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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