Finding all Lines Passing Through a Point Given the Product of Intercepts

Question

How does one find the equation of all lines passing through a point (Ex. $(6, -1)$), satisfying the condition that the product of their $x$ and $y$ intercepts must equal some number $c$ (Ex. $3$)?

As far as I understand this can be conceptualized as finding the equation of the line containing the points $(6,-1), (a,0), (0,b)$ where $a$ is the $x$-intercept, $b$ is the $y$-intercept and $ab=3$.

I've tried finding the slope with $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ and have gotten $m=\frac{-1}{6-a}$ and $m=\frac{-1-b}{6}$. I've also solved for $a$ and $b$ in terms of $m$ and tried susbstituting these values into the equation of a line ($y=mx+b$) but I just cant eliminate enough variables to solve for anything useful. Feel like I'm missing something obvious, I'm not even sure if there is more than one equation that satisfies the conditions.

Answer 1

Intercept equation of a line:

$x/a+y/b=1$;

This line passes through $(x_0,y_0):$

$x_0/a+y_0/b=1$;

Given: $c=ab$;

Then

$x_0/a+(ay_0/c)=1$;

Solve for $a$:

$y_0 a^2-ac+cx_0=0$;

Quadratic in $a$: $a_{1,2}$:

Lines:

$x/a_{1,2} +y/(c/a_{1,2})=1.$

Answer 2

You have the right idea. Since the lines involved in the question must have a $y$-intercept, they can't be vertical and so they have a specific slope. Thus, they can be written in the form

$$y = mx + b \tag{1}\label{eq1A}$$

The $y$-intercept is where $x = 0$, so it's

$$y_i = b \tag{2}\label{eq2A}$$

and the $x$-intercept is where $y = 0$, i.e.,

$$0 = mx_i + b \implies x_i = -\frac{b}{m} \tag{3}\label{eq3A}$$

As you want their product to be some number $c$, you thus get

$$x_i y_i = c \implies \frac{-b^2}{m} = c \implies b^2 = -mc \tag{4}\label{eq4A}$$

Next, as you also require the lines to pass through some point, let's say in general $(x_0, y_0)$, you thus get from \eqref{eq1A} that

$$y_0 = mx_0 + b \implies b = - mx_0 + y_0 \tag{5}\label{eq5A}$$

Substituting this into \eqref{eq4A} gives

$$\begin{equation}\begin{aligned} (- mx_0 + y_0)^2 & = -mc \\ (x_0)^2m^2 - 2(x_0)(y_0)m + y_0^2 & = -mc \\ (x_0)^2m^2 + (c - 2(x_0)(y_0))m + y_0^2 & = 0 \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

If $c = 0$, then you have $x_0 = 0$ and/or $y_0 = 0$. I'll leave it to you to handle that case from \eqref{eq6A}. Otherwise, assuming $c \neq 0$, then $x_0 \neq 0$, so the quadratic formula gives

$$\begin{equation}\begin{aligned} m & = \frac{2(x_0)(y_0) - c \pm \sqrt{(c - 2(x_0)(y_0))^2 - 4(x_0^2)(y_0^2)}}{2x_0^2} \\ & = \frac{2(x_0)(y_0) - c \pm \sqrt{c^2 - 4c(x_0)(y_0) + 4(x_0^2)(y_0^2) - 4(x_0^2)(y_0^2)}}{2x_0^2} \\ & = \frac{2(x_0)(y_0) - c \pm \sqrt{c^2 - 4c(x_0)(y_0)}}{2x_0^2} \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

You can substitute these $m$ values into \eqref{eq5A} to get $b$ and, thus, can then substitute these $2$ values into \eqref{eq1A} to get the line equations. I'll leave it to you to plug into any specific values you want to use.

Answer 3

Points $P,Q,R$ are collinear if and only if they satisfy the expression $m\vec{P}+(1-m)\vec{Q}=\vec{R}$. Substitute $\vec{P}=\{a,0\}, \vec{Q}=\{0,\frac{3}{a}\}, \vec{R}=\{6,-1\}$ To obtain the following expressions

$$ \begin{aligned} am&=6\\ a&=\frac{6}{m}\\ \\ (1-m)\frac{3}{a}&=-1\\ (1-m)\frac{m}{2}&=-1\\ m^{2}-m-2&=0\\ m&\in\{-1,2\} \end{aligned} $$ Substitute $m$ to obtain $a$, we got: $\{-6,0\},\{0,-\frac{1}{2}\}$ or $\{3,0\},\{0,1\}$

For other point $\{x_{i},y_{i}\}$ and $c$ the intercepts are $\{a,0\}, \{0,\frac{c}{a}\}$, $a=\frac{x_{i}}{m}$, $m=\frac{1\pm\sqrt{1-4\frac{x_{i}y_{i}}{c}}}{2}$

Answer 4

So I came across a similar issue and I couldn't find any answers which were satisfactory. So I dug up as much information as I could and I found a solution. Here goes...

