Question

P(−1,4) and Q(11,−8) divide AB harmonically in the ratio 2:3, then A,B

in order are​

Answer 1

Step-by-step explanation:

P (-1, 4) abd Q (11, -8) divides AB harmonically in ration 3:2 then:

Let A has coordinates (a, b)

B has coordinates (x, y)

We know that coordinates of a point dividing a line internally in ration m:n has.

abscissa =

m+n

mx

2

+nx

1

ordinate =

m+n

my

2

+ny

1

Given P has coordinates (-1 , 4)

m = 3, n = 2, we get:

−1=

5

3x+2a

and 4=

5

3y+2b

3x + 2a = -5 -----eq.1

3y + 2b = 20 ----eq.2

According to the question, as P and Q are harmonic conjugate, if P divides in ration m:n, Q will divide in ratio -m : n, hence,

Putting in the above discribed formula, we get:-

11=

−3+2

−3x+2a

−11=−3x+2a

-3x + 2 a = -11 -----eq.3

−3+2

−3y+2b

=−8

-3y + 2b = 8 -----eq.4

solving 1, 2 with 3, 4 we get:-

a = -4 :

b = 7

we get A (a,b) = (-4 , 7)

3x+2a=5

3x=−5−2(−4)

3x=3

x=1

and 3y=20−2b

3y=20−14

3y=6

y=2

We get B (x , y) = (1 , 2)

Hence, ans = (-4, 7) and (1 , 2)