Question

The locus of a point which moves so that the algebraic sum of the perpendiculars let fall from it on two given straight lines is constant,

Answer 1

Answer:

sry, I don't know the answer

Answer 2

Answer:

The locus of the point is a straight line.

Step-by-step explanation:

Let the two lines be L1: ax + by + c = 0 and L2: ex + fy + g = 0

Let the point be P (h,k).

Algebraic distance of P from L1 = (ah + bk + c)/ √(a² + b²)

Algebraic distance of P from L2 = (eh + fk + g)/ √(e² + f²)

The sum should be constant

Hence,

(ah + bk + c)/ √(a² + b²) + (eh + fk + g)/ √(e² + f²)  =  β (any constant)

( a/√(a² + b²) + e/√(e² + f²) ) h + ( b/√(a² + b²) + f/√(e² + f²) ) k  + c/√(a² + b²) - g/√(e² + f²)  - β = 0

(a/√(a² + b²) + e/√(e² + f²)) x + (b/√(a² + b²) + f/√(e² + f²)) y + c/√(a² +b²) - g/√(e² + f²)  - β = 0

It is of form lx + my + n = 0. Therefore, it represents a straight line.

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