We have a known point, $(6,-1)$. For any line that passes through this point, it must have an x-intercept of $(a,0)$ and a y-intercept of $(0,b)$. We don't know what $a$ and $b$ are as of yet, but we do know that their product is 3. By this definition, we also know that any line cannot be horizontal or vertical, since those lines would not have an x-intercept and y-intercept, respectively.

Let's take a look at our line equation: $y=mx+b$. We know that $a⋅b=3$ so with some rearranging we have $b=\frac{3}{a}$. We can sub this in for $b$, such that:

$y=mx+\frac{3}{a}$

We now need to find the slope. We can't explicitly define what the slope is yet, since that's part of what the question is asking. But we can substitute in what we do know. The slope is defined as:

$m=\frac{y_1-y_0}{x_1-x_0}$

This is just the distance between our two points along the graph, in each axis. We only know one point $(6, -1)$ but we can substitute in our x-intercept placeholder $(a,0)$ so our slope becomes:

$m=\frac{0-(-1)}{a-6}=\frac{1}{a-6}$

We can now sub this into our line equation as well, such that:

$y=\frac{1}{a-6}x+\frac{3}{a}$

Then we myltiply by $x$ which is really just $\frac{x}{1}$ such that:

$y=\frac{x}{a-6}+\frac{3}{a}$

We now have a fraction addition, and we do this just like any other: making the denominators the same. Multiply each fraction by the equivalent of $\frac{1}{1}$, using the denominator of the other fraction:

$y=\frac{x(a)}{(a-6)(a)}+\frac{3(a-6)}{a(a-6)}$

$y=\frac{a(x) + 3(a-6)}{a(a-6)}$

After some distribution we come to:

$y=\frac{ax+3a-18}{a^2-6a}$

Multiplying both sides by $(a^2-6a)$ and a bit more distribution we get:

$y(a^2-6a)=ax+3a-18$

$a^2y-6ay=ax+3a-18$

For this next step, we can sub in our known point $(6,-1)$ for $x$ and $y$. Remember that in our original formula $y=mx+b$, we already knew what $(x,y)$ was, as it was given to us as part of the question. The information we're missing are the slope $(m)$ and the y-intercept $(b)$. In fact, we could've substituted in our known point for $(x,y)$ right from the very start.

After some more rearranging and collecting like terms, we can get to an expression in standard polynomial form:

$a^2(-1)-6a(-1)=a(6)+3a-18$

$-a^2+6a=9a-18$

$-a^2-3a+18=0$

It's at this point we can use the quadratic formula to solve for a. But for clarity, let's substitute x for a:

$-x^2-3x+18=0$

For a polynomial in standard form $(ax^2+bx+c=0)$ the quadratic formula is defined as:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Looking at our equation so far, we can sub in our known values of $(a,b,c)$ as $(-1,-3,18)$ into the quadratic formula, such that:

$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(-1)(18)}}{2(-1)} \Rightarrow x=\frac{3\pm\sqrt{9+72}}{-2} \Rightarrow x=\frac{3\pm\sqrt{81}}{-2} \Rightarrow x=\frac{3\pm9}{-2}=-6,3$

At last, we have two points at which any line satisfying our conditions intersects the x-axis: $(-6,0)$ and $(3,0)$. We now have enough information to draw our lines and find the equations. For $y=mx+b$ the two pieces of information we need to find are $m$ and $b$.

For $m$, we just use the slope formula from before:

$m=\frac{y_1-y_0}{x_1-x_0}$

But now we can substitute all values for both our slopes. $(x_0,y_0)$ is our original known point, and $(x_1,y_1)$ are the points we just calculated. Since both points are the x-intercept, their y-coordinate will be 0:

$m_1=\frac{0-(-1)}{-6-6} \Rightarrow m_1=-\frac{1}{12}$

$m_2=\frac{0-(-1)}{3-6} \Rightarrow m_2=-\frac{1}{3}$

Finally, we just need to find $b$. To do so, we just need to sub in a known $(x,y)$ and solve for $b$. Since we know that both lines pass through $(6,-1)$ we can use it for both:

$y=mx+b$

$-1=-\frac{1}{12}(6)+b_1 \Rightarrow b_1=\frac{6}{12}-1 \Rightarrow b_1=-\frac{1}{2}$

$-1=-\frac{1}{3}(6)+b_2 \Rightarrow b_2=\frac{6}{3}-1 \Rightarrow b_2=2-1 \Rightarrow b_2=1$

We now have all the information we need to write our equations, for all lines which pass through $(6,-1)$ and for which the products of their $x/y$ intercepts is 3:

$y=-\frac{1}{12}x-\frac{1}{2}$, and $y=-\frac{1}{3}x+1$

You can see this Desmos graph to get a visualisation.