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Question

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


Solution

$$9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$$

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

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

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

$$(3x+2y)(3x-2x)(x+y-5)=0$$

so the three lines are

$$x=\dfrac{2}{3}y$$, $$x=\dfrac{-2}{3}y$$ 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 $$=\dfrac{1}{2}(x_a-x_c)(y_b-ya)-(x_a-x_b)(y_c-y_a)$$

                                   $$=\dfrac{1}{2}(x_ay_b+x_by_c+x_cy_a-x_ay_c-x_cy_b-x_by_a)$$

                                    $$=\dfrac{1}{2}(0\times15+-10\times3+2\times0-0\times3-2\times15-(-10\times0)$$
  
                                    $$=\dfrac{1}{2}\times-60$$

since, area is non- negativengle

So, the area of the triangle$$=30 unit^2$$

Maths

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