Graphics Reference
In-Depth Information
L
2
:
t
=
−
+
=−
+
2
j
i
j
i
j
2
2
x
v
y
v
x
t
y
t
10
r
2
−
=
−
=
1
0
−
11
The zero discriminant confirms a touch condition. Therefore,
=
1
and the touch point is
x
P
=
1 and y
P
=
2
which is correct.
L
3
:
t
=
0
−
i
+
j
=−
i
−
j
2
2
1
√
2
1
√
2
x
v
y
v
x
t
r
2
−
=
1
−
=
1
y
t
−
1
−
1
The positive discriminant confirms an intersect condition. Therefore,
√
2
and
√
2
2
√
2
±
=
1
=
+
1
−
1
If we use
x
P
=
x
t
3
+
x
v
3
and y
P
=
y
t
3
+
y
v
3
the intersection points are
√
2
2
√
2
1
√
2
1
√
2
=
=
+
1
x
P
=
+
1
+
√
2
2
√
2
1
1
√
2
=
y
P
=
+
1
+
√
2
2
√
2
1
=
√
2
1
√
2
=
−
1
x
P
=
−
1
−
√
2
2
√
2
1
1
√
2
=
y
P
=
−
1
−
The intersection points are
1
2
and
1
2
, which are correct.
√
2
2
1
√
2
√
2
2
1
√
2
+
+
−
−
