Find the values of rr for which the line (r+5)x–(r–1)y+r2+4r−5=0 is 1.parallel to the xx-axis ?
2.parallel to the line 2x−y+4=0 ?
3.parallel to the yy-axis ?
4.perpendicular to the line x−y+32=0?
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Answer:
Given P is equidistant from points A and B
PA=PB .....(1)
and Q is equidistant from points A and B
QA=QB .....(2)
In △PAQ and △PBQ
AP=BP from (1)
AQ=BQ from (2)
PQ=PQ (common)
So, △PAQ≅△PBQ (SSS congruence)
Hence ∠APQ=∠BPQ by CPCT
In △PAC and △PBC
AP=BP from (1)
∠APC=∠BPC from (3)
PC=PC (common)
△PAC≅△PBC (SAS congruence)
∴AC=BC by CPCT
and ∠ACP=∠BCP by CPCT ....(4)
Since, AB is a line segment,
∠ACP+∠BCP=180
∘
Thus, AC=BC and ∠ACP=∠BCP=90
∘
∴,PQ is perpendicular bisector of AB.
Hence proved.
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