Advertisements
Advertisements
प्रश्न
The equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are x − y + 5 = 0 and x + 2y = 0 respectively. If the point A is (1, −2), find the equation of the line BC.
Advertisements
उत्तर
Let the perpendicular bisectors x − y + 5 = 0 and x + 2y = 0 of the sides AB and AC intersect at Dand E, respectively.
Let \[\left( x_1 , y_1 \right) \text { and} \left( x_2 , y_2 \right)\] be the coordinates of points B and C.
\[\therefore \text { Coordinates of D } = \left( \frac{x_1 + 1}{2}, \frac{y_1 - 2}{2} \right) \]
\[\text { and coordinates of E } = \left( \frac{x_2 + 1}{2}, \frac{y_2 - 2}{2} \right)\]
Point D lies on the line x − y + 5 = 0
\[\therefore \frac{x_1 + 1}{2} - \frac{y_1 - 2}{2} + 5 = 0\]
\[\Rightarrow x_1 - y_1 + 13 = 0\] ... (1)
Point E lies on the line x + 2y = 0
\[\therefore \frac{x_2 + 1}{2} + 2 \times \left( \frac{y_2 - 2}{2} \right) = 0\]
\[\Rightarrow x_2 + 2 y_2 - 3 = 0\] ... (2)
Side AB is perpendicular to the line x − y + 5 = 0
\[\therefore 1 \times \frac{y_1 + 2}{x_1 - 1} = - 1\]
\[\Rightarrow x_1 + y_1 + 1 = 0\] ... (3)
Similarly, side AC is perpendicular to the line x + 2y = 0
\[\therefore - \frac{1}{2} \times \frac{y_2 + 2}{x_2 - 1} = - 1\]
\[\Rightarrow 2 x_2 - y_2 - 4 = 0\] ... (4)
Now, solving eq (1) and eq (3) by cross multiplication, we get:
\[\frac{x_1}{- 1 - 13} = \frac{y_1}{13 - 1} = \frac{1}{1 + 1}\]
\[ \Rightarrow x_1 = - 7, y_1 = 6\]
Thus, the coordinates of B are \[\left( - 7, 6 \right)\].
Similarly, solving (2) and (4) by cross multiplication, we get:
\[\frac{x_2}{- 8 - 3} = \frac{y_2}{- 6 + 4} = \frac{1}{- 1 - 4}\]
\[ \Rightarrow x_2 = \frac{11}{5}, y_2 = \frac{2}{5}\]
Thus, coordinates of C are \[\left( \frac{11}{5}, \frac{2}{5} \right)\] .
Therefore, equation of line BC is
\[y - 6 = \frac{\frac{2}{5} - 6}{\frac{11}{5} + 7}\left( x + 7 \right)\]
\[ \Rightarrow y - 6 = \frac{- 28}{46}\left( x + 7 \right)\]
\[ \Rightarrow 14x + 23y - 40 = 0\]
