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

Comparing the equation
`x^(2) - 6xy + 10x - 14y + 9 = 0`
with `ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c = 0`, we get
`a = 1, h = -3, b = 5,g = 5, f = -7 and c = 9`
`therefore D = |{:(a,,"h",,"g"),(h,,b,,f),(g,,f,,c):}|=|{:(1,,"-3",,5),("-3",,5,,"-7"),(5,,"-7",,9):}|`
` = 1(45-49) + (-27 + 35) + 5(21 -25)`
` = -4 + 24 - 20 = 0`
`therefore` the given equation represents a pair a lines.
Let `theta` be the acute angle between the lines.
`therefore tan theta = |(2sqrt((-3)^(2) -1(5)))/(a+b)|`
`= |(2sqrt((-3)^(2) -1(5)))/(1+5)| = (4)/(6) = (2)/(3)`
`therefore theta = tan^(-1) ((2)/(3))`
The point of intersection of the lines is
`((bg-hf)/(h^(2)-ab),(af-hg)/(h^(2)-ab)) = ((25-21)/(9-5),(-7+15)/(9-5))` = (1,2).`