Question

find the distance between the given line and point
1.) x = 7y - 3
(1, -1) Long method
2.) x-y = 1
(6,5) Short method
3.) 3x -4y -10 = 0
(-1 , 3) any method

help.. ;D

Answer 1

hi try graphing the number 1) 7y=x+3...y=(x+3)/7

Answer 2

do you know the distance formula?

Answer 3

using the point(1,-1), at x=1, y=(x+3)/7=4/7 this is point(1,4/7)
distance formula d=sqrt[(x2-x1)^2+(y2-y1)^2]
d=sqrt[(1-1)^2+(-1-4/7)^2]
d=-11/7

Answer 4

2.) x-y = 1 (6,5) set x=6
6-y=1
6-1=y
5=y, therefore this is point(6,5)
distance formula d=sqrt[(x2-x1)^2+(y2-y1)^2]
distance formula d=sqrt[(6-6)^2+(5-5)^2]
d=0

3.) 3x -4y -10 = 0 (-1 , 3) ,,set x=-1
3(-1)-4y-10=0
-3-10=-13=4y
-13/4=y
use point(-1,-13/4) and (-1,3)
d=sqrt[(-1+1)^+(-13/4 -3)^2
d=-25/4

Answer 5

absolute value nang AX BY +C all over 2a (short method)
square root of (X1+X2)(Y1+Y2) (long method)