Question

Determina la ecuación general de la recta que Por (1,3) 3X-5y +8=0 y que Sea perpendicular​

Answer 1

Respuesta:

3x-5y+8 = 0

3x-5y+8-8 = 0-8

3x-5y = -8

3x-5y-3x = -8-3x

-5y = -8-3x

-(-5y) = -(-8-3x)

-(-5y) = -(-8)-(-3x)

5y = 8+3x

(5/5)y = (8+3x)/5

(5/5)y = (3x+8)/5

y = (3x+8)/5

y = (3x/5)+(8/5)

y = (3/5)x+(8/5)

m1×m2 = - 1 ; m1 = 3/5

(3/5)×m2 = - 1

5((3/5)m2) = 5(-1)

3m2 = - 5

(3/3)m2 = - 5/3

m2 = - 5/3

( x1 , y1 ) = ( 1 , 3 )

y-y1 = m(x-x1) ; m2 = m y como m2 = - 5/3 === > m = - 5/3

y-(3) = (-5/3)(x-(1))

y-3 = (-5/3)(x)-(5/3)(-1)

y-3 = (-5/3)x-(-5/3)

y-3 = (-5/3)x+(5/3) ; 3 = 9/3

y-(9/3) = (-5/3)x+(5/3)

y-(9/3)+(9/3) = (-5/3)x+(5/3)+(9/3)

y = (-5/3)x+((5+9)/3)

y = (-5/3)x+(14/3)

3(y) = 3(-5/3)x+3(14/3)

3y = - 5x+14

3y+5x = - 5x+14+5x

3y+5x = 14

5x+3y = 14

5x+3y-14 = 14-14

5x+3y-14 = 0

R// Por consiguiente , la ecuación general de la recta que es perpendicular a la recta " 3x-5y+8 = 0 " y que atraviesa el punto ( 1 , 3 ) es " 5x+3y-14 = 0 " .