Finding distance of point to y-axis.

[V0ODO0CH1LD]

Homework Statement

In a cartesian coordinate system, two lines r and s, with angular coefficients 2 and 1/2, respectively, intercept at the origin. If B \in r and C \in s are two points in the first quadrant such that the line segment BC is perpendicular to r and the area of the triangle OBC is 12x10^-1, then what is the distance from point B to the y axis?

Homework Equations

The Attempt at a Solution

Okay, first I found the relationship between OC and BC using the formula for the area of a triangle:
[OBC] = 12x10^-1 = 12/10 = 6/5 = (OB BC)/2

Now I am stuck. Whatever I do it seems like I am going around in circles..

 

[Simon Bridge]

|OB| and |OC| don't have to be the same length.
But you do know that [OBC] forms a special kind of triangle, and you know one of the apex angles.

You also need a point Q=(0,y) and look at the triangle [OBQ].
You know one of these apex angles too.

 

[V0ODO0CH1LD]

|OB| and |OC| don't have to be the same length.

Yeah, I know. But the relationship between |BC| and |OB| is that |BC| = 12/(5|OB|); right?

But you do know that [OBC] forms a special kind of triangle, and you know one of the apex angles.

I did find all the angles of the triangle OBC, but they're not your usual angles and since this problem was taken from a test where calculators are not allowed, you kind of know you're going the wrong way..

I think I am actually missing some sort of theorem or property that would make this whole problem easier, but I have no idea which.

 

[Simon Bridge]

Reformulating - you want the x-coordinate of point B.

 

[Simon Bridge]

... so, rewrite everything in terms of coordinates:

$y_r=2x, y_s=\frac{1}{2}x$ ... right?
So the position of the ponts B and C can be writen:
$\vec{B}=(x_b, y_b)=(1,2)x_b$
$\vec{C}=(x_c,y_c)=(2,1)x_c$

... and the area of the triangle [OBC] is:
$A=\frac{1}{2}\sqrt{x_b^2+y_b^2}\sqrt{(x_b-x_c)^2+(y_b-y_c)^2}$

... that's three equations and four unknowns: you need one more equation: the line through B perpendicular to r, or Pythagoras on the sides of [OBC]?