find the equation of the line that is equidistant from the points (3,2) and(-4,-3)
write the equation in slope intercept form.
ok what is equidistant?
Raat Key Rani !
You want to find the equation of the line that's made up of all the points that are equidistant from (3,2) and (-4,-3).
This will be the perpendicular bisector of the line between those 2 points.
The gradient between those 2 points is (2-(-3))/(3-(-4)) = 5/7.
The mid-point of the line is (-1/2,-1/2), by averaging each of the two co-ordinates.
So the perpendicular bisector will go through (-1/2,-1/2) and have a gradient of -7/5, so that it is perpendicular.
We can work out the y-intercept as being -1/2 - (-1/2*-7/5) = -6/5
Therefore, the equation of your line is y = -7/5x - 6/5. Or -5y = 7x+6 if you prefer.
Why did the vector cross the road?
It wanted to be normal.