Question

The line joining the points (2,1) and (5, -8) is trisected at the points P and Q. If point P lies on the line

2x – y +k=0. Find the value of k.
NO SCAM PLEASE ​

Answer 1

Step-by-step explanation:

The equation of the line joining (2,1,1) and (1,3,-2) can be written as

x-2/2–1 = y-1/1–3 = z-1/1-(-2)

i.e . x-2 = y-1/-2 = z-1/3 which is equal to some constant 'c'

Hence

Equation of line is

x-2 = y-1/-2 = z-1/3 = c

Any general point on this line is hence

X = c +2

Y = -2c +1

Z = 3c+1

( By equating each fraction equal to c )

Now this point must Lie on the plane

( c +2 , -2c+1 , 3c+1 )

Therefore it must satisfy the equation of plane

i.e. x + 3y + 2z - 6 = 0

(c+2) + 3(-2c+1) + 2(3c+1) - 6 =0

Solving for 'c'

We get c = -1

Hence the point must be

X= c+2 = -1+2 = 1

Y=-2c+1 = 3

Z =3c +1 = -2

Hence the point where plane meets the line PQ is (1,3,-2)