Answer: The given lines are 2x−y=0…(1) 4x+7y+5=0….(2) A(1,2) is point on line (1). Let B be the point of intersection of lines (1) and (2). On solving equation (1) and (2), we obtainx= 18 −5 andy= 9 −5 ∴coordinates of point B are{ 18 −5 , 9 −5 } By using distance formula, the distance between point A and B can be obtained asAB= {1+ 18 5 } 2 +{2+ 9 5 } 2 units = { 18 23 } 2 +{ 9 23 } 2 units = { 2⋅9 23 } 2 +{ 9 23 } 2 units = { 9 23 } 2 { 2 1 } 2 +{ 9 23 } 2 units = ( 9 23 ) 2 ( 4 1 +1) units = 9 23 4 5 units = 9 23 ⋅ 2 5 units = 18 23 5 units Thus, the required distance is= 18 23 5 units Reply
Answer:
The given lines are
2x−y=0…(1)
4x+7y+5=0….(2)
A(1,2) is point on line (1).
Let B be the point of intersection of lines (1) and (2).
On solving equation (1) and (2), we obtainx=
18
−5
andy=
9
−5
∴coordinates of point B are{
18
−5
,
9
−5
}
By using distance formula, the distance between point A and B can be obtained asAB=
{1+
18
5
}
2
+{2+
9
5
}
2
units
=
{
18
23
}
2
+{
9
23
}
2
units
=
{
2⋅9
23
}
2
+{
9
23
}
2
units
=
{
9
23
}
2
{
2
1
}
2
+{
9
23
}
2
units
=
(
9
23
)
2
(
4
1
+1)
units
=
9
23
4
5
units
=
9
23
⋅
2
5
units
=
18
23
5
units
Thus, the required distance is=
18
23
5
units