Here is what we are given. The green line is the line x-y-1=0.
The dotted red line is the line 4x-3y+4=0. The plotted point is
(3,4)
First we find the equation of the left side of the rhombus.
So we find the equation of a line parallel to the red dotted line
that passes through the given point.
By solving the equation of the red dotted line for y, we find its
slope to be 4/3. So we find the equation of the line through (4,3)
with slope 4/3. That turns out to be the line y = 4/3x. That is
the blue line below:
Next we find the lower left vertex of the rhombus by
solving the system of the blue line and the green line
and find that vertex to be (-3,-4).
We now can find the other two vertices simply by counting
blocks. Let's put it on graph paper:
By counting to go from vertex (-3,-4) to vertex (3,4),
we must go 8 blocks up and 6 blocks right. Therefore to
get to the other vertex we must go 8 blocks right and
6 blocks up. That puts us at the point (5,2) which is
the third vertex of the rhombus. Then from (5,2) we
count 6 blocks right and 8 blocks up, which puts us at
the point (11,10), the fourth and final vertex of the
rhombus. So we plot those points, and draw the rhombus:
The reason it works to count blocks is because the
three red right triangles drawn below with blue
hypotenuses which are the sides of the rhombus, are
all congruent and their hypotenuses are all equal
in length making the four blue equal sides of the
rhombus. Note that we could also have found vertex
(11,10) by counting blocks going from (3,4) to
(11,10), counting 8 blocks up and 6 blocks right.
We traveled along the sides of those right triangles
6 units one way and 8 units the other, which is why
the red right triangles with blue hypotenuses
are congruent.
So the vertices of the rhombus are
(3,4), given, (-3,-4), (5,2) and (11,10).
Edwin
