Find the distance of the point (-2,3) from the line x - 3y - 2 = 0, measured parallel to the line 2x + y - 1 = 0.
Answers
The distance of the point (2 3) from the line 2x-3y+9=0 measured along the line x-y+1=0 is 4\sqrt{2}4
2
units.
Step-by-step explanation:
The given point is (2,3).
The given lines are
2x-3y+9=02x−3y+9=0 .... (1)
x-y+1=0x−y+1=0 ....(2)
Multiply equation 2 by 3.
3x-3y+3=03x−3y+3=0 ... (3)
Subtract equation (3) from (1).
2x-3y+9-(3x-3y+3)=02x−3y+9−(3x−3y+3)=0
2x-3y+9-3x+3y-3=02x−3y+9−3x+3y−3=0
-x+6=0−x+6=0
x=6x=6
Put this value in (1).
2(6)-3y+9=02(6)−3y+9=0
21-3y=021−3y=0
y=7y=7
Therefore intersection point of both lines is (6,7).
Distance formula:
D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}D=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
The distance between (2,3) and (6,7) is
D=\sqrt{(6-2)^2+(7-3)^2}=4\sqrt{2}D=
(6−2)
2
+(7−3)
2
=4
2
Therefore the distance of the point (2 3) from the line 2x-3y+9=0 measured along the line x-y+1=0 is 4\sqrt{2}4
2
units.
Answer:
Step-by-step explanation:
line x-y+1=0 is along a point (2,3)
so,x-y+1=0 intersect 2x-3y+9=0 at (6,7)
we get by solving 2x-3y+9 and x-y+1=0
now, we have two point for applying distance formula
d=[(6-2)^2+(7-3)^2]^1/2
final ans is 4{2}^1/2 also write 5.64
thank you