Question

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)`?

Answer 1

Equation of line `l` is `2(c-a)x+2(d-b)y+(a^2+b^2-c^2-d^2)=0`

Explanation:

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

`(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2` or

`x^2-2ax+a^2+y^2-2by+b^2=x^2-2cx+c^2+y^2-2dy+d^2` or

`-2ax+2cx-2by+2dy+(a^2+b^2-c^2-d^2)=0` or

`2(c-a)x+2(d-b)y+(a^2+b^2-c^2-d^2)=0`

Answer 2

`2x(a-c)+2y(b-d)=(a^2+b^2)-(c^2+d^2), if, a!=c,b!=d`.

If, `a=c, b!=d", then eqn. is : "y=(b+d)/2`.

If, `a!=c, b=d", then eqn. is : "x=(a+c)/2`

Explanation:

Let `M` be the mid-pt. of the line segment `bar(PQ)`, where,

`P(a,b) and Q(c,d)`. Hence, `M((a+c)/2,(b+d)/2)`.

The slope of `bar(PQ)=(d-b)/(c-a), c!=a`

`rArr "the slope of the "bot-"bisector line l of "bar(PQ)` is given by,

`(-1)-:(d-b)/(c-a)=(a-c)/(d-b), dneb`.

`"Thus, the slope of line l is "(a-c)/(d-b), and, M in l`. Using, Slope-Pt.

Form for `l`, its eqn. is, `y-(b+d)/2={a-c)/(d-b)(x-(a+c)/2)`, i.e.,

`2y(d-b)+(b+d)(b-d)=2x(a-c)-(a-c)(a+c)`, or,

`2x(a-c)+2y(b-d)=(a^2+b^2)-(c^2+d^2), a!=c,b!=d`.

Case : 1 : `a=c, b!=d` :-

If, `a=c`, then, `bar(PQ)` 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((a+c)/2,(b+d)/2)`, and, as such, its eqn. will

be `l : y=(b+d)/2`.

Case : 2 : `b=d, a!=c` :-

In this case, the eqn. of `l` is `l : x=(a+c)/2.`

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

`2x(a-c)+2y(b-d)=(a^2+b^2)-(c^2+d^2)` by taking either `a=c, or, b=d`.

Finally, the Case `a=c, and, b=d` need not be considered, because, in that case, pts. `P and Q` coincide and hence segment `bar(PQ)` does not exist, so is the case with its perp. bsctr.

Enjoy Maths.!