What is the equation of a line that passes through the point (4,2) and is perpendicular to the line whose equation is y = x3- 1Question 6 options:y = -3x + 14y = x3+ 14y = 3x + 23y = -3x + 23Question 7 (1 point) Question 7 UnsavedWhat is the equation of a line that passes through the point (4,2) and is parallel to the line whose equation is y = x3- 1Question 7 options:y = -3x + 14y = x3+ 14y = 3x - 23y = -3x - 23
Accepted Solution
A:
Part A) the correct question is What is the equation of a line that passes through the point (4,2) and is perpendicular to the line whose equation is y = (x/3)- 1
we have that y = (x/3)- 1
step 1 find the slope m m=(1/3) we know that if two lines are perpendicular so the slopes m1*m2=-1 then (1/3)*m2=-1 m2=-3
step 2 with m=-3 and the point (4,2) find the equation of the line y-y1=m*(x-x1)------> y-2=-3*(x-4)----> y-2=-3x+12----> y=-3x+14
the answer part A) is y=-3x+14
Part B) the correct question is What is the equation of a line that passes through the point (4,2) and is parallel to the line whose equation is y = (x/3)- 1
we have that y = (x/3)- 1
step 1 find the slope m m=(1/3) we know that if two lines are parallel so the slopes m1=m2 then m2=(1/3)
step 2 with m=(1/3) and the point (4,2) find the equation of the line y-y1=m*(x-x1)------> y-2=(1/3)*(x-4)----> y-2=(1/3)x-(4/3) y=(1/3)x-(4/3)+2-----> y=(1/3)x+(2/3)