How to find the slope of a line that is kept at the same proportional distance to two others?

Question

I have been given 4 points which create two separate lines. I have been given another point that falls inside the projection of these two lines. Now I have to draw a line through this point that maintains the same distance from the other lines (as a ratio), but I am having some difficulty working out the slope of this line.

The picture below shows the two lines in red. (1,4) and (3,6) form one line, and (2,1) and (4,2) form the other. The dashed line is where it LOOKS like the line passing through (3,5) should go if it was to be kept at the same proportional distance from the others. I want to know how to calculate the slope and length of it.

Can anyone offer some advice on where I could start on working this out?

Answer 1

The line you need must pass through the intersection point of the two (extended) red lines, i.e. through $(-6,-3)$.

Hence the slope of the line is: $$ {5-(-3)\over3-(-6)}={8\over9}. $$ Now you can find the intersections of this line with the other two sides of the quadrilateral, to compute the length of the segment.

Answer 2

Hint : The distance of a point $(h,k)$ from a line ${\rm L_1}=a_1x+b_1y+c=0$ is given by $$d_1 (h,k)= \frac{|a_1h+b_1k+c_1|}{\sqrt{a_1^2+b_1^2}}$$

What you want is, locus of points such that

$$d_1(h,k)= \lambda \cdot d_2(h,k) $$

Where ${\rm L_1}$ is line passing through $(1,4)$ and $(3,6)$ ; ${\rm L_2}$ is line passing through $(2,1)$ and $(4,2)$.

Answer 3

Bear with me.

Line one is $(y - 1) = \frac 12(x-2)$

Line two is $(y-4) = (x - 1)$

Now your line. The first point $(x, y) = (2,1) + k((1,4) - (2,1))$ the end point is $(x,y) = (4,2) + k((3,6) - (4,2))$ and the slope is $\frac 12 + k(1 - \frac 12)$ where $k$ is the propotion (somewhere between $0$ and $1$). Basical it is the values of the first line plus $k$ times the differences of the values of the two lines.

So first point is $(2 - k, 1 + 3k)$ and the slope is $\frac 12(1+k)$

So formulat for you line is $(y- 1-3k) = \frac 12(1+k)(x + 2+k)$

And it contains the point $(3,5)$.

So $(5 - 1 - 3k) = \frac 12(1+k) (3 + 2 + k)$

$4-3k = \frac 12 (1+k)(5 + k)$

$8 - 6k= 5 + 6k + k^2$ so

$k^2 + 12k +3 = 0$

$k = \frac {-12 \pm \sqrt {144 - 12}}2=$

$-6 + \sqrt{36 - 3}=-6+\sqrt{33}$

The the equation of the line is $(y-19-3\sqrt{33}) = \frac {-5+\sqrt{33}}2(x-4+\sqrt{33})$