Perimeter of a triangle using coordinates

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Perimeter of a triangle using coordinates

Instructional video Archived teaches Common Core State Standards 6.G.A.3 Instructional video Additional materials About this video In this lesson you will learn how to find the length of sides and then use that to find perimeter and area by comparing coordinates. As Andr? Nicolas and Ross Millikan have said there are usually an infinite number of solutions (unless you have an equilateral triangle or there are no solutions). For example, among triangles with perimeter $98$ and area $420$ are those with sides $24,37,37$ and $25,34,39$ and $29,29,40$. The following picture comes from my page called Triangles with the same area and perimeter If you want to find a solution, note that there are usually two isosceles triangles with given perimeter and area , which are easily found. Another related article worth reading is Angles, Area, and Perimeter Caught in a Cubic by George Baloglou and Michel Helfgott, published in Forum Geometricorum, Volume 8 (2008) 13?25 You might be familiar with determining the area and perimeter of twodimensional shapes. However, it may seem like a slightly different task when presented on the coordinate plane. Example #1 Determine the perimeter and area of the rectangle below. Notice that the lengths are not given. Instead, you must use the graph to determine the information. Counting will help you to determine the lengths of the sides. Now that you have the lengths of all the sides, you can add them to get the perimeter. P = 10 + 10 + 11 + 11 P = 42 units You can also use the lengths to calculate the area of the rectangle. For a rectangle, the area is equal to the length times the width. A = lw A =(10 units)(11 units) A = 110 units2 The other option, although quite tedious, would be to count all the squares inside the rectangle. If you were to do so, you would notice that there are 110 squares. Therefore, the area is 110 square units. Example #2 In this case, be sure to count the lengths and not actual squares when determining the lengths of each side. Even though 12 whole squares do not fit across the base of the triangle, there are 12 lengths. It is impossible to determine the length of the longest side from the graph. This is one of the downfalls to being given the information on a coordinate plane. The Pythagorean Theorem can be used to calculate the third side. (Remember that the longest side must be labeled as c in the formula a2 + b2 = c2.) a2 + b2 = c2 122 + 102 = c2 144 + 100 = c2 244 = c2 244 = c 15.6 c This is the approximate length of the third side of the triangle. Now we can determine the approximate perimeter of the triangle. P = 10 + 12 + 15.6 P = 37.6 units For the area, we can use the formula A = ? bh. Be sure to use the base and height that meet at a right angle. A = ? bh A = ? (12 units)(10 units) A = 60 units2 Example #3 Determine the perimeter and area of the irregular figure. Start with the perimeter. First, determine the lengths of all the pieces. Then add the lengths together to get the perimeter. P = 8 + 4 + 3 + 13 + 3 + 2 + 2 + 3 + 6 + 16 P = 60 units For the area, start by chopping up the figure into rectangles. This shape can be divided up many different ways. Here is one possibility. Rectangle #1 A = lw A = (13 units)(3 units) A = 39 units2 Rectangle #2 A = lw A = (3 units)(2 units) A = 6 units2 Rectangle #3 A = lw A = (16 units)(8 units) A = 128 units2 Next, add the areas of all the pieces to get the total area of the shape. Total Area = 39 + 6 + 128 Total Area = 173 units2 Let's Review When two-dimensional figures are shown on the coordinate plane, a mix of counting and the Pythagorean Theorem can be used to determine the lengths of each side. Then add up the lengths to determine the perimeter or use the basic area formulas for triangles and rectangles to determine the area of the figure. Related Links: Math Geometry Area of a Circle Perimeter of Parallelogram To link to this Area and Perimeter on the Coordinate Plane page, copy the following code to your site: Area and Perimeter on the Coordinate Plane In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. In Geometry, a triangle is a three-sided polygon that has three edges and three vertices. The area of the triangle is the space covered by the triangle in a two-dimensional plane. The formula for the area of a triangle is (1/2) ? base ? altitude. Let's find out the area of a triangle in coordinate geometry. What Is the Area of a Triangle in Coordinate Geometry? Coordinate geometry is defined as the study of geometry using coordinate points. The area of a triangle in coordinate geometry can be calculated if the three vertices of the triangle are given in the coordinate plane. The area of a triangle in coordinate geometry is defined as the area or space covered by it in the 2-D coordinate plane. Let us understand the concept of the area of a triangle in coordinate geometry better using the example given below, Consider these three points: A(-2,1), B(3,2), C(1,5). If you plot these three points in