# Let P(a,b) and Q(c,d) be two points in the plane. Find the equation of the line l that is the perpendicular bisector of the line segment bar(PQ)?

Sep 6, 2016

Equation of line $l$ is $2 \left(c - a\right) x + 2 \left(d - b\right) y + \left({a}^{2} + {b}^{2} - {c}^{2} - {d}^{2}\right) = 0$

#### Explanation:

A line which is perpendicular bisector of the line joining $P \left(a , b\right)$ and $Q \left(c , d\right)$ and passes through their midpoint is locus of a point which is equidistant from these two points. Hence, equation is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {\left(x - c\right)}^{2} + {\left(y - d\right)}^{2}$ or

${x}^{2} - 2 a x + {a}^{2} + {y}^{2} - 2 b y + {b}^{2} = {x}^{2} - 2 c x + {c}^{2} + {y}^{2} - 2 \mathrm{dy} + {d}^{2}$ or

$- 2 a x + 2 c x - 2 b y + 2 \mathrm{dy} + \left({a}^{2} + {b}^{2} - {c}^{2} - {d}^{2}\right) = 0$ or

$2 \left(c - a\right) x + 2 \left(d - b\right) y + \left({a}^{2} + {b}^{2} - {c}^{2} - {d}^{2}\right) = 0$

Sep 6, 2016

$2 x \left(a - c\right) + 2 y \left(b - d\right) = \left({a}^{2} + {b}^{2}\right) - \left({c}^{2} + {d}^{2}\right) , \mathmr{if} , a \ne c , b \ne d$.

If, $a = c , b \ne d \text{, then eqn. is : } y = \frac{b + d}{2}$.

If, $a \ne c , b = d \text{, then eqn. is : } x = \frac{a + c}{2}$

#### Explanation:

Let $M$ be the mid-pt. of the line segment $\overline{P Q}$, where,

$P \left(a , b\right) \mathmr{and} Q \left(c , d\right)$. Hence, $M \left(\frac{a + c}{2} , \frac{b + d}{2}\right)$.

The slope of $\overline{P Q} = \frac{d - b}{c - a} , c \ne a$

$\Rightarrow \text{the slope of the "bot-"bisector line l of } \overline{P Q}$ is given by,

$\left(- 1\right) \div \frac{d - b}{c - a} = \frac{a - c}{d - b} , \mathrm{dn} e b$.

$\text{Thus, the slope of line l is } \frac{a - c}{d - b} , \mathmr{and} , M \in l$. Using, Slope-Pt.

Form for $l$, its eqn. is, $y - \frac{b + d}{2} = \frac{a - c}{d - b} \left(x - \frac{a + c}{2}\right)$, i.e.,

$2 y \left(d - b\right) + \left(b + d\right) \left(b - d\right) = 2 x \left(a - c\right) - \left(a - c\right) \left(a + c\right)$, or,

$2 x \left(a - c\right) + 2 y \left(b - d\right) = \left({a}^{2} + {b}^{2}\right) - \left({c}^{2} + {d}^{2}\right) , a \ne c , b \ne d$.

Case : 1 : $a = c , b \ne d$ :-

If, $a = c$, then, $\overline{P Q}$ is vertical , i.e., parallel to the Y-axis,

and, so, the reqd. line $l$ will be horizontal , i.e., parallel to the X-

axis passing through $M \left(\frac{a + c}{2} , \frac{b + d}{2}\right)$, and, as such, its eqn. will

be $l : y = \frac{b + d}{2}$.

Case : 2 : $b = d , a \ne c$ :-

In this case, the eqn. of $l$ is $l : x = \frac{a + c}{2.}$

In both of the Cases, eqn. of $l$ can be derived from

$2 x \left(a - c\right) + 2 y \left(b - d\right) = \left({a}^{2} + {b}^{2}\right) - \left({c}^{2} + {d}^{2}\right)$ by taking either $a = c , \mathmr{and} , b = d$.

Finally, the Case $a = c , \mathmr{and} , b = d$ need not be considered, because, in that case, pts. $P \mathmr{and} Q$ coincide and hence segment $\overline{P Q}$ does not exist, so is the case with its perp. bsctr.

Enjoy Maths.!