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(A) The equations of the given lines are:$ax^2 + 2hxy + by^2 = 0$ ..... $(i)$$a'x^2 + 2h'xy + b'y^2 = 0$ ..... $(ii)$Let the common line be $y = mx$.

Vedclass · Accepted Answer

$(ab' - a'b)^2 = 4(ah' - a'h)(hb' - h'b)$

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(A) The equations of the given lines are:$ax^2 + 2hxy + by^2 = 0$ ..... $(i)$$a'x^2 + 2h'xy + b'y^2 = 0$ ..... $(ii)$Let the common line be $y = mx$. Substituting this into both equations:$a + 2hm + bm^2 = 0$ ..... $(iii)$$a' + 2h'm + b'm^2 = 0$ ..... $(iv)$Using the method of cross-multiplication to eliminate $m$:$\frac{m^2}{2ha' - 2h'a} = \frac{-m}{ab' - a'b} = \frac{1}{2bh' - 2b'h}$From the first and third terms: $m^2 = \frac{2(ha' - h'a)}{2(bh' - b'h)} = \frac{ha' - h'a}{bh' - b'h}$From the second and third terms: $m = -\frac{ab' - a'b}{2(bh' - b'h)}$Squaring the expression for $m$: $m^2 = \frac{(ab' - a'b)^2}{4(bh' - b'h)^2}$Equating the two expressions for $m^2$:$\frac{ha' - h'a}{bh' - b'h} = \frac{(ab' - a'b)^2}{4(bh' - b'h)^2}$$(ab' - a'b)^2 = 4(ha' - h'a)(bh' - b'h) = 4(ah' - a'h)(hb' - h'b)$

If one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is coincident with one of the lines represented by $a'x^2 + 2h'xy + b'y^2 = 0$,then:

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