Advertisements
Advertisements
प्रश्न
A (5, 4), B (–3,–2) and C (1,–8) are the vertices of a triangle ABC. Find the equation of median AD and line parallel to AB passing through point C.
Advertisements
उत्तर
It is given that A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a triangle ABC.
Let the coordinates of point D be (a, b).
Also, D is the mid-point of side BC
So,
\[a = \frac{- 3 + 1}{2}, b = \frac{- 2 + \left( - 8 \right)}{2}\]
\[ \Rightarrow a = \frac{- 2}{2}, b = \frac{- 10}{2}\]
\[ \Rightarrow a = - 1, b = - 5\]
Thus, the coordinates of point D are (−1, −5).
The equation of median AD is
\[y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right)\left( x - x_1 \right)\]
\[ \Rightarrow y - 4 = \left( \frac{- 5 - 4}{- 1 - 5} \right)\left( x - 5 \right)\]
\[ \Rightarrow y - 4 = \left( \frac{- 9}{- 6} \right)\left( x - 5 \right)\]
\[ \Rightarrow y - 4 = \frac{3}{2}\left( x - 5 \right)\]
\[ \Rightarrow 2y = 3x - 7\]
A line parallel to AB will have the same slope as AB.
Slope of AB = \[\frac{y_2 - y_1}{x_2 - x_1} = \frac{- 2 - 4}{- 3 - 5} = \frac{- 6}{- 8} = \frac{3}{4}\]
So, the equation of line parallel to AB and passing through the point C(1, –8) is
\[y - y_1 = m\left( x - x_1 \right)\]
\[ \Rightarrow y - \left( - 8 \right) = \frac{3}{4}\left( x - 1 \right)\]
\[ \Rightarrow 3x - 4y - 35 = 0\]
