HELP PLEASE Consider the line y = 6x-7. Find the equation of the line that is perpendicular to this line and passes through the point (8, 3). Find the equation of the line that is parallel to this line and passes through the point (8, 3).

Step-by-step explanation:

y=m₁x+b₁   ⊥   y=m₂x+b₂   ⇔    m₁×m₂ = -1

{Two lines are perpendicular if the product of theirs slopes is equal -1}

y = 6x - 7   ⇒   m₁=6

6×m₂ = -1        ⇒    m₂ = -¹/₆

The line y=-¹/₆ x+b passes through point (8, 3) so the equation:

3 = -¹/₆ ×8 + b must be true

3 = -⁴/₃ + b

b = 4¹/₃

Therefore equation in  slope-intercept form:

                                                     y = -¹/₆ x + 4¹/₃

y=m₁x+b₁   ║   y=m₂x+b₂   ⇔    m₁ = m₂

{Two lines are parallel if  their slopes are equal}

y = 6x - 7   ⇒   m₁=6   ⇒   m₂=6

The line y=6x+b passes through point (8, 3) so the equation:

3 = 6×8 + b must be true

3 = 48 + b

b = -45

Therefore equation in  slope-intercept form:

                                                     y = 6x - 45

72 is the answer to this problem

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse