Question: Distance Between Intercepts

Comment on Distance Between Intercepts

Is there a quicker way to solve this? I cant imagine one off the top of my head right now.
gmat-admin's picture

I'm not aware of a significantly faster solution, but that doesn't mean there isn't one :-)
Anyone else want to provide an alternate solution?

I used the same steps to find the points of X and Y intercept.After finding the X and Y intercept we can apply the distance formula. The distance formula might be a good thing to remember if the points are not right angled triangle
gmat-admin's picture

The distance formula also works, but I'd like to point out that we can always create a right triangle to find the distance between two points on the coordinate plane (for more, see

The only times we cannot do this are when the two points create either a vertical line or a horizontal line.

Not particularly difficult but insane to imagine having to complete this in <2mins! What level of a question would this be considered?
gmat-admin's picture

I'd say 700+ level


This is how I solved the problem:

STEP 1: calculated x intercept of line l by substituting the following values of

(x1,y1) = (6,8)
(x2,y2) = (x,0) [as we know the y value for the x intercept will be 0]
slope = -2

in the slope formula : slope m = (y2-y1) / (x2-x1)

STEP 2: calculate y intercept of line k by substituting the given values in slope formula:

(x1,y1) = (6,8)
(x2,y2) = (0,y) [as we know the x value for the y intercept will be 0]
slope = 1/2

STEP 3: now we know,

x intercept of line l : (10,0)
y intercept of line k : (0,5)

substitute these values in the formula to find distance between any two points in the coordinate plane:

distance = sqrt{ (x1 - x2)^2 + (y1 - y2)^2 }

by doing this I avoided one step of finding the y intercept of line l. I hope this will work for all values
gmat-admin's picture

That's a perfectly valid approach. Great work!

I solved this question with the point-slope equation. It think it could be a slightly quicker way to directly solve for both intercepts.

y - y1 = m (x - x1)

for line l's x-intercept (y = 0)

0 - 8 = -2*(x - 6) -> x = 10

for line l's y-intercept (x = 0)

y - 8 = 1/2*(0 - 6) -> y = 5
gmat-admin's picture

Nice work!

Hi Brent,

Why did we use Pythagorean Theorem to solve statement 2? If radius is 5 from the origin and (x,y) lie beyond (and are +ve), shouldn't this be enough to conclude on the question?
gmat-admin's picture

Question link:

It would SEEM so. However, notice that the point (4,4) is more than 5 units away from the origin (0,0).
When we apply the distance formula, we see that (4,4) is 4√2 (≈ 5.6) units away from the origin.

So, even though (4,4) is more than 5 units away from the origin, the x-coordinate need not be greater than 4.

Does that help?


Hello, I'm also stuck on how a circle with a radius of 5 with a center at 0,0 would not be sufficient evidence for us to prove that x>5? This would answer the target question. Why would we use the Pythagorean Theorem when we know the arc of the circle would go even further than the hypotenuse of the right triangle, with the right angle situated at the origin? A radius of a circle is the same unit distance from the center in every direction. If the radius is 5, that would make the magnitude of any point beyond the circumference greater than 5.

With regards to your example of (4,4), the hypotenuse of the triangle is ±5.6 units distance from y-intercept 4, and x-intercept 4. Aren't we trying to determine the amount of units from the origin, and not the intercepts?
gmat-admin's picture

I'm not sure what role the x and y-intercepts play in your solution.
A circle with radius 5 and centered at (0,0) will have an x-intercept of 5 and a y-intercept of 5.

Here's the circle with a few points on its lie:

The point made with (4,4) is that this point lies outside a circle of radius 5 centered at the origin. However, we can see that, in the case of (4,4), x is less than 5.

Here's the point (4,4) added to my sketch:
Since the distance from (0,0) to (4,4) is greater than 5, the point (4,4) lies outside the circle with radius 5.

Does that help?


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