Question

Question 20 If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Class X1 - Maths -Straight Lines Page 234

Answer 1

perpendicular distance from point (h,k) to line ,
ax + by + c = 0 is
D = |ah + bk + c|/√(a²+b²)

perpendicular distance of point P(x,y) from line x + y - 5 = 0 is |x + y - 5|/√(1²+1²)

perpendicular distance of point (x, y) from line 3x - 2y + 7 = 0 is |3x - 2y + 7|/√{3²+(-2)²}

A/C to question ,
|x + y - 5 |/√2 + |3x -2 y +7|/√13= 10
√13|x + y - 5| + √2|3x - 2y + 7| = 10√26
√13(x + y - 5) + √2(3x - 2y + 7) = 10√26
x(√13 + 3√2) + y(√13 - 2√2) -5√3+7√2 = 10√26

x(√13+3√2) + y(√13-2√2) -5√3+7√2-10√26 =0
which is the form of ax + by + c = 0
hence, it represents a line .
hence proved
____________

Answer 2

Answer:

blaa blaa

Step-by-step explanation:

it is easy

...

...