Find three points given three lines and some conditions

Question

Let $L$ be the line $2y=x+4$, $M$ be the line $x+y=4$, $N$ be the line $y+8=3x$. Find points $u,v,w\in \mathbb{R}^2$ such that, in the triangle $[u,v,w]$, $L$ is the angle bisector at $u$, $M$ is the altitude from $v$, $N$ is the median from $w$.

My attempt:

• $u$ lies on $L$, so $u=(2t,t+4)$ for some $t\in\mathbb{R}$

• $w$ lies on $N$, so $w=(q, 3q-8)$ for some $q\in\mathbb{R}$

• $v$ lies on $M$, so $v=(s, 4-s)$ for some $s\in\mathbb{R}$

• Some point on $v-w$ lies on $M$, so $3t=4q-8$ (after substituting the coordinates of $v-w$ into the linear equation for $M$.

Now the problem is that I don't know how to utilize the condition for $L$ being the angle bisector at $u$ and the condition $M$ being the altitude from $v$ without introducing additional variables.

Your help would be much appreciated.

Answer 1

As explained bellow, I've found the following points: $$u=(12,8),\qquad v=(-2,6),\qquad w=(2,-2). $$ The associated position vectors are: $$\mathbf{u}=12\,\mathbf{\hat{e}}_{1}+8\,\mathbf{\hat{e}}_{2} ,\quad \mathbf{v}=-2\,\mathbf{\hat{e}}_{1}+6\,\mathbf{\hat{e}}_{2},\quad \mathbf{w}=2\,\mathbf{\hat{e}}_{1}-2\,\mathbf{\hat{e}}_{2} . $$

Triangle $[u,v,w]=[(12,8),(-2,6),(2,-2)]$ and lines $L$ (black), $M$ (blue) and $N$ (red)

As it has already been pointed out, instead of

• $u$ lies on $L$, so $u=(2t,t+4)$ for some $t\in\mathbb{R}$

it should be

• $u$ lies on $L$, so $u=(2t,t+2)$ for some $t\in\mathbb{R}$,

because if we replace $x$ with $2t$ in $2y=x+4$, we obtain $2y=2t+4$, which means that $y=t+2$.

After this correction, your system of three equations becomes

$$u=(2t,t+4)\tag{1} ,$$ $$v=(s,4-s)\tag{2} ,$$ $$w=(q,3q-8)\tag{3} .$$

The position vectors are thus $$\mathbf{u}=(2t)\,\mathbf{\hat{e}_1}+(t+4)\,\mathbf{\hat{e}_2}\tag{1$\mathrm{a}$},$$ $$\mathbf{v}=s\,\mathbf{\hat{e}_1}+(4-s)\,\mathbf{\hat{e}_2}\tag{2$\mathrm{a}$},$$ $$\mathbf{w}=q\,\mathbf{\hat{e}_1}+(3q-8)\,\mathbf{\hat{e}_2}\tag{3$\mathrm{a}$}.$$

Now the problem is that I don't know how to utilize the condition for $L$ being the angle bisector at $u$ and the condition $M$ being the altitude from $v$ without introducing additional variables.

To obtain additional relations between $t,s$ and $q$, such as equations $(5')$, $(7')$ and $(15)$, without introducing additional variables, except the angles between the bisector $L$ and the two sides $[u,v]$ and $[u,w]$, we compute their slopes and use the conditions on $L,M,N$:

(a) $L$ is the angle bisector at $u$,

(b) $M$ is the altitude from $v$,

(c) $N$ is the median from $w$.

