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

definition: a circle is the  set of all points in a plane that are equidistant  from a given  point  called the center of the circle. we use the symbol ⊙ to represent a circle. the a line segment from the center of the circle to  any point  on the circle is a radius of the circle.

step-by-step explanation:

i googled it and this is what i got i don't know how it is. sorry.