Saved Bookmarks
| 1. |
The base of the equilateral triangle has an equation of x+2y=3 and one of the vertex is (1,1).find the equation of other two sides |
|
Answer» Answer: Let ABC be the given EQUILATERAL triangle with side 2a. Accordingly, AB = BC = CA = 2a = Assume that BASE BC lies along the y axis such that the mid-point of BC is at the origin. j.e., Oa, where O is the origin. Now, it is clear that the coordinates of point Care (0, a), while the coordinates of point B are (0, -a). It is known that the line joining a vertex of an equilateral triangle wi f the mid-point of its opposite side perpendicular. Hence, vertex A lies on the y-axis. On applying Pythagoras theorem to AOC, we OBTAIN (AC)2 = (OA)2 + (OC)2 - (2a)2 = (OA)2 + a2 4a2 -a2 = (OA)? (A)2 = 3a2 = OA = 3a = :Coordinates of point A = (=30,0) THUS, the vertices of the given On applying Pythagoras theorem to AOC, we obtain (AC)2 = (OA)2 + (OC)2 = (2a)2 = (A)2 + a2 4a2 -a2 = (OA)2 (A)2 = 3a2 = - OA = 3a :Coordinates of point A = (+V3a,0) Thus, the vertices of the given equilateral triangle are (0, a), (0, -a), (V3a, 0) and or (0, a), (0, -a), and (-13a, 0). Hope it was helpfull |
|