Under which condition, are the points (a, b), (c, d) and (a - c, b - d) collinear?
ad = bc
The question asks for the condition under which three specific points, given by their coordinates as \((a, b)\), \((c, d)\), and \((a - c, b - d)\), lie on the same straight line. Points lying on the same straight line are called collinear points. There are several ways to determine if three points are collinear. One common method involves using the concept of the area of a triangle formed by these points.
If three points are collinear, the area of the triangle formed by them is zero. Let's use this property to find the required condition for the given points.
Let the three points be \(A(x_1, y_1) = (a, b)\), \(B(x_2, y_2) = (c, d)\), and \(C(x_3, y_3) = (a - c, b - d)\).
The formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
Area \(= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
For the points to be collinear, the area must be equal to zero.
Let's substitute the coordinates of points A, B, and C into the area formula:
Area \(= \frac{1}{2} |a(d - (b - d)) + c((b - d) - b) + (a - c)(b - d)|\)
Simplify the terms inside the absolute value:
Now, sum these terms:
Sum \(= (2ad - ab) + (-cd) + (ab - ad - cb + cd)\)
Combine like terms:
Sum \(= 2ad - ab - cd + ab - ad - cb + cd\)
Sum \(= (2ad - ad) + (-ab + ab) + (-cd + cd) - cb\)
Sum \(= ad + 0 + 0 - cb\)
Sum \(= ad - cb\)
So, the area of the triangle is \(\frac{1}{2} |ad - cb|\).
For the points \((a, b)\), \((c, d)\), and \((a - c, b - d)\) to be collinear, the area must be zero:
\(\frac{1}{2} |ad - cb| = 0\)
This equation holds true if and only if the expression inside the absolute value is zero:
\(ad - cb = 0\)
Rearranging this equation gives the condition for collinearity:
\(ad = cb\)
Let's compare the derived condition \(ad = cb\) with the given options:
The derived condition \(ad = cb\) is the same as \(ad = bc\).
Therefore, the points \((a, b)\), \((c, d)\), and \((a - c, b - d)\) are collinear if and only if the condition \(ad = bc\) is satisfied.
| Point | Coordinates |
|---|---|
| A | \((a, b)\) |
| B | \((c, d)\) |
| C | \((a - c, b - d)\) |
| Concept | Description | Condition for Collinearity |
|---|---|---|
| Area of Triangle | If three points form a degenerate triangle (a line segment), the area is zero. | Area \(= 0\) |
| Slope Method | The slope between any two pairs of points is the same. | Slope of AB \(=\) Slope of BC (assuming distinct points) |
| Vector Method | Two vectors formed by the points (e.g., \(\vec{AB}\) and \(\vec{AC}\)) are parallel, meaning one is a scalar multiple of the other. | \(\vec{AB} = k \cdot \vec{AC}\) for some scalar \(k\) |
Besides the area method, another common way to check for collinearity is using slopes. For points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) to be collinear, the slope of line segment AB must be equal to the slope of line segment BC, provided the denominators are not zero.
Slope of AB = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{d - b}{c - a}\) (if \(c \neq a\))
Slope of BC = \(\frac{y_3 - y_2}{x_3 - x_2} = \frac{(b - d) - d}{(a - c) - c} = \frac{b - 2d}{a - 2c}\) (if \(a-c \neq c\), i.e., \(a \neq 2c\))
Setting slopes equal:
\(\frac{d - b}{c - a} = \frac{b - 2d}{a - 2c}\)
Cross-multiplying (assuming denominators are non-zero):
\((d - b)(a - 2c) = (b - 2d)(c - a)\)
\(ad - 2cd - ab + 2bc = bc - ab - 2cd + 2ad\)
\(ad - 2cd - ab + 2bc = bc - ab - 2cd + 2ad\)
Notice that \(-2cd\), \(-ab\), and \(2ad\) appear on both sides (with opposite signs for \(2ad\) relative to the original positions but they are on the right side as \(2ad\)). Let's move all terms to one side:
\(ad - 2cd - ab + 2bc - (bc - ab - 2cd + 2ad) = 0\)
\(ad - 2cd - ab + 2bc - bc + ab + 2cd - 2ad = 0\)
Combine like terms:
\((ad - 2ad) + (-2cd + 2cd) + (-ab + ab) + (2bc - bc) = 0\)
\(-ad + 0 + 0 + bc = 0\)
\(-ad + bc = 0\)
\(bc = ad\) or \(ad = bc\).
This confirms the same condition obtained using the area method. The slope method requires careful handling of cases where denominators are zero (vertical or horizontal lines), while the area method handles these cases naturally as the area formula works universally. The condition \(ad = bc\) encompasses all possibilities for the collinearity of these specific points.
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