If (2, 9) (−2, 1), (6, 3) and the area of ∆ is 28, then find the altitude drawn from A on .
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Answer:
Solution:-
Let ABC be the triangle with vertices A(2,−2),B(1,1) and C(−1,0)&AD be the altitude of △ABC drawn from A.
Let m
1
&m
2
be the slope of line AD and BC respectively.
Now, AD⊥BC
∴m
1
×m
2
=−1
⇒m
1
=
m
2
−1
⟶(i)
Slope of line BC-
Slope of a line joining points (x
1
,y
1
)&(x
2
,y
2
)=
x
2
−x
1
y
2
−y
1
∴ Slope of BC joining B(1,1)&C(−1,0)=
−1−1
0−1
=
−2
−1
=
2
1
On substituting the value of m
2
in eq
n
(i), we get
m
1
=
(
2
1
)
−1
=−2
The equation of line passing through the point (x
1
,y
1
) with slope m is-
y−y
1
=m(x−x
1
)
∴ Equation of altitude AD passing through A(2,−2) with slope 2 is-
y−(−2)=−2×(x−2)
⇒y+2=−2x+4
⇒y=−2x+2
solution
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