HELP ME OUT PLEASE SOLVE IT..............

Question

Find the equation of a line that meets the given condition : perpendicular to 6x-2y=10,passes through (6,-3)

Answer 1

 Perpendicular to L1: 6X -2y =10  through   ( 6, -3) 

 

 

    L1:      2Y = -6X +10      Y = -3X +5

 

 

    L2:   Y=  1/3 X + b                Equation of each line is Y = mx +b   , Since L1  l  L2  m_{2 = -}1/m₁

 

           ( 6,3 ) must satisfy the equation:

 

         - 3  = 6/3  + b    , b =-5

 

          L ₂ :   Y = 1/3 X -5

 

 

 

 

Answer 2

6x-2y=10

put the equation in the form y=mx+b, where m is the slope and b is the y-intercept

6x-2y=10

-2y=-6x+10

y=3x+(-5) or y=3x-5

the slope is 3 and the y-intercept is -5

if two lines are perpendicular to each other their slopes are negative reciprocals of each other

the slope of the perpendicular line is -1/3

y=(-1/3)x+b

substitute 6 for x and -3 for y to find b

-3=(-1/3)6 +b

-3=-2+b

-1=b

therefore the equation of the line is y=(-1/3)x-1

Answer 3

First, you need to put this equation into the format y=mx+b to figure out what the slope, or m, is.  To do this, subtract 6x from both sides and then divide by -2.

 

Now that you have the slope, remember that if two lines are perpendicular, their slopes are negative reciprocals of each other.  For example, if your slope is 2 - let's call it "2/1" so you can see when the fraction flips - the slope of the perpendicular line is -1/2.  So take the slope you got when you put the equation in y=mx+b form and take its negative reciprocal to find the slope, m, of your new line.  You can then plug in x=6 and y=-3 to find out what b should be in your new line.