Let $A(-6,9)$ and $B(10,-3)$ be points in the plane Find an equation of the perpendicular bisector of $AB$ the answer I get is $y=(4/3)x+(1/3)$ but it is not correct.
$\endgroup$ 54 Answers
$\begingroup$Here's how I'd do it (with details omitted):
- Find the midpoint $P$ of the line segment. (How?)
- Find the slope of the line segment. (How?)
- Find the slope of a line perpendicular to the line segment. (How?)
- Find the line of that slope that goes through $P$. (Using what equation?)
Your answer is correct. (Maybe, the answer you saw is wrong.)
The midpoint of the line segment $AB$ is $((-6+10)/2,(9-3)/2)=(2,3)$.
The slope of the line $AB$ is $(-3-9)/(10-(-6))=-3/4$. Hence, the slope of the line we want is $4/3$. (Why?)
These lead that the answer is $y-3=(4/3)(x-2)\iff y=(4/3)x+(1/3)$, which is what you wrote as your answer.
$\endgroup$ 1 $\begingroup$Say $C$ is the mid-point of line segment $AB$. So $C=(2,3)$
And slope of $AB$ is $\frac{-3}{4}$
Hence, slope of line-segment perpendicular to $AB$ is $\frac{4}{3}$
And hence, equation of perpendicular bisector of $AB$ is: $(y-3)=\frac{4}{3}$$(x-2)$
Solving this, $3y-4x=1$ is the required equation.
$\endgroup$ $\begingroup$Slope....midpoint ..... egad! Just find an equation for the set of points that are equidistant from both points $A$ and $B$.
hint: its easier to find the square of the distance than the distance, much easier than figuring out this diagram.