1. As for $M: x+y=4$, since its slope is $-1$ and it is is perpendicular to the vector
   $$\mathbf{w}-\mathbf{u}=(q-2t)\,\mathbf{\hat{e}}_{1}+(3q-t-10)\,\mathbf{\hat{e}}_{2}\tag{4},$$ by $(\mathrm{D})$ bellow, the slope of $\mathbf{w}-\mathbf{u}$ should be $1$. Then $$ 3q-t-10=q-2t\Leftrightarrow 2q+t-10=0.\tag{5} $$ Solving $(5)$ for $q$, we obtain $$q=-\frac{1}{2}t+5.\tag{5'}$$ Substituting $(5')$ in $(3)$, we obtain $$w=\left(-\frac{1}{2}t+5,-\frac{3}{2}t+7\right)\tag{3'}$$
2. As for $N:$ $y+8=3x$, its intersection with the side $\left[ u,v\right] $ is the point $$\frac{u+v}{2}=\frac{(2t,t+2)+(s,4-s)}{2}=\left( \frac{1}{2}s+t,-\frac{1}{2}s+ \frac{1}{2}t+3\right).\tag{6} $$ Replacing $x$ with $\frac{1}{2}s+t$ and $y$ with $-\frac{1}{2}s+\frac{1}{2}t+3$ in the equation $y+8=3x$, we obtain $$\left( -\frac{1}{2}s+\frac{1}{2}t+3\right) +8=3\left( \frac{1}{2}s+t\right).\tag{7}$$ Solving for $s$, we obtain $$s=\frac{11}{2}-\frac{5}{4}t.\tag{7'}$$ Now, substituting $(7')$ in $(2)$, we obtain $$v=\left(-\frac{5}{4}t+\frac{11}{2},\frac{5}{4}t-\frac{3}{2}\right).\tag{2'}$$
3. As for $L:2y=x+4$, we now use this equation to find the acute angles $\theta_1$ and $\theta_2$ between $L$ and, respectively, the sides $[u,v]$ and $[u,w]$. These angles satisfy the condition $$\theta_1=\theta_2.\tag{8} $$ The slope $m$ of $L$ is $$m=\frac{1}{2}.\tag{9}$$ Also, let $m_1$ and $m_2$ be, respectively, the slopes of the sides $[u,v]$ and $[u,w]$. The vectors $\mathbf{v}-\mathbf{u}$ and $\mathbf{w}-\mathbf{u}$ are parallel, respectively, to $[u,v]$ and $[u,w]$. Combining $(2'),(1), (2\mathrm{a}), (1\mathrm{a})$, we have $$ \mathbf{v}-\mathbf{u}=\left( -\frac{13}{4}t+\frac{11}{2}\right)\, \mathbf{\hat{e}}_{1}+\left( \frac{1}{4}t-\frac{7}{2}\right)\, \mathbf{\hat{e}}_{2}\tag{4'} $$ Let's denote its slope by $$ m_{1}=\frac{t/4-7/2}{-13t/4+11/2}=\frac{14-t}{13t-22}\tag{10}. $$ Similarly, from $(4),(5')$, we obtain $$ \mathbf{w}-\mathbf{u}=\left( -\frac{5}{2}t+5\right)\, \mathbf{\hat{e}}_{1}+\left( -\frac{5}{2}t+5\right)\, \mathbf{\hat{e}}_{2},\tag{11} $$ whose slope is denoted by $$ m_{2}=\frac{-5t/2+5}{-5/2t+5}=1.\tag{12} $$ We now apply twice the formula $(\mathrm{C})$ bellow, one for the lines with slopes $m$ and $m'=m_1$, and another for the lines with slopes $m$ and $m'=m_2$: \begin{equation*} \tan \theta_1 =\left|\frac{m-m_1}{1+mm_1}\right|,\tag{$\mathrm{C_1}$} \end{equation*} \begin{equation*} \tan \theta_2 =\left|\frac{m-m_2}{1+mm_2}\right|.\tag{$\mathrm{C_2}$} \end{equation*} From $(8)$ we should have \begin{equation*} \tan \theta_1 =\tan \theta_2 \Leftrightarrow\left|\frac{m-m_1}{1+mm_1}\right| =\left|\frac{m-m_2}{1+mm_2}\right|.\tag{$\mathrm{C_3}$} \end{equation*} Equivalently, \begin{equation*} \left\vert \left( m-m_{1}\right) \left( 1+mm_{2}\right) \right\vert =\left\vert \left( m-m_{2}\right) \left( 1+mm_{1}\right) \right\vert .\tag{13} \end{equation*} For $m=\frac{1}{2}$ and $m_{2}=1$, we obtain \begin{equation*} \frac{3}{2}\left\vert \frac{1}{2}-m_{1}\right\vert =\frac{1}{2}\left\vert 1+ \frac{1}{2}m_{1}\right\vert .\tag{14} \end{equation*} The possible solutions are \begin{equation*} \frac{14-t}{13t-22}=m_{1}=1\qquad \text{or}\qquad \frac{14-t}{13t-22}=m_{1}= \frac{1}{7};\tag{15} \end{equation*} the corresponding possible values of $t$ being \begin{equation*} t=\frac{18}{7}\qquad \text{or}\qquad t=6.\tag{16} \end{equation*} Using equations $\left( 1\right) ,\left( 2^{\prime }\right) $ and $\left( 3^{\prime }\right) $, for $t=\frac{18}{7}$, we obtain the points $$u=\left( \frac{36}{7},\frac{32}{7}\right) ,\qquad v=\left( \frac{16}{7},\frac{12}{7}\right) ,\qquad w=\left( \frac{26}{7},\frac{22}{7}\right) ,\tag{17}$$ which are colinear; for $t=6$, we obtain the vertices $u,v$ and $w$ indicated in the beginning: $$u=(12,8),\qquad v=(-2,6),\qquad w=(2,-2). \tag{18}$$

