Question

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

Answer 1

y = 2x + 6

y = −1/2x + 3

y = −1/2x + 6

y = 2x + 3

Answer 2

so noone wants to help? thankss everyone

Answer 3

ur given equation of line is y=2x+2

Answer 4

i m now helping u

Answer 5

so the slope of ur given line is m=2

Answer 6

u know \[m _{1}m _{2}=-1\]

Answer 7

when 2 lines r perpendicular with each other

Answer 8

so the slope of ur required line will b 2m=-1

Answer 9

m=-1/2

Answer 10

r u following me

Answer 11

y=2x+2 is the given equation
here the slope = 2
now the line which will be perpendicular to this line will have the slope = -1/2
now we got the slope
the point is (6,3)
using point slope form we can get the equation
y-y1 = m(x-x1) where m=-1/2 x1 = 6 , y1 = 3
y-3 = -1/2(x-3)

Answer 12

so the equation of ur required line will \[y=-\frac{ 1 }{ 2 }x+c\]

Answer 13

sorry its y-3 = -1/2(x-6)
=> y= -1/2x +6 -----> required equation

Answer 14

or,\[3=-\frac{ 1 }{ 2 } \times 6 +c\]

Answer 15

because ur required line will go through the point 6,3

Answer 16

c=6

Answer 17

\[y=-\frac{ 1 }{ 2 }x+6\]

Answer 18

i think u understood

Answer 19

ok

Answer 20

yes i did thank you

Answer 21

Write the equation of the line that is parallel to the line 3x − y = −3 and passes through the point (4, −2).
y = −1/3x − 6
y = −1/3x − 14
y = 3x − 6
y = 3x − 14
solution:
3x-y= -3----> given
writing in term of x we get
y= 3x+3
here the slope = 3
now the line parallel to this line will have have the same slope
thus slope = 3
point = (4, -2)
equation is
y+2 = 3(x-4)
y= 3x-14----> right ans