![]() ![]() If the triangle defined by the three vertices is a right triangle, the hypotenuse is the side with the largest size and that is D2 = √ (10). We now use the distance formula to find these distances.ĭ2 = √ = √ (10) Let D1 be the distance from (-2, 1) to (1, 1), D2 be the distance from (-2, 1) to (1, 2) and D3 the distance from (1, 1) to (1, 2). Let us find the sum of the squares of D1 and D2 and the square of D3ĭ1, D2 and D3 satisfy Pythagorean theorem and therefore the triangle, whose vertices are the three points given above, is a right and isosceles triangle. For the triangle to be a right triangle, D3 the largest of the three distances must be the length of the hypotenuse and all three sides must satisfy Pythagorean theorem. Note that two sides of the triangle have equal lengths so the triangle is isosceles. We now use the distance formula to find the above distances. Let us first define the distances D1, D2 and D3 as follows The point that equidistant from the two given points is given by Use distance formula to find distance the D1 from (0, y) to (4, -9) and the distance D2 from (0, y) to (0, -2). Simplify the expression under the square root Use the distance formula to write an equation in x.ĥ = srqt ![]() We now use the distance formula to find the distance between the points (-1, -3) and (3, 5) We first find the coordinates of the midpoint M of the segemnt joining (2, 4) and (4, 6) The formula for the distance D between two points (a, b) and (c, d) is given byĪpply the formula given above to find distance D between the points (2, 3) and (0, 6) as follows Show that the triangle that has (0, 1), (2, 3) and (2, -1) as vertices is right isosceles.įind the length of the hypotenuse of the right triangle whose vertices are given by the points (-2, 1), (1, 1) and (1, 2).įind a relationship between x and y so that the distance between the points (x, y) and (-2, 4) is equal to 5.įind a relationship between x and y so that (x, y) is equidistant from the two points (-3, 4) and (0, -3).įind a relationship between x and y so that the triangle whose vertices are given by (x, y), (1, 1) and (5, 1) is a right triangle with the hypotenuse defined by the points (1, 1) and (5, 1). A set of distance problems with detailed solutions (at the bottom of this page) are presented.įind the distance between the points (2, 3) and (0, 6).įind the distance between point (-1, -3) and the midpoint of the line segment joining (2, 4) and (4, 6).įind x so that the distance between the points (-2, -3) and (-3, x) is equal to 5.įind x and y if (2, 5) is the midpoint of points (x, y) and (-5, 6).įind the point (0, y) that is equidistant from (4, -9) and (0, -2). The concept of the distance between two points is an important one in mathematics. ![]()
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