Question

find the distance of a point (2,3,1/y+z_1=0=2x_3y_2z+4​

Answer 1

Answer:

2x−3y+9=0 ……..(i),

x−y+1=0 ………(ii)

Multiply equation (ii) by 2 and subtract it from equation (1)

2x−3y+9=0

2x−2y+2=0

_____________

−y+7=0

_____________

y=7,x=6

Point of intersection (6,7)

Distance between (2,3) and (6,7)=

(6−2)

2

+(7−3)

2

=4

2

units

∴ Distance of the point (2,3) from the line 2x−3y+9=0 measured along line x−y+1=0 is 4

2

units.