Find the distance of the line 4x + 7y + 5 = 0 from the point(1, 2) along the line 2x - y =0
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Answer:
The equation of line AB is
4x+7y+5=0 ....(i)
and equation of line PQ is
2x−y=0 ...(ii)
On solving equations (i) and (ii), we get
x=−
18
5
and y=−
9
5
∴ Coordinates of Q=(−
18
5
,−
9
5
)
Length of QP=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(1+
18
5
)
2
+(2+
9
5
)
2
=
(
18
23
)
2
+(
9
23
)
2
=
9
23
4
1
+1
=
9
23
×
2
5
=
18
23
5
sq units.
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