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If the transversal y = m(r)x: r = 1,2,3 ...

If the transversal `y = m_(r)x: r = 1,2,3` cut off equal intercepts on the transversal `x +y = 1` then `1 +m_(1),1 +m_(2),1+m_(3)` are in

A

A.P.

B

G.P.

C

H.P.

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Solving `y = m_(r)x` and `x+y = 1`, we get `x = (1)/(1+m_(r))` and `y = (m_(r))/(1+m_(r))`.
Thus the points of intersection of the three lines on the transversal are
`P ((1)/(1+m_(1)),(m_(1))/(1+m_(1))), Q ((1)/(1+m_(2)),(m_(2))/(1+m_(2)))` and `R((1)/(1+m_(3)),(m_(3))/(1+m_(3)))`
According to question
`PQ = QR`
`((1)/(1+m_(1))-(1)/(1+m_(2)))^(2)+((m_(1))/(1+m_(1))-(m_(2))/(1+m_(2)))^(2)`
`= ((1)/(1+m_(2))-(1)/(1+m_(3)))^(2) +((m_(2))/(1+m_(2))-(m_(3))/(1+m_(3)))^(2)`
`rArr (m_(2)-m_(1))/(1+m_(1)) = (m_(3)-m_(2))/(1+m_(3))`
`rArr (1+m_(2))/(1+m_(1)) - 1 = 1 -(1+m_(2))/(1+m_(3))`
`rArr (1+m_(2))/(1+m_(1)) +(1+m_(2))/(1+m_(3)) =2`
`rArr 1+m_(2) =(2(1+m_(1))(1+m_(3)))/((1+m_(1))+(1+m_(3)))`
`rArr 1+m_(1), 1 +m_(2), 1 +m_(3)` are in H.P.
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