Find the angle between two lines 6x + 2y - 7 = 0 and 3x

+ 6y + 5 = 0

slope of line 6x + 2y – 7 = 0: m1 = -5/2 = -2.5

slope of line 3x + 6y + 5; m2 = -3/6 = -0.5Angle = 450 and 1350

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

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) = (x1 , y1 ) and (3, -1) = (x2 , y2 ) is

• y – y1 = (y2 – y1 ).(x - x1)/(y2 – y1 )• 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 = 0or, k = -9

• Then from equation (ii), requires equation of straight line is 2x + y – 7 = 0

•NEXT WAY

• Slope of the line joining the points (2, 3) and (3, -1) ; m1 = -4

• Let, the slope of the line parallel to this line is m2 . Then m1 . m2 = - 4

• So the equation of the line having slope m2 and passing through the point (2,1) = (x1 ,y1 ) is

• y – y1 = m2 (x – x1 )• 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

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

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) = (x1 , y1 ) and (1, 3) = (x2 , y2 ) is

• y – y1 = (y2 – y1 ).(x - x1)/(y2 – y1 )• 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