Question

Write an equation of a line in slope-intercept form that passes through (-3,5) and is parallel to y = -6x +1.

Answer 1

Answer:

a) The equation of the Parallel line to the given straight line is

       6 x + y + 13 =0

b) Slope - intercept form

        y = - 6 x - 13

c) The intercept - form

            $\frac{x}{\frac{-13}{6} } + \frac{y}{-13} = 1$

x - intercept  =  $\frac{-13}{6}$

y - intercept = - 13

Step-by-step explanation:

Step(i):-

Given the equation of the straight line

                             y = -6x +1

                  6 x + y - 1 = 0

The equation of the Parallel line to the given straight line is

                6x + y + k=0 and it passes through the point (-3, 5 )

           ⇒ 6 (-3 ) + 5 + k =0

          ⇒  - 18 + 5 + k=0

         ⇒   -13 + k = 0

        ⇒     k = 13

The equation of the Parallel line to the given straight line is

       6 x + y + 13 =0

Step(ii):-

Slope - intercept form

                y = m x + C

                y = - 6 x - 13

Step(iii):-

Intercept - form

              6 x + y + 13 =0

              6 x + y = - 13

              $\frac{6x + y}{-13} = \frac{-13}{-13}$

             $\frac{6x}{-13} + \frac{y}{-13} = 1$

             $\frac{x}{\frac{-13}{6} } + \frac{y}{-13} = 1$

The intercept - form

            $\frac{x}{\frac{-13}{6} } + \frac{y}{-13} = 1$

x - intercept  =  $\frac{-13}{6}$

y - intercept = - 13