Geometry: Distance Formula
This is a free lesson from our course in Geometry
 
   
Example 4: Given- the vertices of a triangle are, (3, 8), (5, 3), (4, –2).. Classify the triangle as isosceles, equilateral or scalene.
You’ll determine first the distances between the vertices.
Step 1: Distance between (3, 8) and (5, 3) is given by:
            D =
            = [2 2 + (-5)2] = (4 + 25) = 29 i.e. the length of one side is 29
Step 2: Similarly the distance for second and third side can be found by plugging given values of coordinates and it equals to, 26 and 82 respectively.
Step 3: Thus the lengths of the sides are 29, 26, 82.
Since lengths of the sides are unequal, the triangle is scalene, as the final answer. (More text below video...)
<h2> Distance Formula - Watch video</h2> <p> triangle, calculate, pythagorean theorem, geometry, formula, point, hypotenuse, line segment, video, right triangle, coordinates, example, pythagrean, distance formula, theorem, practice questions, quizzes </p> <p> The distance between two points can be calculated by drawing a right triangle in the coordinate plane using the two points as the endpoints of the hypotenuse. Length of the line segment whose coordinates are (3, -2) and (7, 3), is 6.4 units. </p>
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(Continued from above) Example 5: Given- find the radius of a circle, given that the coordinates of center are (2, –3) and the point (–1, –2) lies on the circle.
Recall the definition of radius- it is the distance between the center and any point on the circle. To calculate radius, therefore, find the distance between center and the point (say D):
Step 1: D =  
               = (52 + 0) = 52. , it equals to 5.
The radius of circle is 5, as the final answer.
Example 6: Given- find the points (5, x) that are 8 units from the point (–3, –1).
Step 1: Plug in the values of coordinates in the distance formula. D =
                    = [(-8)2 + (-1-x)2]
                     = [64 + (1 + 2x + x2)]
                     = (65 + 2x + x2) 
Step 2: D = 8 (given). Therefore, 8 = (65 + 2x + x2 )
Step 3: Square both the sides, and solve equation for variable x.
Step 4: 64 = 65 + 2x + x2
Step 5: x2 + 2x +1 = 0
Step 6: Solving the quadratic equation gives: (x +1)2 = 0
Step 7: Simplifying it gives x = -1.
The coordinates of point are: (5, -1), as the final answer.
Example 7: Given- An observer; while surveying notes, that from point A (Figure: below) the angle of elevation of the top of a tower (CD) is (degrees) and from point B the angle of elevation is If points A, B and C (the bottom of the tower at ground level) are collinear, and the distance between A and B is D. Find the height h of the tower in terms of D and angles and .
CD is a vertical tower making right angle with the level ground. Assume that distance BC is 'A'.
Step 1: In the right angle ACD: tan () = h ('D' + 'A')
Step 2: In the right angle BCD: tan () = h /('A').
On simplifying it, 'A' = h / tan ()
Step 3: Solve tan ()= h/('D' + 'A')
      for h, i.e. h = tan ( )* ('D' + 'A')
              By substituting value of  'A', as h/ tan (), it gives 
              h =[ 'D' +h/tan ()] * tan ()
Step 4: Solving this for h gives,
            h = 'D' * tan () tan ()/ [tan () - tan ()]
Since 'D', () and () are known, their tangent value can be picked from the Trigonometric table and calculate h by plug in the known values, is the final answer. 		 
Example 8: Given- Two trees stand opposite one another, at points P and Q, on opposite banks of a river. PR is perpendicular to PQ and the distance PR is 150 ft. The measure of angle PRQ is 75, find the width of the river.
As the trees are on opposite banks, width of river is same as distance between the trees. Since PR is perpendicular to PQ, and distance PR measures 150 ft. Measure of angle PRQ is 75.
Step 1: Say width of river is w, PQR being right angle triangle:
            tan 75 = w /150
Step 2: From the trigonometric table,
            tan 75= 3.73205.
Step 3: w = 150 * tan 75 = 150 * 3.73205 = 560 ft (rounded whole number) Width of the river is 560 ft, as the final answer. 		 
Remember that you simplify inside the parentheses ‘before you square’; not after, and that the square is on everything inside the parentheses, ‘including the minus sign’, i.e. the square of a negative is a positive.
The video above will explain more in detail about, Distance Formula and how to apply the concepts in solving real-world problems. This is explained with the help of several examples and done watching video. This helps you to deal with solving problems and help doing the Geometry home work.
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