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.

Explanation

Equation `x^(2)-6xy+5y^(2)+10x-14y+9=0`
Compare with `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`
Then `a=1, h=-3, b=5, g=5, f=-7, c=9`
If equation represents a pair of line then
`abc+2fgh-af^(2)-bg^(2)-ch^(2)=0`
`therefore" "abc+2fgh-af^(2)-bg^(2)-ch^(2)`
`=1xx5xx9+2(-7)xx5(-3)-1(49)-5xx25-9xx9`
`=45+120-49-125-81`
`=255-255=0`
Hence, the given equaiton represents a pair of lines.
Acute angle, `tan theta=(2sqrt(h^(2)-ab))/(a+b)`
`therefore" "tan theta=(2sqrt(-5))/(1+5)=(2xx2)/(6)`
`therefore" "theta=tan^(-1)((2)/(3))`
Point of intersection `=((hf-bg)/(ab-h^(2)),(hg-af).(ab-h^(2)))`
`=(((-3)xx(-7)-5xx5)/(5-9),((-3)xx5-1xx(-7))/(5-9))`
`=((21-25)/(-4),(-15+7)/(-4))`
`=((-4)/(-4),(-8)/(-4))`
`therefore ` Intersection point is (1, 2)