Find the value of p so that the lines (11 - x)p = (3y - 3)/2 = (17 -

1.	Find the value of p so that the lines (11 - x)p = (3y - 3)/2 = (17 - z)/5 and (x - 22)/3p = (2y - 7)/27p = (z - 100)/(6/5).
Answer» Given the equation, (11 - x)/p = (3y - 3)/2 = (17 - 2)/5 ...(i) (x - 11)/-p = (y - 1)/(2/3) = (2 - 17)/-5 (Middle term dividing numerator denominator by 3) Again linear equation, (x - 22)/3p = (2y - 7)/(2 + p) = (2 - 100)/(6/5) (x - 22)/3p = (y - (7/2))/((2 + p)/2) = ((2 - 100)/(6/5) ...(ii) line (i) and (ii) perpendicular if a₁a₂ + b₁b₂ + c₁c₂= 0 -p(3p) + (2/3) x (27p/2) + (-5 x (6/5)) = 0 -3p² + 9p - 6 = 0 3p² - 9p + 6 = 0 (take sign common) 3p² - 6p - 3p + 6 = 0 3p(p - 2) - 3(p - 2) = 0 (p - 2) (3p - 3) = 0 when p - 2 = 0 then p = 2 when 3p - 3 = 0 then 3p = 3 p = 3/3 = 1 p = 1 ∴ p = 1, 2

Find the value of p so that the lines (11 - x)p = (3y - 3)/2 = (17 - z)/5 and (x - 22)/3p = (2y - 7)/27p = (z - 100)/(6/5).

Answer»

Given the equation,

(11 - x)/p = (3y - 3)/2 = (17 - 2)/5 ...(i)

(x - 11)/-p = (y - 1)/(2/3) = (2 - 17)/-5

(Middle term dividing numerator denominator by 3)

Again linear equation,

(x - 22)/3p = (2y - 7)/(2 + p) = (2 - 100)/(6/5)

(x - 22)/3p = (y - (7/2))/((2 + p)/2) = ((2 - 100)/(6/5) ...(ii)

line (i) and (ii) perpendicular if

a₁a₂ + b₁b₂ + c₁c₂= 0

-p(3p) + (2/3) x (27p/2) + (-5 x (6/5)) = 0

-3p² + 9p - 6 = 0

3p² - 9p + 6 = 0 (take sign common)

3p² - 6p - 3p + 6 = 0

3p(p - 2) - 3(p - 2) = 0

(p - 2) (3p - 3) = 0

when p - 2 = 0 then p = 2

when 3p - 3 = 0

then 3p = 3

p = 3/3 = 1

p = 1

∴ p = 1, 2