Acute angle $\theta$ between two lines with slopes $m$ and $m'$ using trigonometry (angle difference identity for tangent) $[1]$

If $\alpha $ is the angle that a line with slope $m$ makes with the $x$-axis and $\alpha ^{\prime }$ is the angle that another line with slope $m^{\prime }$ makes with the $x$-axis, then, assuming that $\alpha\ge\alpha'$, the acute angle $\theta $ between the two lines is given by

\begin{equation*} \theta =\alpha -\alpha ^{\prime }\qquad \text{or}\qquad \theta =\pi -\left( \alpha -\alpha ^{\prime }\right),\quad 0\le\theta\le \frac{\pi}{2}.\tag{$\mathrm{A}$} \end{equation*}

Consequently,

\begin{equation*} \tan \theta =\tan \left( \alpha -\alpha ^{\prime }\right) \qquad \text{ or }\qquad\tan \theta =-\tan \left( \alpha -\alpha ^{\prime }\right). \tag{$\mathrm{B}$} \end{equation*}

From trigonometry, we know that

\begin{equation*} \tan \theta =\tan \left( \alpha -\alpha ^{\prime }\right) =\frac{\tan \alpha -\tan \alpha ^{\prime }}{1+\tan \alpha \tan \alpha ^{\prime }} \end{equation*}

Since

\begin{equation*} m=\tan \alpha \qquad\text{ and }\qquad m^{\prime }=\tan \alpha \end{equation*}

we obtain

\begin{equation*} \tan \theta =\left|\frac{m-m^{\prime }}{1+mm^{\prime }}\right|,\qquad 0\le\theta \le\frac{\pi}{2}.\tag{$\mathrm{C}$} \end{equation*}

If $\alpha <\alpha'$, this equation is still valid, because its right-hand side is non-negative.

In particular, if $\theta=\frac{\pi}{2}$,the denominator should satisfy the well known equation

\begin{equation*} mm^{\prime }=-1.\tag{$\mathrm{D}$} \end{equation*}

$[1]$ J. Sebastião e Silva, "Geometria Analítica Plana", pp. 76-77, 1967

Answer 2

NOTE: Your point $u$ is incorrect. It should be $u=(2t,t+\color{red}{2})$.

Treating the points $u,v,w$ as position vectors, we can get $$\overline{uv}=(s,4-s)-(2t,t+\color{red}{2})=(s-2t, \color{red}{2-s-t}).$$ Likewise $$\overline{uw}=(q,3q-8)-(2t,t+\color{red}{2})=(q-2t, \color{red}{3q-t-10}).$$ The midpoint of the side represented by the vector $\overline{uv}$, call it $A$, is given by $\left(\frac{s+2t}{2},\frac{\color{red}{6-s+t}}{2}\right)$.

Now observe the following:

1. The mid-point $A$ should lie on the line $N$ because $N$ is the median from $w$. So it's coordinates should satisfy the equation for $N$.
2. The side represented by $\overline{uw}$ is perpendicular to the line $M$ because $M$ is the altitude friom $v$. Therefore the product of the slopes should be $-1$.
3. For $L$ being the bisector we need: angle between $\overline{uv}$ and $L$ should be equal to the angle between $\overline{uw}$ and L.

If you can set up these three equations, you will have the values of the three unknowns.

Answer 3

Just to better clarify the vectorial approach in writing down the last two conditions given in AnuragA answer, hoping you have a basic knowledge of vectors.

Given a line written as $ax+by+c=0$, you know that you can rewrite it as $a(x-x_0)+b(y-y_0)=0$, where $P_0=(x_0,y_0)$ is a point on the line.
Then indicating by $X=(x,y)$ a generic point on the line, the line equation becomes $(a,b)\cdot \mathop {P_0 X}\limits^ \to=0$. So all segments on the line are, and therefore the line is, normal to $(a,b)$, i.e. parallel to $(b,-a)$.

