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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 |
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Answer» A.P. 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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