Help: tricky geometry question (1 Viewer)

[movie_detective]

Hello,

Require help with this tough question: The equations of two adjacent sides of a rhombus are y=2x +4 and y=-1/3x +4. If A( 12,0 ) is one vertex and all vertices have positive coordinates, find the coordinates of the other three vertices.*


Thx to all who provide support

 

[Kurosaki]

Graph it first. You also know that the opposite sides are parallel and all sides are equal (rhombus).
Also where they intersect should be a vertex I think (the two lines)

Use the distance formula and that kind of stuff.
And parallel lines have same gradient

 

[movie_detective]

I graphed it and found another vertex where they intersect as u said . However could not determine the other two.

 

[Kurosaki]

The opposite sides are parallel and they are all equal in length.
Think about it

 

[movie_detective]

Yep, I'm taking all that into account however still unable to determine how to find the other 2 vertices. By the way, thx for responding and helping out

 

[Kurosaki]

No worries bro. I enjoy doing maths and helping people .find the distance between (0,4) and (12,0).
Then taking into account that the lengths if the sides are all the same, and that parallel lines have the same gradient, find the vertices

 

[Drongoski]

The other 3 vertices are:

 

[Drongoski]

One solution:

Let the 4 vertices, in anti-clockwise order, be A,B,C,D with co-ords of A & D being (0,4) & (0,4) resp.

Let the unknown co-ords be: B(p,q), C(r,s)

We'll use fact: AD = AB and DC // AB to solve for p & q.

gradient of AB = (q-0)/(p-12) = 2 (the gradient of DC)

.: q = 2(p-12) . . . . . [1]

Since AB² = da²

(p-12)^2 + (q-0)^2 = (12-0)^2 + (0-4)^2 = 160

.: (p-12)^2 = 160 - q^2

.: [2(p-12)]^2 = 4[160 - q^2]

by [1], q^2 = 4[160 - q^2]

Solving we get q = 8sqrt(2), noting that q > 0.

by [1], p = 12 + q/2 = 12 + 4sqrt(2).


Instead of using fact: grad of AC * grad of BD = -1 (diags of rhombus are perpendicular)

maybe it's easier to use fact that mid-point of AC = mid-point of BD.

.: (r+12)/2 = (p+0)/2 and (s+0)/2 = (q+4)/2

Solving, we get: r = 4sqrt(2) and s = 4 + 8sqrt(2)