Question

The equation $9x^3+9x^{2}y-45x^2=4y^3+4x{y}^2-20y^2$ represents $3$ straight lines, two of which passes through origin. Then find the area of the triangle formed by these lines.

Answer 1

Given: $9x^3+9x^{2}y-45x^2=4y^3+4x{y}^2-20y^2$

$=9x^2(x+y-5)=4y^2(y+x-5)$

$=(9x^2-4y^2)(x+y-5)=0$

So, $3$ lines are $\left(i\right)x=\frac{2}{3}y,\left(ii\right)x=-\frac{2}{3}y,\left(iii\right)x+y-5=0.$

Now we can easily find intersection of all lines.

Intersection points are

$A(0,0)($Between i and ii$)$

$B(-10,15)($Between ii and iii$)$

$C(2,3)($Between i and iii$)$

Now, if $(x_A,y_A),(x_B,y_B),(x_C,y_C)$ are the vertices of a triangle then its area is given by

$Area=\frac{1}{2}\mid (x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y)A)\mid =\frac{1}{2}\mid x_A{y}_B+x_B{y}_C+x_C{y}_A-x_A{y}_C-x_C{y}_B-x_B{y}_A|$

Putting values of all coordinates, we get area $=\frac{1}{2}|0\times 15+(-10)\times 3+2\times 0-0\times 3-2\times 15-(-10)\times 0|$

$=\frac{1}{2}|-60|=30$

Hence, area of the triangle formed by these lines is $30$.