What is the area of the triangle formed by the lines $x+y=3$ and angle bisectors of pair of straight lines $x^2-y^2+2y=1$ . I found the intersection point of these equations $(1,2)$ but not getting any idea of how to find the area . Any way to solve this problem. Thanks!
[Math] Area of the triangle determined by the line $x+y=3$ and the bisector of angle between the lines $x^2-y^2+2y=1$
analytic geometryareageometrytriangles
Related Solutions
We'll derive the equation using the fact: $$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$ Using the angle bisector theorem we get: $$BP=\frac{ac}{a+b},\quad (1)$$ $$BR=\frac{ac}{b+c}, \quad (2)$$ $$CR=\frac{ab}{b+c},\quad (3)$$ $$CQ=\frac{ab}{a+c},\quad (4)$$ $$AQ=\frac{bc}{a+c},\quad (5)$$ and $$AP=\frac{bc}{a+b}. \quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$ $$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$ $$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$ and $$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$
Let $R$ be the circumradius, we know that: $$\sin \alpha = \frac{a}{2R}, \quad (11)$$ $$\sin \beta = \frac{b}{2R}, \quad (12)$$ $$\sin \gamma = \frac{c}{2R}, \quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get: $$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$
$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$
$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$ $$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$ Using Heron's formula we are done.
$$x^2 - y^2 + 2y = 1\iff x^2 = y^2 -2y+1 = (y-1)^2 \iff y-1 = \pm x \iff y = 1\pm x$$
Thus the two intersecting lines are given by $y = 1+x$ and $y = 1-x$. The point of intersection is given by $(0, 1)$. One bisector is the y-axis: the line $x=0$. The other bisector is given by $y = 1$.
Can you take it from here?
Best Answer
First, observe that just like @Nicholas said, the equation $\,x^2-y^2+2y=1\,$ defines two lines:
\begin{align} x^2-y^2+2y=1 \iff x^2 = (y-1)^2 \implies \begin{cases} l_1: & y = x + 1 \\ l_2: & y = -x + 1 \end{cases} \end{align}
The slope of the first one is $\,\dfrac{\pi}{4} = 45º,\,$ and for the second one the slope is $\,\dfrac{3\pi}{4} = 135º,\,$ i.e. they are perpendicular.
Equations of bisecting lines can be found by adding and subtracting equations of original lines:
\begin{align} \begin{cases}b_1:&x=0\\b_2:&y=1\end{cases} \end{align}
Points of intersection of these lines with each other and with the line $\,l_0: \ \; y = 3-x\,$ are
\begin{align} A & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ x = 0\end{cases} &\implies&& A &= \big(\,0,\,3\,\big) \\ B & := b_1 \cap b_2 &\iff& &&\begin{cases}x = 0\\y=1\end{cases} &\implies&& B &= \big(\,0,\,1\,\big) \\ C & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ y=1\end{cases} &\implies&& C &= \big(\,2,\,1\,\big) \\ \end{align} Observe that $\;b_1\perp b_2\;$ so that $\;\triangle\, ABC\;$ is right triangle. Therefore the area $\,S\,$ of $\;\triangle\, ABC\;$ is just a half of product of legs $\, AB\,$ and $\,BC\,$: \begin{align} S_{\triangle ABC} = \dfrac{1}{2}\,\big\|\left(0,\,1\right) - \left(0,\,3\right)\big\|\, \big\|\left(2,\,1\right) - \left(0,\,1\right)\big\| = 2 \end{align}