Equation of line solution12



In this page equation of line solution12 we are going to see solution of each problem with detailed explanation of the worksheet slope of the line.

(8) Find the equation of the straight line parallel to the line 3x – y + 7 =0 and passing through the point (1,-2)

Solution:

The required line is parallel to the line 3x – y + 7 = 0 and passing through the point (1,-2)

Slope of required line = -3/(-1)

                                   = 3

Equation of required line:

               (y - y1)= m (x - x1)

              (y – (-2)) = 3 (x – 1)

               y + 2 = 3 x – 3

             3x – y – 3 – 2 = 0

             3x – y – 5 = 0


(9)Find the equation of the straight line perpendicular to the straight line x – 2 y + 3 = 0 and passing through the point (1,-2).

Solution:

The required line is perpendicular to the straight line x – 2 y + 3 = 0 and passing through the point (1,-2)

Slope of the given line (m1) = -1/(-2)

                                              = 1/2

Slope of the required line = -1/(1/2)

                                         = -2

Equation of required line:

               (y - y1) = m (x - x1)

              (y – (-2)) = -2 (x – 1)

               y + 2 = -2 x + 2

             2x + y + 2 - 2 = 0

             2x + y = 0


(10) Find the equation of the perpendicular bisector of the straight line segment joining the points (3, 4) and (-1, 2).

Solution:

The required line being a perpendicular bisector of the straight line segment joining the points (3, 4) and (-1, 2)

Midpoint of the line segment = (x₁+x₂)/2 , (y₁+y₂)/2

                                              = (3 + (-1))/2,(4 + 2)/2

                                              =  2/2 , 6/2

                                              = (1,3)

Slope of the line segment joining the points (3, 4) and (-1, 2)

                                              = (y₂ - y₁)/(x₂ – x₁)

                                              = (2 – 4)/(-1-3)

                                              = -2/(-4)

                                              = ½

Equation of the required line:

(y – y1) = m (x – x1)

(y – 3) = (½)(x – 1)

2(y – 3) = 1(x – 1)

2 y – 6 = x – 1

x – 2 y – 1 + 6 = 0

x – 2 y + 5 = 0

equation of line solution12  equation of line solution12