What is an equation of the line that is perpendicular to y+1=-3(x-5)

 3y - x - 42 = 0

Step-by-step explanation:

The equation                         y  +  1 = -3(x - 5)

Now expand the expression by opening the brackets

                                              y + 1 = -3x + 15

Recall,

Equation of line , y = mx + c , where m = slope

rearranging the equation above in order to determine the  ( m)

                                             y = -3x -1 + 15

                                             y = -3x + 14

Recall again  condition for perpendicularity

                                             m₁m₂ = -1

From the equation given, m₁ = -3, therefore , to get m₂

                                           -3 x m₂ = -1

                                            -3m₂    = -1

Therefore                               m₂    = -1/-3

                                               m₂   = 1/3

Now , to find the equation of the second line parallel to the first one,

Just replaced the -3 in the equation above by  m₂ = 1/3 in

               y = -3x + 14

               y = x/3 + 14

Now , multiply both the RHS and LHS of the equation by 3, to make it a linear.

              3y = x + 3 x 14

               3y = x + 42

              3y - x - 42 =0

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

she gotta workk that a 2

step-by-step explanation: