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find the area of the isocele triangle formed by the points of intersection of parabolas y=-x^2+9 and y=2x^2-3 and the origin

Sagot :

Answer:

[tex]10\ \text{sq. units}[/tex]

Step-by-step explanation:

The parabola's are

[tex]y=-x^2+9[/tex]

[tex]y=2x^2-3[/tex]

So

[tex]-x^2+9=2x^2-3\\\Rightarrow 12=3x^2\\\Rightarrow \\\Rightarrow x=\sqrt{\dfrac{12}{3}}\\\Rightarrow x=\pm 2[/tex]

[tex]y=-x^2+9=-(2)^2+9\\\Rightarrow y=5[/tex]

[tex]y=-(-2)^2+9=5[/tex]

So, the points at which the parabola's intersect each other is [tex](2,5)[/tex] and [tex](-2,5)[/tex]

The three points of the triangle are [tex](2,5)[/tex], [tex](-2,5)[/tex] and [tex](0,0)[/tex]

Area of a triangle is given by

[tex]A=\dfrac{1}{2}[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]\\\Rightarrow A=\dfrac{1}{2}[2(5-0)+(-2)(0-5)+0(5-5)]\\\Rightarrow A=10\ \text{sq. units}[/tex]

Area of the triangle formed is [tex]10\ \text{sq. units}[/tex].

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