Calculating the area of Triangle

Question

Find the area of triangle formed by the lines $x^2+4xy+y^2=0$ and $x+y=1$.

I know that the equation $x^2+4xy+y^2=0$ represents a pair of straight lines but how do i factorize it to get the two lines represented by it.

Answer 1

Note that $x^2+4xy+y^2=x^2+2xy+y^2+2xy$ so, taking into account the other equation, you get $1+2xy=0$.

So you are reduced to finding two numbers $x$ and $y$ of which you know the sum, because $x+y=1$, and the product, because $xy=-1/2$.

Such numbers are the roots of the quadratic equation $$ z^2-z-\frac{1}{2}=0 $$ that is, $$ \frac{1+\sqrt{3}}{2} \quad\text{or}\quad \frac{1-\sqrt{3}}{2} $$ So you have two points $$ \left(\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2}\right) \qquad \left(\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}\right) $$ and the third vertex is the unique point of intersection of the pair of lines represented by $x^2+4xy+y^2$, which is the origin. Then the area of the triangle is $$ \frac{1}{2}\left|\det\begin{bmatrix} \dfrac{1+\sqrt{3}}{2} & \dfrac{1-\sqrt{3}}{2}\\[6px] \dfrac{1-\sqrt{3}}{2} & \dfrac{1+\sqrt{3}}{2} \end{bmatrix} \right|= \frac{1}{2}\left(\frac{4+2\sqrt{3}}{4}-\frac{4-2\sqrt{3}}{4}\right) =\frac{\sqrt{3}}{2} $$

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You may enjoy proving the fact that, if $(a,b)$ and $(c,d)$ are two points, then the area of the triangle having those points and the origin as the vertices is $$ \frac{1}{2}\left|\det\begin{bmatrix}a & c\\b&d\end{bmatrix}\right| = \frac{1}{2}|ad-bc| $$

You can prove it by using the fact that the line passing through $(a,b)$ and $(c,d)$ has equation $$ (b-d)(x-c)-(a-c)(y-d)=0 $$ that can also be written as $$ (b-d)x-(a-c)y+(ad-bc)=0 $$ and so the distance of this line from the origin is $$ h=\frac{|(b-d)\cdot0-(a-c)\cdot0+(ad-bc)|}{\sqrt{(b-d)^2+(a-c)^2}} = \frac{|ad-bc|}{\sqrt{(b-d)^2+(a-c)^2}} $$ and that the denominator is exactly the distance between the two points. So, taking this segment as the basis, and multiplying by the height $h$, we get that the area is $$ \frac{1}{2}|ad-bc| $$ What for a general triangle having vertices at $(x_0,y_0)$, $(x_1,y_1)$ and $(x_2,y_2)$? Just do a translation and consider the congruent triangle having one vertex at the origin and the other two at $(x_1-x_0,y_1-y_0)$ and $(x_2-x_0,y_2-y_0)$. Then the area is $$ \frac{1}{2}\left|\det \begin{bmatrix} x_1-x_0 & x_2-x_0 \\ y_1-y_0 & y_2-y_0 \end{bmatrix} \right|= \frac{1}{2}\left|\det \begin{bmatrix} x_0 & x_1 & x_2 \\ y_0 & y_1 & y_2 \\ 1 & 1 & 1 \end{bmatrix} \right| $$

For the last formula, compute the determinant of the $3\times3$ matrix by first adding to the first row the third one multiplied by $-x_0$, then adding to the second row the third one multiplied by $-y_0$, getting the matrix $$ \begin{bmatrix} 0 & x_1-x_0 & x_2-x_0 \\ 0 & y_1-y_0 & y_2-y_0 \\ 1 & 1 & 1 \end{bmatrix} $$ These operations don't change the determinant. Now use Laplace on the first column.

Answer 2

Hint The symmetry in $x, y$ of the quadratic expression suggests that we can eliminate the cross term by substituting $x = s - t, y = s + t$. Indeed, substituting and simplifying gives in these coordinates that the quadratic equation is $$6 s^2 - 2 t^2 = 0 ,$$ but this is a difference of perfect squares, and so is easy to factor.

Answer 3

HINT: Let $y-m_1x=0$ & $y-m_2x=0$ be the straight lines passing through the origin represented by $x^2+4xy+y^2=0$ then it can be factorize as follows $$x^2+4xy+y^2=(y-m_1x)(y-m_2x)$$ $$m_1m_2x^2-(m_1+m_2)xy+y^2=x^2+4xy+y^2$$ comparing the corresponding coefficients on both the sides, one should get $$m_1m_2=1\tag 1$$ $$m_1+m_2=-4\tag 2$$ solving (1) & (2), one should get $m_1=-2-\sqrt 2$ & $m_2=-2+\sqrt 3$

hence, the pair of lines is $\color{blue}{y=(-2-\sqrt3)x}$ & $\color{blue}{y=(-2+\sqrt3)x}$

Now, solve the three equations: $\color{}{y=(-2-\sqrt3)x}$, $\color{}{y=(-2+\sqrt3)x}$ & $x+y=1$ to find coordinate of the vertices & then apply formula for area of triangle having vertices $(x_1, y_1)$, $(x_2, y_2)$ & $(x_3, y_3)$ $$=\frac 12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$

Answer 4

Hint:

solving $x^2+4xy+y^2=0$ with respect to y you find: $$ y=-2x\pm x\sqrt{3} $$ so you can factorize.

Answer 5

As usual, complete the square:

$$x^2+4xy+y^2=(x+2y)^2-3y^2=(x+2y-\sqrt3y)(x+2y+\sqrt3y).$$

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More generally,

$$ax^2+bxy+cy^2=a(x^2+\frac b{2a})^2+\left(c-\frac{b^2}{4a}\right)y^2=a\left(x+\frac{b-\sqrt{b^2-4ac}}{2a}y\right )\left(x+\frac{b+\sqrt{b^2-4ac}}{2a}y\right).$$