Question 129732
3.  Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 2y = 8 and passing through (1, -6).


First Step put in standard form which is y = MX + B

X + 2Y = 8

TO DO THIS YOU WANT TO GET Y BY INSELF

x + 2Y = 8
-X      -X     SUBTRACT x fROM EACH SIDE ( * DO SAME THING TO BOTH SIDES)

2Y = 8-X      NOW TO YOU MUST GET RID OF THE 2
/2    /2      DIVDE 2 

Y = 4 - (1/2)X  NOW YOU MUST ORDER THEM IN PROPER ORDER ( y = MX + B)
Y = -(1/2)X + 4  NOW YOU HAVE IT STANDARD FORM


NoW MAKE IT PERPENDICULAR BY GET RECPICROL OF THE SLOPE 
WHICH IS   1

SO NOW YOUR EQUATION IS Y= X + 4
ORDER PAIR IT IS PASSING THROUGH ( 1,- 6 ) SO 1 IS x VALUE AND -6 IS Y VALUE

NEXT FORMULA Y -Y1  = M ( X - X1) M  IS SLOPLE (1) AND Y1 AND X2 IS ORDER PAIR

PLUG IN Y - (6) = 1(X- (-6))        DISTRIBUTE
        Y - (6) = 1X + 6             (* REMBEMBER TWO - - EQUALS A PLUS)
          +6                          ADD 6
 fINAL ANSWER =       Y = 1X + 6 OR Y = X + 6 ( YOU DONT NEED 1 BECAUSE THE LONE X STANDS FOR IT)