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Chapter 18: Problem 34

\(A(0,0), B(2,1)\) and \(C(3,0)\) are the vertices of a \(\triangle A B C\) and \(B D\) is its altitude. If the line through \(D\) parallel to the side \(A B\) intersects the side \(B C\) at a point \(K\) then the product of the areas of the triangles \(A B C\) and \(B D K\) is (A) 1 (B) \(\frac{1}{2}\) (C) \(\frac{1}{4}\) (D) none of these

Short Answer

Expert verified
The product of the areas of triangles ABC and BDK is 1, answer (A).

Step by step solution

01

Determine Equation of BC

We first find the equation of line segment BC. Since B and C have coordinates \((2,1)\) and \((3,0)\), we calculate the slope \(m = \frac{0-1}{3-2} = -1\). Thus, the equation of line BC is \(y - 1 = -1(x-2)\). Simplifying gives \(y = -x + 3\).
02

Find Point D

Since BD is the altitude from B onto AC and point C is on the x-axis, D must be directly below B on the x-axis. So, point D has coordinates \((2, 0)\).
03

Find Equation of Line Parallel to AB through D

The slope of AB using points A(0,0) and B(2,1) is \(\frac{1 - 0}{2 - 0} = \frac{1}{2}\). So the line through D parallel to AB has the equation \(y - 0 = \frac{1}{2}(x - 2)\), simplifying to \(y = \frac{1}{2}x - 1\).
04

Determine Intersection Point K on BC

Set the equations of BC and the parallel line equal: \(-x + 3 = \frac{1}{2}x - 1\).Solving this, we combine terms to get \(3 + 1 = \frac{1}{2}x + x\), leading to \(x = \frac{8}{3}\). Substitute into either line's equation to find y, giving \(y = -\frac{8}{3} + 3 = \frac{1}{3}\). So, point K is \(\left(\frac{8}{3}, \frac{1}{3}\right)\).
05

Calculate Areas of Triangles ABC and BDK

First, the area of \(\triangle ABC\) using vertices at \((0,0), (2,1), (3,0)\) is \(\frac{1}{2} \left| 0(1-0) + 2(0 - 0) + 3(0-1) \right| = \frac{1}{2} \times 3 = \frac{3}{2}.\)Next, the area of \(\triangle BDK\) with vertices at \((2,1),(2,0),\left(\frac{8}{3},\frac{1}{3}\right)\) is:\(\frac{1}{2} \left| 2(0-\frac{1}{3}) + 2(\frac{1}{3}-1) + \frac{8}{3}(1-0)\right|\)Calculating, we get:\(= \frac{1}{2} (\frac{-2}{3} + \frac{-2}{3} + \frac{8}{3}) = \frac{1}{2}(\frac{4}{3}) = \frac{2}{3}\).
06

Evaluate Product of Areas

The product of the areas of \(\triangle ABC\) and \(\triangle BDK\) is:\(\frac{3}{2} \times \frac{2}{3} = 1.\)
07

Conclusion

The product of the areas of the triangles is \(1\), so the correct answer is (A) 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of a Line
Understanding the equation of a line is fundamental in coordinate geometry. It helps us describe the exact path that a line traverses on a coordinate plane. A line's equation can typically be expressed in the slope-intercept form: \(y = mx + c\), where \(m\) represents the slope and \(c\) is the y-intercept, or the point where the line crosses the y-axis. For example, if we have two points, B(2,1) and C(3,0), the first step is to calculate the slope \(m\), using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). This provides us a measure of how steep the line is. Once we have the slope, we can formulate the line's equation by substituting one of the points into the formula to solve for \(c\). This process enables us to pinpoint any location along the line, making it crucial for analyzing geometric figures like triangles.
Area of a Triangle
The area of a triangle in coordinate geometry can be determined using the formula: \( \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \). This formula calculates the actual space enclosed by a triangle when its vertices are plotted on the coordinate plane. In the context of the exercise, using vertices A(0,0), B(2,1), and C(3,0), substituting into the formula yields a straightforward calculation. Each term of the formula corresponds to a segment of the triangle which, when taken together and multiplied by \(\frac{1}{2}\), provides the whole area. This technique is especially useful for triangles constructed in coordinate geometry, as it avoids direct measurement and instead relies on purely numerical calculations.
Slope of a Line
The slope of a line fundamentally defines the line's direction and steepness. It's calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). The result tells us how far up (or down) we move on the y-axis for a positive movement along the x-axis. In simple terms, if the slope is positive, the line rises from left to right; if negative, it falls. For instance, in our exercise, calculating the slope between points A (0,0) and B (2,1) with \(m = \frac{1-0}{2-0} = \frac{1}{2}\), tells us that for every 2 units we move horizontally, we move 1 unit vertically. The slope is not only pivotal for constructing line equations but also for ensuring parallel lines, as parallel lines share identical slopes. So, having a concrete understanding of slopes can greatly enhance one鈥檚 ability to navigate and analyze coordinate plane problems efficiently.

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