The vertices of triangle ABC are A (4, 6), B(1,5)
and C(7, 2). A line is drawn to intersect
sides AB and AC at D and E, respectively such that AD upon AB = AE upon AC = 1 upon 4. Determine the coordinates of D and E.
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Answer:
Given,
AB
AD
=
AC
AE
=
4
1
Thus,
DB
AD
=
EC
AE
=
3
1
For D, x=
m+n
mx
2
+nx
1
=
1+3
1.1+3.4
=
4
13
y=
m+n
my
2
+ny
1
=
1+3
1.5+3.6
=
4
23
∴D=(
4
13
,
4
23
)
for E, x=
1+3
1.7+3.4
=
4
19
y=
1+3
1.2+3.6
=5
∴E=(
4
19
,5)
ar(ΔABC)=
2
1
[4(5−2)+1(2−6)+7(6−5)]=
2
15
ar(ΔADE)=
2
1
[4(
4
23
−5)+
4
13
(5−6)+
4
19
(6−
4
23
)]=
4
9
∴
ar(ΔABC)
ar(ΔADE)
=
2
15
4
9
=
10
3
solution
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