Points (-1, 1) and (2, -4)
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
\( y-1=\dfrac{-4-1}{2-(-1)}(x-(-1))\)
y – 1 = \( \dfrac{-5}{3}\) (x + 1)
3 (y – 1) = (-5) (x + 1)
3y – 3 = -5x – 5
3y – 3 + 5x + 5 = 0
5x + 3y + 2 = 0
The equation of the line is 5x + 3y + 2 = 0.
Answered by Abhisek | 1 year agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.
Find the angles between pairs of straight lines 3x – y + 5 = 0 and x – 3y + 1 = 0
Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0