Given pair of lines is

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0 ...\left(i\right)$

For option $\left(1\right)$, If the slope of one line is negative of the other.

Let the lines represented by $\left(i\right)$ are

$lx+my+n_1=0 \text{\&} lx-my+n_2=0$

$\therefore a{x}^2+2hxy+b{y}^2+2gx+2fy+c=\left(lx+my+n_1\right)\left(lx-my+n_2\right)$

Comparing the coefficient, we get

$l^2=a, -m^2=b, lm-lm=2h,$

$l{n}_2+l{n}_1=2g, m{n}_2-m{n}_1=2f, c=n_1{n}_2$

$\therefore h=0$

Hence, option $\left(1\right)$ is correct.

For option $\left(2\right)$

As, we know that if two lines are parallel then

$\frac{a}{h}=\frac{h}{b}=\frac{g}{f}$

$\therefore af=gh$ so, $2f\left(gh+af\right)=0$if $f=0$

So, option $\left(2\right)$ is wrong

For option $\left(3\right)$,

Let the two lines intersecting at origin be $l_{1}x+m_{1}y=0 \& l_{2}x+m_{2}y=0$

$\therefore a{x}^2+2hxy+b{y}^2+2gx+2fy+c=\left(l_{1}x+m_{1}y\right)\left(l_{2}x+m_{2}y\right)$

Comparing, we get

$g=f=c=0$

$\therefore$Option $\left(3\right)$ is also wrong.

For option $\left(4\right)$,

X-coordinate of intersecting pair of lines in $\left(i\right)$ is given by $\frac{bg-hf}{h^2-ab}$

If $\frac{bg-hf}{h^2-ab}>0$ then either $bg-hf>0 \& h^2-ab>0$ or $bg-hf<0 \& h^2-ab<0$

$\therefore$Option $\left(4\right)$ is also wrong.