find the equation of the straight line passing through the point (1, 2) and having inclination π/3
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Answer:
Inclination, θ=
3
π
⇒tanθ=
3
⇒ Slope =
3
⇒ Equation y−2=
3
(x+1)
⇒y=
3
x+
3
+2
⇒p:
3
x+
3
+2=5−x
⇒x=
3
+1
3−
3
=
3
+1
3
(
3
−1)
⇒x=
2
3
(
3
−1)
2
=
2
3
(4−2
3
)
=
3
(2−
3
)
=2
3
−3
y=
3
(
3
(2−
3
)+1)+2
=
3
(2
3
−2)+2=6+2−2
3
=8−2
3
⇒AP=
(2
3
−3+1)
2
+(8−2
3
−2)
2
=
(2
3
−2)
2
+(6−2
3
)
2
=2
(
3
−1)
2
+(3−
3
)
2
=2
3+1−2
3
+9+3−6
3
=2
16−8
3
=4
4−2
3
AP=4(
3
−1).
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