A pair of straight lines is represented by #2y^2 +5xy -3x^2 = 0#. Another line having an equation #x+y = k#. These lines form a triangle whose centroid is #(1/18,11/18)#. Find the value of k?

Question

2076 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-20T04:08:57

#k=1#

#2y^2+5xy-3x^2=(2 y-x) (3 x + y) = 0#

so the three lines are

#{(2 y-x=0),(3 x + y=0),(x+y=k):}#

and given

#{(2 y_1-x_1=0),(3 x_1 + y_1=0),(3x_2+y_2=0),(x_2+y_2=k),(x_3+y_3=k),(2y_3-x_3=0),((x_1+x_2+x_3)/3=1/18),((y_1+y_2+y_3)/3=11/18):}#

where

#p_1 = (x_1,y_1)#
#p_2=(x_2,y_2)#
#p_3=(x_3,y_3)#

are the intersection points or the vertices.

Solving for #x_1,x_2,x_3,y_1,y_2,y_3,k# we obtain

#x_1 = 0, x_2 = -1/2, x_3 = 2/3, y_1 = 0, y_2 = 3/2, y_3 = 1/3, k = 1#

and then #k = 1#