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Problem
Problem
A straight line through
A
(
−
15
,
−
10
)
A(-15, -10)
A
(
−
15
,
−
10
)
meets the lines
x
−
y
−
1
=
0
x - y - 1 = 0
x
−
y
−
1
=
0
,
x
+
2
y
=
5
x + 2y = 5
x
+
2
y
=
5
and
x
+
3
y
=
7
x + 3y = 7
x
+
3
y
=
7
respectively at
A
,
B
A, B
A
,
B
and
C
C
C
. If
A
B
A
C
=
12
40
=
3
10
\frac{AB}{AC} = \frac{12}{40} = \frac{3}{10}
A
C
A
B
=
40
12
=
10
3
, prove that the line passes through the origin.
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