Question

The graph of the equations 5.r - 3y + 9 = 0 and
10x - 6y + 18 = 0, represents ... lines.
(A) Parallel
(B) Perpendicular
(CVCoincident (D) Intersecting​

Answer 1

Answer:

(C) coincident

Step-by-step explanation:

They will be coincident as a1/a2=b1/b2= c1/c2

Answer 2

$\tt The\:lines\:are\: \\ \\ \tt 5x - 3y + 9 = 0 \\ \tt 10x - 6y + 18 = 0 \\ \\ \tt Here \\ \tt {a}_{1} = 5 \:\: {b}_{1} = -3 \:\: {c}_{1} = 9 \\ \tt {a}_{2} = 10 \:\: {b}_{2} = -6 \:\: {c}_{2} = 18 \\ \\ \tt \implies \frac{{a}_{1}}{{a}_{2}} = \frac{5}{10} = \frac{1}{2} \\ \\ \tt \implies \frac{{b}_{1}}{{b}_{2}}= \frac{-3}{-6} = \frac{1}{2} \\ \\ \tt \implies \frac{{c}_{1}}{{c}_{2}} = \frac{9}{18} = \frac{1}{2} \\\\ \tt \therefore \boxed{\bold{\underline{\green{\tt \frac{{a}_{1}}{{a}_{2}} = \frac{{b}_{1}}{{b}_{2}} = \frac{{c}_{1}}{{c}_{2}} }}}} \\ \\ \tt We\:know\:that \\ \\ \tt If \\ \\ \tt \frac{{a}_{1}}{{a}_{2}} = \frac{{b}_{1}}{{b}_{2}} = \frac{{c}_{1}}{{c}_{2}} \implies Coincident\:lines \\ \\ \tt \frac{{a}_{1}}{{a}_{2}} = \frac{{b}_{1}}{{b}_{2}} ≠ \frac{{c}_{1}}{{c}_{2}} \implies Parallel\:lines \\ \\ \tt \frac{{a}_{1}}{{a}_{2}} ≠ \frac{{b}_{1}}{{b}_{2}} ≠ \frac{{c}_{1}}{{c}_{2}} \implies Intersecting\: lines \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt Correct\:option\:is\:(C)\: Coincident }}}}$