# Equation of the straight line

• Mar 26th 2010, 02:01 AM
lindros
Equation of the straight line
The coordinates of $E,F$ and $H$ of an equilateral are $(3,3),(0,-1)$and $(6,2)$ respectively. $FH$ is perpendicular to $EG$ whereas the straight line $FG$ is parallel to $x-axis$.Find

a)the equation of straight line $EG$

b)the coordinates of point $G$

c)the equation of locus $R$ such that the distance of $R$ from point $E$ and $F$ are equal.
• Mar 26th 2010, 02:22 AM
sa-ri-ga-ma
• Mar 26th 2010, 02:23 AM
sa-ri-ga-ma
• Mar 26th 2010, 05:29 AM
HallsofIvy
Quote:

Originally Posted by lindros
The coordinates of $E,F$ and $H$ of an equilateral are $(3,3),(0,-1)$and $(6,2)$ respectively. $FH$ is perpendicular to $EG$ whereas the straight line $FG$ is parallel to $x-axis$.Find

a)the equation of straight line $EG$

Since EG is perpendicular to FH, which has slope 1/2, EG has slope -2. That means that EG has equation y= -2(x- 3)+ 3.

Quote:

b)the coordinates of point $G$
Since FG is parallel to the x-axis, G must have the same y coordinate as F: -1. Solve -1= -2(x- 3)+ 3 to find the x coordinate.

Quote:

c)the equation of locus $R$ such that the distance of $R$ from point $E$ and $F$ are equal.
This is the straight line perpendicular to EF and passing through its midpoint. EF has slope (3-(-1))/(3- 0)= 4/3 so this line has slope -3/4. The midpoint of EF is ((3+(-1))/2, (3+ 0)/2).