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The points represent the vertices of a triangle. (a) Draw triangle \(A B C\) in the coordinate plane, (b) find the altitude from vertex \(B\) of the triangle to side \(A C,\) and \((c)\) find the area of the triangle. \(A(-3,0), B(0,-2), C(2,3)\)

Short Answer

Expert verified
To solve this task, we first draw a triangle with the given vertices. Then we calculate the altitude from vertex B to side AC using the distance formula. With the base and altitude obtained, we then compute the area of the triangle using the formula for the area of a triangle, which is \(\frac{1}{2} base * height\).

Step by step solution

01

Draw Triangle ABC

Plotting the given vertices A(-3,0), B(0,-2), and C(2,3) on a coordinate plane, it's clear to see that these points form triangle ABC.
02

Find the Altitude from Vertex B to side AC

The altitude of a triangle can be found using the distance formula. In this case, the altitude from vertex B to side AC is given by the perpendicular distance from point B to the line passing through A and C. The equation of the line AC is given by \(y-y1 = m(x-x1)\), where m is the slope of AC, which is \(\frac{(y2-y1)}{(x2-x1)} = \frac{(3-0)}{(2--3)} = \frac{3}{5}\). After finding the equation of line AC, substitute the coordinate of B into the equation to find the distance d, which is our desired altitude. Recall that the distance between a point and a line in 2D is given by the formula \(d = \frac{|Ax_{0}+By_{0}+C|}{\sqrt{A²+B²}}\), where (x₀, y₀) is the point and Ax+By+C=0 is the line.
03

Find the Area of The Triangle

After we have located the altitude, we can then proceed to calculate the area. The area of a triangle is given by half the product of the base and the height (altitude). In this problem, we consider AC as our base and the distance we calculated in step 2 as our altitude. So, Area = \(\frac{1}{2} base * height\).

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Key Concepts

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

Coordinate Geometry
Coordinate geometry is a branch of mathematics that explores the relationship between geometry and algebra using a coordinate system. When dealing with triangles on a coordinate plane, we locate each vertex using an ordered pair of numbers, a coordinate. These numbers pinpoint the exact location of a point, such as a vertex of a triangle, on a plane.
To solve problems involving triangles in coordinate geometry, you often plot the points representing the triangle's vertices on the Cartesian plane. This involves identifying the horizontal (x-axis) and vertical (y-axis) positions of each vertex. For example, given points A(-3,0), B(0,-2), and C(2,3), these coordinates are used to plot triangle ABC. Note how each coordinate gives us the x and y value of each point, essential for visual and analytical purposes.
Triangle Altitude
In geometry, an altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. This segment represents the shortest distance from the vertex to the base of a triangle. In the context of coordinate geometry, finding the altitude involves calculations using coordinates and line equations.
For triangle ABC, the altitude from vertex B to side AC requires determining the perpendicular distance from point B to the line formed by A and C. First, we need to find the equation of line AC, using its slope which can be calculated as \( m = \frac{y2-y1}{x2-x1} \), giving us \( \frac{3}{5} \). With the slope, the line equation AC is determined, enabling us to substitute B's coordinates into a formula to find this perpendicular distance (the altitude).
Triangle Area Calculation
Calculating the area of a triangle on a coordinate plane requires knowing the base and the height (altitude) of the triangle. The formula is straightforward: the area is half of the product of the base and the height. When the triangle is plotted on a coordinate plane, any side can be designated as the base.
In the case of triangle ABC, after determining the length of AC (the base), and the altitude from B, you can apply the formula: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \.\) This formula ensures you consider the two-dimensional space the triangle occupies, allowing you to find its specific area by integrating both its dimensions.
Distance Formula
The distance formula is a formula used to find the distance between two points on a coordinate plane. It is derived from the Pythagorean theorem and is essential for various calculations in coordinate geometry, including finding the altitude of triangles.
Given two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance, d, between these points is calculated by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].
This formula enables us to determine not only the length of a side of a triangle but is also adapted into the formula to find the perpendicular distance from a point to a line, crucial for determining altitudes in triangles such as the altitude from vertex B to side AC in triangle ABC.

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Most popular questions from this chapter

Each cable of the Golden Gate Bridge is suspended (in the shape of a parabola) between two towen that are 1280 meters apart. The top of each tower is 152 meters above the roadway. The cables touch the roadway midway between the towers. (a) Draw a sketch of the bridge. Locate the origin of a the origin of a rectangular coordinate system at the center of the roadway. Label the coordinates of the known points. (b) Write an equation that models the cables. (c) Complete the table by finding the height y of the suspension cables over the roadway at a distance of \(x\) meters from the center of the bridge.

Using a Graphing Utility to Find Rectangular Coordinates In Exercises \(35-42,\) use a graphing utility to find the rectangular coordinates of the point given in polar coordinates. Round your results to two decimal places. $$(8.25,3.5)$$

Satellite Dish The parabolic cross section of a satellite dish can be modeled by a portion of the graph of the equation $$x ^ { 2 } - 2 x y - 27 \sqrt { 2 } x + y ^ { 2 } + 9 \sqrt { 2 } y + 378 = 0$$ where all measurements are in feet. (a) Rotate the axes to eliminate the \(x y\) -term in the equation. Then write the equation in standard form. (b) A receiver is located at the focus of the cross section. Find the distance from the vertex of the cross section to the receiver.

Projectile Motion A projectile is launched at a height of \(h\) feet above the ground at an angle of \(\theta\) with the horizontal. The initial velocity is \(v_{0}\) feet per second, and the path of the projectile is modeled by the parametric equations $$ \begin{array}{l}{x=\left(v_{0} \cos \theta\right) t} \\ {\text { and }} \\\ {y=h+\left(v_{0} \sin \theta\right) t-16 t^{2}}\end{array} $$ In Exercises 93 and \(94,\) use a graphing utility to graph the paths of a projectile launched from ground level at each value of \(\theta\) and \(v_{0 .}\) . For each case, use the graph to approximate the maximum height and the range of the projectile. \(\begin{array}{ll}{\text { (a) } \theta=60^{\circ},} & {v_{0}=88 \text { feet per second }} \\ {\text { (b) } \theta=60^{\circ},} & {v_{0}=132 \text { feet per second }} \\ {\text { (c) } \theta=45^{\circ},} & {v_{0}=88 \text { feet per second }} \\ {\text { (d) } \theta=45^{\circ},} & {v_{0}=132 \text { feet per second }}\end{array}\)

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