Question

If sum of the perpendicular distances of a variable point P ( x , y ) from the lines x + y – 5 = 0 and 3 x – 2 y + 7 = 0 is always 10. Show that P must move on a line.

Answer 1

The sum of the perpendicular distances of the point ( x,y ) from the lines x+y−5=0 and 3x−2y+7=0 is always 10.

The equation of the lines is

x+y−5=0 (1)

3x−2y+7=0(2)

The formula for the perpendicular distance d of a line Ax+By+C=0 from a point ( x 1 , y 1 ) is given by,

d= | A x 1 +B y 1 +C | A 2 +B 2 (3)

Let A 1 ,  B 1 ,  C 1 and A 2 ,  B 2 ,  C 2 be the values for the line of equation (1) and equation (2).

Compare equation (1) and equation (2) with the general form of equation of line Ax+By+C=0.

A 1 =1,  B 1 =1,  C 1 =−5 A 2 =3,  B 2 =−2,  C 2 =7

Let d 1 and d 2 be the perpendicular distances of the point ( x,y ) from the lines x+y−5=0 and 3x−2y+7=0.

Substitute the values of A 1 ,  B 1 ,  C 1 in equation (1).

d 1 = | x+y−5 | 1 2 +1 2 d 1 = | x+y−5 | 2

Substitute the values of A 2 ,  B 2 ,  C 2 in equation (2).

d 2 = | 3x−2y+7 | 3 2 +2 2 d 2 = | 3x−2y+7 | 9+4 d 2 = | 3x−2y+7 | 13

The sum of the distances is 10 units.

d 1 + d 2 =10

Substitute the values of d 1 and d 2 in the above expression.

| x+y−5 | 2 + | 3x−2y+7 | 13 =10 13 ×| x+y−5 |+ 2 ×| 3x−2y+7 | 26 =10 13 ×| x+y−5 |+ 2 ×| 3x−2y+7 |=10× 26

Further simplify the above expression assuming ( x+y−5 ) and ( 3x−2y+7 ) to be positive.

13 ×( x+y−5 )+ 2 ×( 3x−2y+7 )=10× 26 13 x+ 13 y−5 13 +3 2 x−2 2 y+7 2 =10 26 ( 13 +3 2 )x+( 13 −2 2 )y+( 7 2 −5 13 −10 26 )=0

Thus, the point P will move on the line.