Question

20. Find the equation of the line perpendicular to the line x-2y+3=0 and passing through
the point (1,-2)
21. If (1,1,1)is the centroid of the triangle with (3.–5.7) and (-1,7,-6) as the two vertices, find
the third vertex.​

Answer 1

Answer:

Answer:-2x

Answer:-2xStep-by-step explanation:

Answer:-2xStep-by-step explanation:Let the required line be y = mx +c

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0are

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicular

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1.m =-2

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1.m =-2line passe through (1,-2)

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1.m =-2line passe through (1,-2)y=-2x +c -2=-2(1)+c

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1.m =-2line passe through (1,-2)y=-2x +c -2=-2(1)+cc= 0

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1.m =-2line passe through (1,-2)y=-2x +c -2=-2(1)+cc= 0y= -2x

Answer:-2xStep-by-step explanation:Let the required line be y = mx +cy = mx +candx - 2y -3 = 0areperpendicularm1 X m2 = -1mi= m m2 y=: x-3)m2 = 1/2m(1/2) = -1.m =-2line passe through (1,-2)y=-2x +c -2=-2(1)+cc= 0y= -2xThe required line equation is y = -2x