the plane, you will find that they are non-collinear, which means that they can be the vertices of a triangle, as shown below: The area covered by the triangle ABC in the x-y plane is the region marked in blue. Now, with the help of coordinate geometry, we can find the area of this triangle. Let us learn more about it in the following section. How Do You Calculate the Area of A Triangle in Coordinate Geometry? In coordinate geometry, if we need to find the area of a triangle, we use the coordinates of the three vertices. Consider ABC as given in the figure below with vertices A(x\(_1\), y\(_1\)), B(x\(_2\), y\(_2\)), and C(x\(_3\), y\(_3\)). In this figure, we have drawn perpendiculars AE, CF, and BD from the vertices of the triangle to the horizontal axis. Notice that three trapeziums are formed: BAED, ACFE, and BCFD. We can express the area of a triangle in terms of the areas of these three trapeziums. Area(ABC) = Area(Trap.BAED) + Area(Trap.ACFE) - Area(Trap.BCFD) Now, the area of a trapezium in terms of the lengths of the parallel sides (the bases of the trapezium) and the distance between the parallel sides (the height of the trapezium): Trapezium Area = (1/2) ? Sum of bases ? Height Consider any one trapezium, say BAED. Its bases are BD and AE, and its height is DE. BD and AE can easily be seen to be the y coordinates of B and A, while DE is the difference between the x coordinates of A and B. Similarly, the bases and heights of the other two trapeziums can be easily calculated. Thus, we have: Area(Trap.BAED) = (1/2) ? (BD + AE) ? DE = (1/2) ? (y\(_2\) + y\(_1\)) ? (x\(_1\) - x\(_2\)) Area(Trap.ACFE) = (1/2) ? (AE + CF) ? EF = (1/2) ? (y\(_1\) + y\(_3\)) ? (x\(_3\) - x\(_1\)) Area(Trap.BCFD) = (1/2) ? (BD + CF) ? DF = (1/2) ? (y\(_2\) + y\(_3\)) ? (x\(_3\) - x\(_2\)) The expression for the area of the triangle in terms of the coordinates of its vertices can thus be given as, Area(ABC) = Area(Trap.BAED) + Area(Trap.ACFE) - Area(Trap.BCFD) = (1/2) ? [(y\(_2\) + y\(_1\)) ? (x\(_1\) - x\(_2\))] + (1/2) ? [(y\(_1\) + y\(_3\)) ? (x\(_3\) - x\(_1\))] - (1/2) ? [(y\(_2\) + y\(_3\)) ? (x\(_3\) - x\(_2\))] However, we should try to simplify it so that it is easy to remember. For that, we simplify the product of the two brackets in each terms: = (1/2) (x\(_1\)y\(_2\) - x\(_2\)y\(_2\) + x\(_1\)y\(_1\) - x\(_2\)y\(_1\)) + (1/2) (x\(_3\) y\(_1\) - x\(_1\)y\(_1\) + x\(_3\)y\(_3\) - x\(_1\)y\(_3\)) - (1/2)(x\(_3\)y\(_2\) - x\(_2\)y\(_2\) + x\(_3\)y\(_3\) - x\(_2\)y\(_3\)) Take the common term 1/2 outside the bracket. =(1/2) (x\(_1\)y\(_2\) - x\(_2\)y\(_2\) + x\(_1\)y\(_1\) - x\(_2\)y\(_1\) - x\(_3\) y\(_1\) - x\(_1\)y\ (_1\) + x\(_3\)y\(_3\) - x\(_1\)y\(_3\) - x\(_3\)y\(_2\) + x\(_2\)y\(_2\) - x\(_3\)y\(_3\) + x\(_2\)y\(_3\)) Thus, Area(ABC) = (1/2){x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\(_2\))} As the area is always positive. (ABC) = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\(_2\))| This is a symmetric expression, and there is an easy technique to remember it, which we will now discuss as Determinants Method. Area of a Triangle Using Determinant Method To calculate the area of a triangle using determinants, we use the formula as shown below, Area = 1/2 \(\begin{bmatrix}{{x_1}}&{{y_1}}&{{1}}\\{{x_2}}&{{y_2}}&{{1}}\\{x_3}& {y_3}&1\end{bmatrix}\) Let us solve the above expression in order to obtain the formula for the area of a triangle using coordinates. We will solve the determinant along the first column. Now, the first term in the expression for the area is \({x_1}\left( {{y_2} - {y_3}} \right)\). To obtain this, we solve determinant for the first term in the first column. Ignore the terms in the first row and column other than the first term and proceed according to the following visual representation (the cross arrows represent multiplication). Solving determinant we get, x\(_1\)(y\(_2\) - y\(_3\)). The second term in the expression for the area is x\(_2\)(y\(_3\) - y\(_1\)). To obtain this, we solve determinant for the second term in the first column. Ignore the terms in the second row and first column other than the first term in the second column. Solving determinant, we get -x\(_2\)(y\(_1\) - y\(_3\)) = x\(_2\)(\({y_3} - {y_1}\)): Next, the third term in the expression for the area is \({x_3}\left( {{y_1} - {y_2}} \right)\). To obtain this, we solve determinant for the third term in the first column. Ignore the terms in the first row and third column other than the first term in the third column: Finally, we add these three terms to get the area (and divided by a factor of 2, because we had this factor in the original expression we determined): Area = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\ (_1\) - y\(_2\))| Note that we have put a modulus sign (vertical bars) around our algebraic expression, and removed the negative sign because the area is always positive, which we obtained in the original expression. So even if we get a negative value through the algebraic expression, the modulus sign will ensure that it gets converted to a positive value. We can write the above expression for area compactly as follows: \(A = \frac{1}{2}\;\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&{{1}}\\{{x_2}}&{{y_2}}&{{1}}\\{x_3}&{y_3}&1\end{array}} \right|\) Important Notes: The area of a triangle cannot be negative. In case we get the answer in negative terms, we should consider the numerical value of the area, without the negative sign. To find the area of a triangle in coordinate geometry, we need to find the length of three sides of a triangle using the distance formula. If three points A(x\(_1\),y\(_1\)), B(x\(_2\),y\(_2\)), and C(x\(_3\),y\(_3\)) are collinear, then x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\ (_2\)) = 0. Challenging Questions Solved Examples on Area of a Triangle in Coordinate Geometry Example 1: Consider a triangle with the following vertices: A(-1,2), B(2,3), C(4,-3). Find the area of this triangle in coordinate geometry? Solution: To illustrate, we will calculate each of the three terms in the formula for the area separately, and then put them together to obtain the final value. \(\left| {\begin{array}{*{20}{c}}{ - 1}&2&4\\2&3&{ - 3}\\1&1&1\end{array}} \right|\) We will use the determinant formula to find the area of the given triangle. Area of ABC = 1/2 \(\left| {\begin{array}{*{20}{c}}{ - 1}&2&4\\2&3&{ - 3}\\1&1&1\end{array}} \right|\) Area of ABC = 1/2 |-1(3 - (-3)) - 2(2 (-3)) + 4(2 - 3)| Area(ABC) = (1/2) |(-6) - (10) + (-4)| = (1/2) ? 20 = 10 sq.units The area of a triangle is 10 unit square. Example 2: Find the area of a triangle with the vertices: A(3,4), B(4,7), and C(6,-3). Solution: We have: (ABC) = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\))+ x\(_3\)(y\(_1\) - y\(_2\))| (ABC) = (1/2) |3(7 - (-3)) + 4((-3) - (-4)) + 6(4 - (7))| = 12|30 + 4 - 18| = (1/2) ? 16 = 8sq.units Example 3: Find the area of the triangle in coordinate geometry, whose vertices are: A(1,-2), B(-3,4), C(2,3) Solution: Area of ABC = 1/2 \(\left| {\begin{array}{*{20}{c}}1&{-2}&1\\{-3}&4&1\\2&3&1\end{array}} \right|\) Area of ABC = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\ (_2\)(y\(_3\) - y\(_1\))+ x\(_3\)(y\(_1\) - y\(_2\))| (1/2)|1 ? (4 - 3) - 3 ? ((3) + 2)) + 2(-2 - 4)| = (1/2) |1 -15 - 12| =13 sq.units The area of a triangle is 13 sq.units go to slidego to slidego to slide Breakdown tough concepts through simple visuals. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Book a Free Trial Class FAQs on Area of a Triangle in Coordinate Geometry The distance formula is used to find the length of a triangle using coordinates. Distance formula can be used to find the length of any side given the coordinates of the triangle's vertices. What Is the Formula of the Area of a Triangle in Coordinate Geometry? The formula of area of triangle formula in coordinate geometry the area of triangle in coordinate geometry is: A = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\(_2\))|, where (x\(_1\),y\(_1\)),(x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the coordinates of vertices of triangle. How Do You Find the Area and Perimeter of a Triangle With Coordinates? For the area and perimeter of a triangle with coordinates first, we have to find the distance between each pair of points by distance formula and then we apply the formula for area and perimeter. How Do You Find the Area of a Triangle With 3 Coordinates? Area of triangle with 3 points is: A = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\) (y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\(_2\))|, where (x\(_1\),y\(_1\)),(x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the coordinates of vertices of triangle. How Do You Find the Area of Triangle Using Vertices? The formula of the area of triangle in coordinate geometry is: A = (1/2)|x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\(_2\))|, where (x\(_1\),y\(_1\)), (x\(_2\),y\(_2\)), and (x\(_3\),y\(_3\)) are the vertices of triangle. How Do You Find the Area of an Isosceles Triangle Using Coordinates? First, we use the distance formula to calculate the length of each side of the triangle. If two sides are equal then it's an isosceles triangle. We can apply the area of an isosceles triangle formula using the side lengths. How Do You Find the Area of a Right-Angled Triangle Using Coordinates? First, we use the distance formula to calculate the length of each side of the triangle. If the squares of the smaller two distances equal the square of the largest distance, then these points are the vertices of a right triangle. or we can use the Pythagoras theorem. We can apply the area of a right triangle formula using side lengths. How Do You Calculate the Area of a Triangle on a Graph? The area of a triangle on a graph is calculated by the formula of area which is: A = (1/2) |x\(_1\)(y\(_2\) - y\(_3\)) + x\(_2\)(y\(_3\) - y\(_1\)) + x\(_3\)(y\(_1\) - y\(_2\))|, where (x\(_1\),y\(_1\)), (x\(_2\),y\(_2\)), and (x\ (_3\),y\(_3\)) are the vertices of triangle. How Do You Find the Missing Coordinate of a Right-Angled Triangle? We use the distance formula and Pythagoras theorem to calculate the missing coordinate of a right-angled triangle.

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