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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 is 1.

Step by step solution

01

Calculate the Area of Triangle ABC

The vertices of triangle \(\triangle ABC\) are given as \(A(0,0), B(2,1), \text{ and } C(3,0)\). We use the formula for the area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\):\[\text{Area} = \frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |.\]Substituting the given points:\[\text{Area}_{ABC} = \frac{1}{2} | 0(1-0) + 2(0-0) + 3(0-1) | = \frac{1}{2} | -3 | = \frac{3}{2}.\]
02

Identify the Altitude BD

The altitude \(BD\) is perpendicular to the base \(AC\). The base \(AC\) is the x-axis (since both \(A\) and \(C\) have a y-coordinate of 0), so the altitude drawn from \(B(2,1)\) will be a vertical line parallel to the y-axis with coordinates \((2, 0)\). Thus, coordinate of \(D\) is \((2,0)\).
03

Consider Line through D Parallel to AB

The slope of line \(AB\) is calculated as \(\frac{1-0}{2-0} = \frac{1}{2}\). A line through \(D(2,0)\) parallel to \(AB\) will also have a slope of \(\frac{1}{2}\). The equation of this line is: \(y = \frac{1}{2}(x - 2)\).
04

Find Intersection Point K

We need to find the point \(K\) where the line through \(D\) intersects \(BC\). The equation for \(BC\) can be formed using points \(B(2,1)\) and \(C(3,0)\). The slope \(m\) of \(BC\) is \(-1\), so its equation is \(y = -1(x - 3)\) or \(y = -x + 3\). Setting the two line equations equal to find intersection:\[\frac{1}{2}(x - 2) = -x + 3.\]Solving for \(x\): Multiply both sides by 2: \(x - 2 = -2x + 6\) leading to \(3x = 8\), so \(x = \frac{8}{3}\).Substitute in \(y = -x + 3\):\(y = -\frac{8}{3} + 3 = \frac{1}{3}\).Thus, \(K\left(\frac{8}{3}, \frac{1}{3}\right)\).
05

Calculate the Area of Triangle BDK

Using vertices \(B(2,1), D(2,0), K\left(\frac{8}{3}, \frac{1}{3}\right)\), compute the area using the triangle area formula:\[\text{Area}{_{BDK}} = \frac{1}{2} | 2(0 - \frac{1}{3}) + 2(\frac{1}{3} - 1) + \frac{8}{3}(1 - 0) |.\]Simplifying:\[= \frac{1}{2} |-\frac{2}{3} - \frac{2}{3} + \frac{8}{3}| = \frac{1}{2} |\frac{8}{3} - \frac{4}{3}| = \frac{1}{2} * \frac{4}{3} = \frac{2}{3}.\]
06

Compute the Product of the Areas

Multiply the areas of \(\triangle ABC\) and \(\triangle BDK\):\[\text{Area}_{ABC} \times \text{Area}_{BDK} = \frac{3}{2} \times \frac{2}{3} = 1.\]

