Equation of a Line | Geometry

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

$y - y_1 = m \times (x - x_1)$ [where $m$ is the slope and $(x_1, y_1)$ is a point on the line]

We also know that $m = \frac{y_2 - y_1}{x_2 - x_1}$ – this is how we find the slope given two points that lie on a line. The variables are $x_1$ , $y_1$ and $x_2$ , $y_2$ , and they represent specific values.

But think about it, is $m = \frac{y_2 - y}{x - x_1}$ 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

(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.

$\frac{4 - y}{x - 3} = 2$

$(4 - y) = 2\times(x - 3)$

Thus, $2x + y = 10$

Here, test-takers will use the two equations to solve for $x$ and $y$ and get $x = \frac{5}{2}$ 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 $m = \frac{y_2 - y}{x - x_1}$ . 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 $\frac{4 - a}{b - 3} = 2$ 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 the $y$ -coordinate, $a$ , will be $y = 2\times3 = 6$ . 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.

We know that 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, $x + y = 2 + 6 = 8$ .

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