Absurdly simply, what is the slope of the line?

Geometry Level 3

A straight, infinite line passes through only one integer lattice point on an Euclidean plane. A integer lattice point is a point that has integer coordinates. What is the slope of the line? Select from the available choices. A numeric answer is not expected.

There is insufficient data to select an answer. Such a slope is impossible. The slope is an integer. The slope is rational. The slope is none of the other answers. The slope is a complex number with a nonzero imaginary part.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

4 solutions

There is sufficient data. If the slope is rational, which includes integers, than it would pass through an infinite number of integer lattice points. The problem environment was specified as the Euclidean plane therefore the slope is not a complex number with a non-zero imaginary part. Such a slope is possible any irrational number qualifies. Therefore the correct answer is none of the other answers.

4 Helpful 3 Interesting 0 Brilliant 0 Confused

The answer 'there is insufficient data' is just very confusing: once one realizes the other answers are incorrect it just makes the question unclear. There IS insufficient data to say if it's, say, a number. I admit the problem attempts to clarify it, but the answer choices are nonetheless just very bad.

Alex Li - 2 years, 4 months ago

Log in to reply

All straight lines have slopes, that is, amount of change of rise (or fall) or a given amount horizontal motion (run), that is, a slope is a ratio. In the case of a vertical line, that slope would be infinite, e.g., 1 0 \frac10 . That is not the case here as such a vertical line, if the straight line passed through one integer lattice point then the straight line would pass through an infinite number of integer lattice points. Now, read Peter Macgregor's comment below. He establishes that the slope exists and must be irrational, which eliminates the answers 'such a slope is impossible,' 'the slope is an integer' as an integer is a particular type of rational number and the answer 'the slope is rational' as rational is the opposite of irrational. On the Euclidean plane, both the rise and the run are real numbers, therefore their ratio is a real number or infinite as real numbers are closed under the operations of multiplication and multiplicative inverse. Therefore the slope can not be a complex number with a non-zero imaginary part. Note, a complex number with a zero imaginary part could be considered to be a real; but, that choice was not offered.

Peter Macgregor's comment below has established that such a slope exists and is irrational. Therefore, there is sufficient data to answer the question.

Therefore, 'the slope is none of the other answers.'

A Former Brilliant Member - 2 years, 4 months ago

Log in to reply

You don't understand my comment. I was saying that the answer choices are misleading.

Alex Li - 2 years, 4 months ago

All straight lines have slopes, that is, amount of change of rise (or fall) or a given amount horizontal motion (run), that is, a slope is a ratio. In the case of a vertical line, that slope would be infinite, e.g., 1 0 \frac10 . That is not the case here as such a vertical line, if the straight line passed through one integer lattice point then the straight line would pass through an infinite number of integer lattice points. Now, read Peter Macgregor's comment below. He establishes that the slope exists and must be irrational, which eliminates the answers 'such a slope is impossible,' 'the slope is an integer' as an integer is a particular type of rational number and the answer 'the slope is rational' as rational is the opposite of irrational. On the Euclidean plane, both the rise and the run are real numbers, therefore their ratio is a real number or infinite as real numbers are closed under the operations of multiplication and multiplicative inverse. Therefore the slope can not be a complex number with a non-zero imaginary part. Note, a complex number with a zero imaginary part could be considered to be a real; but, that choice was not offered.

Peter Macgregor's comment below has established that such a slope exists and is irrational. Therefore, there is sufficient data to answer the question.

Therefore, 'the slope is none of the other answers.'

A Former Brilliant Member - 2 years, 4 months ago
Prince Zarzees
Jan 24, 2019

y=(5+√2)x-√2 is such a straight line

0 Helpful 1 Interesting 2 Brilliant 0 Confused
Peter Macgregor
Jan 26, 2019

Such a line must have an irrational gradient.

To answer the question fully we need to show that a line with an irrational gradient cannot pass through two integer lattice points, and we should also show that if a line with a rational gradient passes through one integer lattice point then it passes through another one as well. (In fact it will pass through an infinite number of integer lattice points, but that is not needed to answer the problem!)

First lets prove that a line with an irrational gradient cannot pass through two integer lattice points .

Suppose it did, passing through (a,b) and (c,d) where all four coordinates are integers. Then

m = b − d a − c m=\frac{b-d}{a-c}

but this is a rational number, and we can complete the proof by contradiction. In the awkward case a = c, the line passes through every integer lattice point (a,y)

Now for the second part if a line with a rational gradient passes through one integer lattice point then it passes through another one as well

Suppose the rational gradient is m = v h m=\frac{v}{h} where v and h are integers. Starting from the given point move horizontally by h units and vertically by v units. This leads to another lattice point on the line.

Taking these two results, and recognising the last option as a meaningless distractor, lets us answer the question.

2 Helpful 1 Interesting 1 Brilliant 0 Confused

Possible with any irrational slope.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

1 pending report

Vote up reports you agree with

×

Problem Loading...

Note Loading...

Set Loading...