On comparing the ratios a1a2, b1b2 and c1c2, and without drawing them,

Solution:

(i) Given equation are: $5 x+4 y+8=0$

7x + 6y − 9 = 0

Where, $a_{1}=5, b_{1}=-4, c_{1}=8$

$a_{2}=7, b_{2}=6, c_{3}=-9$

We have $\frac{a_{1}}{a_{2}}=\frac{5}{7}, \frac{b_{1}}{b_{2}}=\frac{-4}{6}=\frac{-2}{3}$ And $\frac{c_{1}}{c_{2}}=\frac{8}{-9} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Thus the pair of linear equation is intersecting.

(ii) Given equation are: $9 x+3 y+12=0$

18x + 6y + 24 = 0

Where, $a_{1}=9, b_{1}=3, c_{1}=12$

$a_{2}=18, b_{2}=6, c_{2}=24$

We have $\frac{a_{1}}{a_{2}}=\frac{9}{18}, \frac{b_{1}}{b_{2}}=\frac{3}{6} \frac{c_{1}}{c_{2}}=\frac{12}{24}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}$

Thus the pair of linear is coincident lines.

(iii) Given equation are: $6 x-3 y+10=0$

$2 x-y+9=0$

Where, $a_{1}=6, b_{1}=-3, c_{1}=10$

$a_{2}=2, b_{2}=-1, c_{2}=9$

We have $\frac{a_{1}}{a_{2}}=\frac{6}{2}, \frac{b_{1}}{b_{2}}=\frac{-3}{-1} \frac{c_{1}}{c_{2}}=\frac{10}{9}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=3$

Thus the pair of line is parallel lines.