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Question
Show that the equation `x^2-6xy+5y^2+10x-14y+9=0 ` represents a pair of lines. Find the acute angle between them. Also find the point of intersection of the lines.
Solution
`x^2-6xy+5y^2+10x-14y+9=0`
comaparing with `ax^2+2hxy+by^2+2gx+2fy+c=0`
we get a=1, h=-3, b=5, g=5, f=-7, c=9
Consider ` |[a,h,g],[h,b,f],[g,f,c]|`
`|[1,-3,5],[-3,5,-7],[5,-7,9]|`
=1(45-49)+3(-27+35)+5(21-25)
=(-4)+3(8)+5(-4)
=-4+24-20=0
Given equation represents a pair of lines
`"Now " tan theta =|(2sqrt(h^2-ab))/(a+b)|=|(2sqrt(9-5))/(1+5)|=2/3`
`theta =tan^(-1)(2/3)`
The point of intersection= `((hf-bg)/(ab-h^2),(gh-af)/(ab-h^2))`
`=((21-25)/(5-9),(-15+7)/(5-9))`
`=(1,2)`
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