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Hint: Here, first we will consider the roots of the cubic equation as the slopes of three lines and then apply the formula of sum of roots of the equation and product of roots of the equation.

First we will consider the statement-1.

The given equation of line \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\] is of third degree, so it represents family of three straight lines through the origin.

We know the roots of the cubic equation in one variable, so let us convert this equation into a cubic equation in one variable.

Let one of the line of the family be \[y=mx\], substituting this values in the given equation, we get

\[a{{x}^{3}}+b{{x}^{2}}(mx)+cx{{(mx)}^{2}}+d{{(mx)}^{3}}=0\]

\[a{{x}^{3}}+mb{{x}^{3}}+{{m}^{2}}c{{x}^{3}}+{{m}^{3}}d{{x}^{3}}=0\]

Dividing through out by \[{{x}^{3}}\], we get

\[a+bm+c{{m}^{2}}+d{{m}^{3}}=0.......(i)\]

This is a cubic equation.

Now let the roots of the equations be \[{{m}_{1}},{{m}_{2}},{{m}_{3}}\].

Then we know the product of all the three roots of cubic equation is,

\[{{m}_{1}}{{m}_{2}}{{m}_{3}}=-\dfrac{a}{d}........(ii)\]

Now let us consider two lines are perpendicular to each other, then we know the product of their slopes is \['-1'\]. So we have

\[{{m}_{1}}{{m}_{2}}=-1\]

So, equation (ii) becomes,

\[\begin{align}

& {{m}_{3}}(-1)=-\dfrac{a}{d} \\

& \Rightarrow {{m}_{3}}=\dfrac{a}{d} \\

\end{align}\]

So this is one of the root of equation (i), substituting this in equation (i), we get

\[a+\left( \dfrac{a}{d} \right)b+{{\left( \dfrac{a}{d} \right)}^{2}}c+{{\left( \dfrac{a}{d} \right)}^{3}}d=0\]

\[a+\dfrac{ab}{d}+\dfrac{c{{a}^{2}}}{{{d}^{2}}}+\dfrac{{{a}^{3}}}{{{d}^{2}}}=0\]

Taking the LCM, we get

\[\begin{align}

& \dfrac{a{{d}^{2}}+ab{{d}^{2}}+c{{a}^{2}}+{{a}^{3}}}{{{d}^{2}}}=0 \\

& \Rightarrow a{{d}^{2}}+ab{{d}^{2}}+c{{a}^{2}}+{{a}^{3}}=0 \\

\end{align}\]

Dividing through out by \['a'\], we get

\[{{d}^{2}}+b{{d}^{2}}+ca+{{a}^{2}}=0\]

So, if two lines are at right angles in the family of lines \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\], then \[{{a}^{2}}+ac+bd+{{d}^{2}}=0\].

So, Statement-1 is true.

Now consider Statement-2.

We know for a cubic equation\[p{{x}^{3}}+q{{x}^{2}}+rx+s=0\], if \[\alpha ,\beta \] and \[\gamma \] are the roots, then \[\alpha \beta \gamma \text{ }=-\dfrac{s}{p}\].

So statement -2 is also true and it is the correct explanation for statement-1.

Hence, the correct answer is option (1).

Note: We can solve the statement-1 by considering two lines perpendicular to each other is represented by equation \[{{x}^{2}}+2hxy-{{y}^{2}}=0\].

Now converting it to third degree homogeneous equation, we get

\[\left( {{x}^{2}}+2hxy-{{y}^{2}} \right)\left( ax-dy \right)=0\]

Now solving we get statement-1 as true. But proving statement-2 is the correct reason for statement-1 will be a lengthy process.

__Complete step-by-step solution -__First we will consider the statement-1.

The given equation of line \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\] is of third degree, so it represents family of three straight lines through the origin.

We know the roots of the cubic equation in one variable, so let us convert this equation into a cubic equation in one variable.

Let one of the line of the family be \[y=mx\], substituting this values in the given equation, we get

\[a{{x}^{3}}+b{{x}^{2}}(mx)+cx{{(mx)}^{2}}+d{{(mx)}^{3}}=0\]

\[a{{x}^{3}}+mb{{x}^{3}}+{{m}^{2}}c{{x}^{3}}+{{m}^{3}}d{{x}^{3}}=0\]

Dividing through out by \[{{x}^{3}}\], we get

\[a+bm+c{{m}^{2}}+d{{m}^{3}}=0.......(i)\]

This is a cubic equation.

Now let the roots of the equations be \[{{m}_{1}},{{m}_{2}},{{m}_{3}}\].

Then we know the product of all the three roots of cubic equation is,

\[{{m}_{1}}{{m}_{2}}{{m}_{3}}=-\dfrac{a}{d}........(ii)\]

Now let us consider two lines are perpendicular to each other, then we know the product of their slopes is \['-1'\]. So we have

\[{{m}_{1}}{{m}_{2}}=-1\]

So, equation (ii) becomes,

\[\begin{align}

& {{m}_{3}}(-1)=-\dfrac{a}{d} \\

& \Rightarrow {{m}_{3}}=\dfrac{a}{d} \\

\end{align}\]

So this is one of the root of equation (i), substituting this in equation (i), we get

\[a+\left( \dfrac{a}{d} \right)b+{{\left( \dfrac{a}{d} \right)}^{2}}c+{{\left( \dfrac{a}{d} \right)}^{3}}d=0\]

\[a+\dfrac{ab}{d}+\dfrac{c{{a}^{2}}}{{{d}^{2}}}+\dfrac{{{a}^{3}}}{{{d}^{2}}}=0\]

Taking the LCM, we get

\[\begin{align}

& \dfrac{a{{d}^{2}}+ab{{d}^{2}}+c{{a}^{2}}+{{a}^{3}}}{{{d}^{2}}}=0 \\

& \Rightarrow a{{d}^{2}}+ab{{d}^{2}}+c{{a}^{2}}+{{a}^{3}}=0 \\

\end{align}\]

Dividing through out by \['a'\], we get

\[{{d}^{2}}+b{{d}^{2}}+ca+{{a}^{2}}=0\]

So, if two lines are at right angles in the family of lines \[a{{x}^{3}}+b{{x}^{2}}y+cx{{y}^{2}}+d{{y}^{3}}=0\], then \[{{a}^{2}}+ac+bd+{{d}^{2}}=0\].

So, Statement-1 is true.

Now consider Statement-2.

We know for a cubic equation\[p{{x}^{3}}+q{{x}^{2}}+rx+s=0\], if \[\alpha ,\beta \] and \[\gamma \] are the roots, then \[\alpha \beta \gamma \text{ }=-\dfrac{s}{p}\].

So statement -2 is also true and it is the correct explanation for statement-1.

Hence, the correct answer is option (1).

Note: We can solve the statement-1 by considering two lines perpendicular to each other is represented by equation \[{{x}^{2}}+2hxy-{{y}^{2}}=0\].

Now converting it to third degree homogeneous equation, we get

\[\left( {{x}^{2}}+2hxy-{{y}^{2}} \right)\left( ax-dy \right)=0\]

Now solving we get statement-1 as true. But proving statement-2 is the correct reason for statement-1 will be a lengthy process.

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