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Key Concepts

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

Area of Triangle
Determining the area of a triangle using its vertices is a fundamental concept in coordinate geometry. In this approach, you use the coordinates of the triangle's vertices to compute the area using a specific formula: \[\text{Area} = \frac{1}{2} | x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) |.\]
  • This formula simplifies the process of finding the area without relying on a base and height, especially when dealing with axes.
  • For triangle \(\triangle ABC\) with vertices \( A(0,0), B(2,1), C(3,0)\), substitute these points into the formula to find the area.
  • Perform the calculations: \(\text{Area}_{ABC} = \frac{1}{2} | -3 | = \frac{3}{2}\).
The result of \(\frac{3}{2}\) gives the area of the triangle \(\triangle ABC\) in square units calculated using coordinates.
Slope of Line
The slope of a line is a measure of its steepness. It is calculated as the "rise" over the "run" between two points on the line. The formula for the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[m = \frac{y_2 - y_1}{x_2 - x_1}.\]
  • In the problem, we calculate the slope of line \(AB\) using points \(A(0,0)\) and \(B(2,1)\):
  • \(m = \frac{1 - 0}{2 - 0} = \frac{1}{2}\).
This slope tells us that for every unit the line moves horizontally, it moves 0.5 units vertically. Knowing the slope is essential to finding parallel lines and calculating equations of lines.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves studying geometry using a coordinate system. This branch of mathematics allows us to use algebraic techniques to handle geometric problems. In the context of this problem:
  • We use coordinates to define vertices of triangles, making it easier to perform calculations like finding distances, midpoints, and areas.
  • Coordinate geometry helps solve problems where geometry meets algebra, such as finding intersection points of lines or determining parallel and perpendicular lines.
By employing coordinate geometry, we're able to systematically find intersection points and compute areas, enhancing our problem-solving ability in geometry.
Triangle Altitude
An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side (or the line that contains the opposite side). Altitudes are crucial because they help in calculating areas of triangles where height is involved:
  • For \(\triangle ABC\), the altitude \(BD\) is perpendicular to the base \(AC\), simplifying the calculation as it lies on the x-axis.
  • Since \(AC\) is on the x-axis, the altitude runs vertically from \(B(2,1)\) straight down to the x-axis at \(D(2,0)\).
Understanding altitudes allows us to break down complex geometric shapes into simpler parts, especially when dealing with triangular regions on a coordinate plane.

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Most popular questions from this chapter

The line parallel to the \(x\)-axis and passing through the intersection of the lines \(a x+2 b y+3 b=0\) and \(b x-2 a y\) \(-3 a=0\), where \((a, b) \neq(0,0)\) is (A) below the \(x\)-axis at a distance of \(\frac{3}{2}\) from it (B) below the \(x\)-axis at a distance of \(\frac{2}{3}\) from it (C) above the \(x\)-axis at a distance of \(\frac{3}{2}\) from it (D) above the \(x\)-axis at a distance of \(\frac{2}{3}\) from it

The equation of the line passing through the point (2, 3) and making intercept of length 2 units between the lines \(y+2 x=3\) and \(y+2 x=5\), is (A) \(x=2\) (B) \(3 x+4 y=18\) (C) \(4 x+3 y=18\) (D) none of these

A line cuts the \(x\)-axis at \(A(7,0)\) and \(y\)-axis at \(B(0,-5)\). A variable line \(P Q\) is drawn \(\perp\) to \(A B\) cutting the \(x\)-axis in \(P\) and the \(y\)-axis in \(Q .\) If \(A Q\) and \(B P\) intersect at \(R\), then the locus of \(R\) is (A) \(x(x-7)+y(y+5)=0\) (B) \(x(x-7)-y(y+5)=0\) (C) \(x(x+7)+y(y-5)=0\) (D) none of these

The four points \(A(p, 0), B(q, 0), C(r, 0)\) and \(D(s, 0)\) are such that \(p, q\) are the roots of the equation \(a x^{2}+2 h x+\) \(b=0\) and \(r, s\) are those of equation \(a^{\prime} x^{2}+2 h^{\prime} x+b^{\prime}=0\). If the sum of the ratios in which \(C\) and \(D\) divide \(A B\) is zero, then (A) \(a b^{\prime}+a^{\prime} b=2 h h^{\prime}\) (B) \(a b^{\prime}+a^{\prime} b=h h^{\prime}\) (C) \(a b^{\prime}-a^{\prime} b=2 h h^{\prime}\) (D) none of these

A straight line through the origin \(O\) meets the parallel lines \(4 x+2 y=9\) and \(2 x+y+6=0\) at points \(P\) and \(Q\), respectively. The point \(O\) divides the segment \(P Q\) in the ratio (A) \(1: 2\) (B) \(3: 4\) (C) \(2: 1\) (D) \(4: 3\)
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