Hi All,

We're asked for the equation of a straight line that contains the points (10, 100) and (15, 60). This question can be solved by simply 'plugging' in the two co-ordinates and finding the equation that includes both among its solutions. The first coordinate is arguably the easiest (since X=10, Y=100 should stand out when it's a solution). As it stands, there's only one Answer that includes that option...

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorLU wrote:The equation of a straight line containing the points (10, 100) and (15, 60) is?

$$A. y=-8x+80$$
$$B. y=8x-180$$
$$C. \dfrac{x}{8} + 7.5$$
$$D. -8x-180$$
$$E. -\dfrac{x}{8} + 22.5$$

The OA is A.

We can use the equation,

$$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}\cdot(x-x_1)$$
Then,
$$y-100=\dfrac{60-100}{15-10}\cdot(x-10)$$
$$y-100=\dfrac{-40}{5}\cdot(x-10)$$
$$y-100=-8x +80$$
$$y=-8x + 180$$
Option A.

Please, can anyone explain another way to solve this PS question? Thanks!

We know the equation has to be of the form y = mx + b where m and b are constants.

If we let y = 100 and x = 10, we get 10m + b = 100.

If we let y = 60 and x = 15, we get 15m + b = 60.

Subtracting the second equation from the first, we get -5m = 40 and thus m = -8. If we substitute either of the solutions to y = -8x + b, for instance, if we let x = 10 and y = 100, we get b = 180. Thus, the equation of the line is y = -8x + 180.

Answer: A