Show that the two pairs of lines $12x^2+7xy-12y^2=0, 12x^2+7xy-12y^2-x+7y-1=0$ form a square. I know that both the equations represent a pair of straight lines. Also the first equation represents a pair of straight lines through origin which are as follows $(4x-3y)(3x+4y)=0$ But how should I calculate for the second equation and after calculating how should I prove.
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Working on the assumption that the second equation represents a pair of lines parallel to the first pair, you could try factorising the expression as $(4x-3y+a)(3x+4y+b)=0$. – Frentos Feb 04 '16 at 03:07
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The second pair of equations can be factored as (4x-3y+1)(3x+4y-1)
Thus, you have two pairs of parallel lines-
Pair 1: 4x-3y=0 and 4x-3y+1=0 Pair 2: 3x+4y=0 and 3x+4y-1=0
Two pairs of parallel lines intersect to make a square! Hope this helps :)
spandan madan
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