Find the value of 32 ^ y_{3} * x * (1/2) ^ (2/3) = 4 ^ k;
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Given: A(a,3),B(2,1) and C(5,a) are collinear.
∴ Slope of AB= Slope of BC
⇒
2−a
1−3
=
5−2
a−1
⇒
2−a
−2
=
3
a−1
⇒−6=(2−a)(a−1)
⇒−6=2a−2−a
2
+a
⇒a
2
−3a−4=0
⇒a
2
−4a+a−4=0
⇒(a−4)(a+1)=0
a=4,−1
For a=4
Slope of BC=
5−2
a−1
=
3
4−1
=
3
3
=1
Equation of BC;(y−1)=1(x−2)
⇒y−1=x−2
⇒x−y=1
For a=−1
Slope of BC=
5−2
a−1
=
3
−1−1
=−
3
2
Equation of BC:(y−1)=−
3
2
(x−2)⇒3y−3=4−2x
⇒2x+3y=7
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