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Q.

Show that the equation 2x2 – 13xy – 7y2 + x +23y – 6 = 0 represents a pair of straight lines. Also find the angle between them and the coordinates of the point of intersection of the lines.

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Detailed Solution

Given equation is
2x213xy7y2+x+23y6=0
comparing the equation with we get
a=2,2h=13h=132,b=7,2g=1g=12,2f=23f=232,c=6
 Consider i) abc+2fghaf2bg2ch2
=2(7)(6)+22321213222322(7)122(6)1322 =84129942×5294+714+6×1694=3362991058+7+10144=0abc+2fghaf2bg2ch2=0
 Now ii) h2ab=1322(2)(7)>0h2>ab
g2ac=1222(6)>0g2>acf2bc=2322(7)(6)>0f2>bc
From (i) (ii) the given equation represents a pair of lines
Point of intersection=hfbgabh2,ghafabh2
=132×232(7)122(7)1322,12×1322×2322(7)1322
=2994+72141694,13423141694=299+1456169,139256169=285225,105225=1915,715
If θ is the acute angle between the lines then
cosθ=|a+b|(ab)2+4h2            =|27|(2+7)2+41322               =|5|(9)2+41694                 =|5|81+169                                    =|5|250                     =5510θ=cos1110

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