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Using determinants, are the points zero, one; two, one-half; and four, zero collinear?
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In this question, we need to determine whether three given points are collinear.
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That means, do they lie on the same straight line?
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And there’s a few different ways we could go about doing this.
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For example, we could find an equation of a straight line between a pair of these two points and then determine whether the third point lies on this line.
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However, the question wants us to do this by using determinants.
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And to do this, we need to recall the following fact about determinants.
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If we have three distinct points 𝑥 sub one, 𝑦 sub one; 𝑥 sub two, 𝑦 sub two; and 𝑥 sub three, 𝑦 sub three, then we can determine whether these points are collinear by calculating the determinant of the three-by-three matrix 𝑥 sub one, 𝑦 sub one, one, 𝑥 sub two, 𝑦 sub two, one, 𝑥 sub three, 𝑦 sub three, one.
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If this determinant is zero, then the three points are collinear.
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If this determinant is nonzero, then the three points are noncollinear.
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And it is worth noting the statement does work in both directions.
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If the determinant is zero, then the points are collinear.
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And similarly, if the points are collinear, then the determinant is zero, where we assume we have three distinct points.
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And to see why this is true, we can recall the absolute value of this determinant gives us the area of a parallelogram with these three points as vertices.
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And the only way the area of a parallelogram can be zero is if its vertices are collinear.
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Therefore, we can determine whether these three points are collinear by substituting the three points given to us in the question into this equation.
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We need to determine whether the determinant of the matrix zero, one, one, two, one-half, one, four, zero, one is equal to zero.
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And we can evaluate the determinant of this matrix in any way we choose.
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We’re going to expand over the first row because it includes the number zero.
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To expand over the first row in this matrix, we need to find all of the matrix minors of this row.
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And remember, the sign of this expansion will change depending on the parity.
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In this case, the second term will be negative.
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This gives us zero times the determinant of the matrix one-half, one, zero, one minus one times the determinant of the matrix two, one, four, one plus one times the determinant of the matrix two, one-half, four, zero.
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Now, all that’s left to do is evaluate this expression.
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The first term has a factor of zero, so it’s equal to zero.
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And remember, to evaluate the determinant of a two-by-two matrix, we need to find the difference in the products of the diagonals.
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In the second term, that’s two times one minus four times one, which is two minus four.
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And in the third term, that’s two times zero minus four times one-half, which is zero minus two.
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This gives us negative one times two minus four plus one times zero minus two.
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And if we evaluate this expression, we get two minus two, which is equal to zero.
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And since this determinant is equal to zero, the three points are either not distinct or collinear.
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And we can see that they are distinct points, so they must be collinear.
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Therefore, we were able to show by using the determinants, the points zero, one; two, one-half; and four, zero are collinear.