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## Comment on

Distance Between Intercepts## Is there a quicker way to

## I'm not aware of a

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 distance formula also

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 https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...)

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

## Not particularly difficult

## I'd say 700+ level

I'd say 700+ level

## Hi,

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

## That's a perfectly valid

That's a perfectly valid approach. Great work!

## Hello,

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

## Nice work!

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?

https://www.beatthegmat.com/mba/2011/04/19/manhattan-gmat-challenge-problem-of-the-week-19-april-2011

## Question link: https://www

Question link: https://www.beatthegmat.com/mba/2011/04/19/manhattan-gmat-challenge-prob...

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?

Cheers,

Brent

## Hello, I'm also stuck on how

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?

## I'm not sure what role the x

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: https://imgur.com/of44vua

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: https://imgur.com/hkeUy8P

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?

Cheers,

Brent

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