Question

Write the equation of the line that is perpendicular to the line y = 2x + 2 and passes through the point (6, 3).

y = 2x + 6

y = −one halfx + 3

y = −one halfx + 6

y = 2x + 3

Answer 1

@DebbieG @Luigi0210

Answer 2

slope of required line is -1/2 so equation of line (y-3)=-1/2(x-6) so 2(y-3)=-(x-6)
that is 2y=-x+6 that is y=-1/2x+6

Answer 3

How do you get that ?

Answer 4

Since, is perpendicular to the line y = 2x + 2. therefore the slope of the r3equired line is given as \[m=\frac{-1}{2}\]
Since required line passes through the point (6, 3). therefore x1= 6 and y1= 3
Equation of line in slope point form is given as:
(y-y1)=m(x-x1)
(y-3)=-1/2(x-6)
2(y-3)=-1(x-6)
2y-6=-x+6
2y = -x+12
y= -1/2x+6
is the required eq of the line
Thus y = −one halfx + 6 is the correcxt Answer

Answer 5

@Sceenie

Answer 6

Thank you .