That premised,
1) $N$ median from $w \quad \Rightarrow \quad w \in N ,\; (u+v)/2 \in N$

2) $\mathop {uw}\limits^ \to \bot M \quad \Rightarrow \quad (q-2t,\;3q-t-10) \cdot(1,\;-1)=0$
3) $L$ bisects $\angle (\mathop {uv}\limits^ \to,\mathop {uw}\limits^ \to)\quad \Rightarrow \quad (-2,\,-1)=\lambda(\mathop {uv}\limits^ \to /|\mathop {uv}\limits^ \to| +\mathop {uw}\limits^ \to / |\mathop {uw}\limits^ \to|)$

As you see no further variable is introduced, once you get rid of $\lambda$ by dividing the components (if not null).

To complete the task as to check the results, let's proceed.
You already rightly found that for each point to lie on the respective line, we shall have this parametric representation (corrected concerning $u$, as already indicated by Américo) $$ \bbox[lightyellow] { \left\{ \matrix{ u = \left( {2t,t + 2} \right) \hfill \cr w = \left( {q,3q - 8} \right) \hfill \cr v = \left( {s,4 - s} \right) \hfill \cr} \right. } \tag{1}$$

The three conditions indicated above, then translates into (the condition $w \in N$ is already satisfied) $$ \bbox[lightyellow] { \left\{ \matrix{ {{t + 2 + 4 - s} \over 2} + 8 = 3{{2t + s} \over 2}\quad \Rightarrow \quad 5t + 4s = 22 \hfill \cr q - 2t - 3q + t + 10 = 0\quad \Rightarrow \quad t + 2q = 10 \hfill \cr \left( { - 2, - 1} \right) = \lambda \left( {{{\left( {s - 2t,\; - t - s + 2} \right)} \over {\sqrt {\left( {s - 2t} \right)^{\,2} + \left( { - t - s + 2} \right)^{\,2} } }} + {{\left( {q - 2t,\;3q - t - 10} \right)} \over {\sqrt {\left( {q - 2t} \right)^{\,2} + \left( {3q - t - 10} \right)^{\,2} } }}} \right) \hfill \cr} \right. } \tag{2}$$

Using the first two to express $s$ and $q$ in terms of $t$, and replacing that in the third equation after some simple manipulations we get $$ \left( { - 2, - 1} \right) = \lambda \left( {{{\left( {22 - 13t,\; - 14 + t} \right)} \over {\sqrt {\left( {22 - 13t} \right)^{\,2} + \left( { - 14 + t} \right)^{\,2} } }} + {{\left( {1,\;1} \right)} \over {\sqrt 2 }}} \right) $$ that is $$ \eqalign{ & 2 = \left( {{{{{\left( {22 - 13t} \right)} \over {\sqrt {\left( {22 - 13t} \right)^{\,2} + \left( { - 14 + t} \right)^{\,2} } }} + {1 \over {\sqrt 2 }}} \over {{{\left( {\; - 14 + t} \right)} \over {\sqrt {\left( {22 - 13t} \right)^{\,2} + \left( { - 14 + t} \right)^{\,2} } }} + {1 \over {\sqrt 2 }}}}} \right) = \cr & = \left( {{{\sqrt 2 \left( {22 - 13t} \right) + \sqrt {170\,t^{\,2} - 600t + 680} } \over {\sqrt 2 \left( {\; - 14 + t} \right) + \sqrt {170\,t^{\,2} - 600t + 680} }}} \right) \cr} $$

then $$ \sqrt {5\left( {17\,t^{\,2} - 60t + 68} \right)} = \left( {22 - 13t} \right) - 2\left( {\; - 14 + t} \right) = 5\left( {10 - 3t} \right) $$ and finally $$ \bbox[lightyellow] { 7t^{\,2} - 60t + 108 = 0 } $$ which provides two values for $t$ $$ t = 18/7,\quad t = 6 $$

The first is to be discarded, since when replaced in the third equation gives $2=1$. So we are left with $t=6$.

Consequently, $q=2$ and $s=-2$ and finally we get the three points as: $$ \bbox[lightyellow] { \left\{ \matrix{ u = \left( {12,\;8} \right) \hfill \cr w = \left( {2,\; - 2} \right) \hfill \cr v = \left( { - 2,\;6} \right) \hfill \cr} \right. } \tag{3}$$

which confirms Américo's answer .