The angle between two lines

Equation of lines in different conditions

1. Find the angle between two lines 6x + 2y - 7 = 0 and 3x + 6y + 5 = 0 slope of line 6x + 2y – 7 = 0: m 1 = -5/2 = -2.5 slope of line 3x + 6y + 5; m 2 = -3/6 = -0.5 Angle = 45 and 135

2. Find the equation of straight line which is parallel to the line with equation 5x + 7y = 14 and passes through the point (-2, -3) 2068R-I Equation of the line parallel to the line 5x + 7y = k……………….( i ) Since this line passes through the point (-2, -3), we have 5.(-2) + 7.(-3) = k Or, K = -31 From equation ( i ) requires equation of straight line is 5x + 7y = 31 5x + 7y – 31 = 0

3. Find the equation of straight line passing through the point (2, 1) and is parallel to the line joining the points (2, 3) and (3, -1). SLC2065S Equation of the straight line joining two points (2, 3) = (x 1 , y 1 ) and (3, -1) = (x 2 , y 2 ) is y – y 1 = (y 2 – y 1 ).(x - x 1 )/(y 2 – y 1 ) Using this formula, 4x + y – 11 = 0……………( i ) Equation of the line parallel to 4x + y – 11 = 0 is 4x + y + k = 0………..(ii) Since the requires line (ii) passes through the point (2, 1), we have 4.(2) + 1+ k = 0 or, k = -9 Then from equation (ii), requires equation of straight line is 2x + y – 7 = 0

4. NEXT WAY

5. Slope of the line joining the points (2, 3) and (3, -1) ; m 1 = -4 Let, the slope of the line parallel to this line is m 2 . Then m 1 . m 2 = - 4 So the equation of the line having slope m 2 and passing through the point (2,1) = (x 1 ,y 1 ) is y – y 1 = m 2 (x – x 1 ) y – 1 = -4(x – 2) 4x + y – 9 = 0 Find the equation of straight line passing through the point (2, 1) and is parallel to the line joining the points (2, 3) and (3, -1). SLC2065S

6. Find the equation of straight line passing through the point (2, 3) and perpendicular to the line 4x - 3y = 10. SLC2057R Equation of the line perpendicular to 4x – 3y = 10 is 3x + 4y = k-----------------( i ) Since this line passes through the point (2, 3), we have 3.(2) + 4.(3) = k Or, K = 18 Equation ( i ) becomes 3x + 4y = 18 3x + 4y – 18 = 0 Which is the required equation of the line

7. Find the equation of the straight line through (2, -1) and is perpendicular to the line joining the two points (3, -1) and (1, 3). Equation of the straight line joining two points (3, -1) = (x 1 , y 1 ) and (1, 3) = (x 2 , y 2 ) is y – y 1 = (y 2 – y 1 ).(x - x 1 )/(y 2 – y 1 ) Using above formula, 2x + y – 5 = 0……….( i ) Equation of the line perpendicular to 2x + y = 0 is x – 2y + k = 0…………..(ii) Since the line requires line (ii) passes through the point (2, -1), we have 2 – 2.(-1) + k = 0 Or, k = - 4 Then from equation (ii), requires equation of straight line is x – 2y – 4 = 0