Question

Unit 3 homework 6 Gina Wilson

Answer 1

Answer:

5) The equation of the straight line is   2 x - y + 1 =0

6) The equation of the straight line is   x + y -5 =0

7) The equation of the straight line is   5 x + 6 y - 24 =0

8) The equation of the straight line is  x - 4 y -4 =0

9) The equation of the parallel line is 3x + y -19 =0

Step-by-step explanation:

5)

The equation of the straight line is

                         $y - y_{1} = m ( x - x_{1} )$

          Slope of the line

                      $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

        Given points are (1,3) , ( -3,-5)

         $m = \frac{-5-3 }{-3-1 } = \frac{-8}{-4} = 2$

The equation of the straight line is

                         $y - y_{1} = m ( x - x_{1} )$

                        $y - 3 = 2 ( x - 1 )$

                        y = 2x - 2 +3

                       2 x - y + 1 =0

The equation of the straight line is   2 x - y + 1 =0

  6)

The equation of the straight line is

                         $y - y_{1} = m ( x - x_{1} )$

          Slope of the line

                      $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

        Given points are (1,4) , ( 6,-1)

         $m = \frac{-1-(4) }{6-1 } = \frac{-5}{5} = -1$

The equation of the straight line is  

                         $y - y_{1} = m ( x - x_{1} )$

                        $y - 1 = -1 ( x - 4 )$

                        y - 1 = - x +4

The equation of the straight line is   x + y -5 =0

7)

The equation of the straight line is

                         $y - y_{1} = m ( x - x_{1} )$

          Slope of the line

                      $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

 Given points are (-12 , 14) , ( 6,-1)

         $m = \frac{-1-(14) }{6+12 } = \frac{-15}{18} = \frac{-5}{6}$

         $y - 14 = \frac{-5}{6} ( x - (-12) )$

       6( y - 14 ) = - 5 ( x +12 )

      6 y - 84 = - 5x -60

       5 x + 6 y  -84 + 60 =0

      5 x + 6 y - 24 =0

8)

The equation of the straight line is

                         $y - y_{1} = m ( x - x_{1} )$

          Slope of the line

                      $m = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Given points are (-4 , -2) , ( 4 , 0)

              $m = \frac{0+2}{4 +4} = \frac{2}{8} = \frac{1}{4}$

           $y - (-2) =\frac{1}{4} ( x - (-4) )$

             4 ( y + 2) = x + 4

                x - 4 y -4 =0

  9)

The equation of the line y = 3x + 6  is parallel to the line

3x + y + k =0 is passes through the point ( 4,7 )

⇒   3x + y + k =0

⇒    12 + 7 + k =0

⇒    k = -19

The equation of the parallel line is 3x + y -19 =0