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: m1 = -5/2 = -
2.5
slope of line 3x + 6y + 5; m2 = -3/6 = -0.5
Angle = 450 and 1350

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) =
(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 = 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) ; 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

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) = (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