0 – 5)^2) = 5√(5). Therefore, our answer is C.

Method 2: Using the Slope Concept
Although the using the traditional method is effective, we can answer this question much quicker using the concept we discussed above.

Line L has a slope of -2, which means that for every 1 unit the x-coordinate increases, the y-coordinate decreases by 2. Line L also passes through the point (6, 8). We know the line must intersect the x-axis at y = 0, which is a decrease of 8 y-coordinates from the given point (6,8). If y increases by 8, according to our slope concept, x will increase by 4 to give 6 + 4 = 10. So the x-intercept of line L is at (10, 0).

Line K has slope of 1/2 and also passes through (6, 8). We know the this line must intersect the y-axis at x = 0, which is a decrease of 6 x-coordinates from the given point (6,8). This means y will decrease by 1/2 of that (6*1/2 = 3) and will become 8 – 3 = 5. So the y-intercept of line K is at (0, 5).

The distance between the two points can now be found using the Pythagorean Theorem – √(10^2 + 5^2) = 5√(5), therefore our answer is, again, C.

Using the slope concept makes solving this question much less tedious and saves us a lot of precious time. That is the advantage of using holistic approaches over the more traditional approaches in tackling GMAT questions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Coordinate Geometry: Solving GMAT Problems With Lines Crossing Either the X-Axis or the Y-Axis

Today let’s learn about the cases in which lines on the XY plane cross, or do not cross, the x- or y-axis. Students often struggle with questions such as this:

Does the line with equation ax+by = c, where a,b and c are real constants, cross the x-axis?

What concepts will you use here? How will you find whether or not a line crosses the x-axis? What conditions should it meet? Think about this a little before you move ahead.

We know that most lines on the XY plane cross the x-axis as well as the y-axis. Even if it looks like a given line doesn’t cross either of these axes, eventually, it will if it has a slope other than 0 or infinity.

Note that by definition, a line extends infinitely in both directions – it has no end points (otherwise it would be a line “segment”). We cannot depict a line extending infinitely, which is why we will only show a small section of it. Ideally, a line on the XY plane should be shown with arrowheads to depict that it extends infinitely on both sides, but we often omit them for our convenience. For instance, if we try to extend the example line above, we see that it does, in fact, cross the x-axis:

So what kind of lines do not cross either the x-axis or the y-xis? We know that the equation of a line on the XY plane is given by ax + by  + c = 0. We also know that if we want to find the slope of a line, we can use the equation y = (-a/b)x – c/b, where the slope of the line is -a/b.

A line with a slope of 0 is parallel to the x-axis. For the slope (i.e. -a/b) to be 0, a must equal 0. So if a = 0, the line will not cross the x-axis – it is parallel to the x-axis. The equation of the line, in this case, will become y = k. In all other cases, a line will cross the x-axis at some point.

Similarly, it might appear that a line doesn’t cross the y-axis but it does at some point if its slope is anything other than infinity. A line with a slope of infinity is parallel to the y-axis. For -a/b to be infinity, b must equal 0. So if b = 0, the line will not cross the y-axis. The equation of the line in this case will become x = k. In all other cases, a line will cross the y-axis at some point.

Now, we can easily solve this official question:

Does the line with equation ax+by = c, where a, b and c are real constants, cross the x-axis?

Statement 1: b not equal to 0

Statement 2: ab > 0

As we discussed earlier, all lines cross the x-axis except lines which have a slope of 0, i.e. a = 0.

Statement 1: b not equal to 0

This tells statement us that b is not 0 – which means the line is not parallel to y-axis – but it doesn’t tell us whether or not a is 0, so we don’t know whether the line is parallel to the x-axis or crosses it. Therefore, this statement alone is not sufficient.

Statement 2: ab>0

If ab > 0, it means that neither a nor b is 0 (since any number times 0 will equal 0). This means the line is parallel to neither x-axis nor the y-xis, and therefore must cross the x-axis. This statement alone is sufficient and our answer is B.

Hopefully this has helped clear up some coordinate geometry concepts today.