Question 833797
Encontrar las expresiones algebraicas y graficarlas:

e.Recta cuya pendientes es ¾   y pasa por el punto (-5,-7)


<pre>

Utilizamos la ecuación punto-pendiente:

{{{y-y[1]}}}{{{""=""}}}{{{m(x-x[1])}}}

{{{y-(-7)}}}{{{""=""}}}{{{expr(3/4)(x-(-5))}}}

{{{y+7}}}{{{""=""}}}{{{expr(3/4)(x+5)}}}

Multiplique ambos lados por 4:


{{{4(y+7)}}}{{{""=""}}}{{{(4)(3/4)(x+5)}}}

{{{4y+28}}}{{{""=""}}}{{{3(x+5)}}}

{{{4y+28}}}{{{""=""}}}{{{3x+15}}}

{{{4y+13}}}{{{""=""}}}{{{3x}}}

{{{13}}}{{{""=""}}}{{{3x-4y}}}

{{{3x-4y}}}{{{"=""}}}{{{13}}}

o

{{{y}}}{{{""=""}}}{{{expr(3/4)x-13/4}}}

 x| y
-1|-4
 3|-1
 7| 2


{{{drawing(400,400,-7,9,-10,6,graph(400,400,-7,9,-10,6),
circle(-1,-4,0.15),circle(-1,-4,0.13),circle(-1,-4,0.11),circle(-1,-4,0.09),circle(-1,-4,0.07),circle(-1,-4,0.05),circle(-1,-4,0.03),circle(-1,-4,0.01),locate(-3.2,-3.5,"(-1,-4)"),locate(3,-1,"(3,-1)"), locate(7,2,"(7,2)"),

locate(-7.2,-6.5,"(-5,-7)"),

circle(3,-1,0.15),circle(3,-1,0.13),circle(3,-1,0.11),circle(3,-1,0.09),circle(3,-1,0.07),circle(3,-1,0.05),circle(3,-1,0.03),circle(3,-1,0.01),

circle(-5,-7,0.15),circle(-5,-7,0.13),circle(-5,-7,0.11),circle(-5,-7,0.09),circle(-5,-7,0.07),circle(-5,-7,0.05),circle(-5,-7,0.03),circle(-5,-7,0.01),



circle(7,2,0.15),circle(7,2,0.13),circle(7,2,0.11),circle(7,2,0.09),circle(7,2,0.07),circle(7,2,0.05),circle(7,2,0.03),circle(7,2,0.01),
line(-13,-13,15,8) )}}}

Edwin</pre>