The equation 9x3+9x2y−45x2=4y3+4xy2−20y2 represents 3 straight lines, two of which passes through origin. Then find the area of the triangle formed by these lines.
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Solution
9x3+9x2y−45x2=4y3+4xy2−20y2
9x2(x+y−5)=4y2(x+y−5)
9x2(x+y−5)−4y2(x+y−5)=0
(9x2−4x2)(x+y−5)=0
(3x+2y)(3x−2x)(x+y−5)=0
so the three lines are
x=23y, x=−23y and x+y−5
from equation of lines , we get the vertices of triangle formed by the lines
A(0,0),B(−10,15) and C(2,3)
Area of the triangle =12(xa−xc)(yb−ya)−(xa−xb)(yc−ya)