A line 2x+3y=6 meets x axis at A and
y axis at B.The line OC divide the line
segment AB in the ratio 2:3 where O is
the origin. Find the equation of line OC.
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ANSWER
2x+3y=12
6
x
+
4
y
=1
Using point form,
A≡(6,0);B=(0,4)
also, slope of line AB=
3
−2
So, slope of perpendicular =
m
−1
=
2
+3
So, equation will be,
y−5=
2
3
(x−5)
3x−2y=5
5
3x
−
5
2y
=1
Using point form, C≡(
3
5
,0);D≡(0,
2
−5
)
and solving with 2x+3y=12, we get E≡(3,2).
So, O≡(0,0)
C≡(
3
5
,0)
E≡(3,2)
B≡(0,4)
Ar(OCEB)=∣Ar(OCE)∣+∣Ar(OEB)∣
Ar(OCEB)=
2
1
×
3
5
+
2
1
×12=
6
41
Step-by-step explanation:
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