Question

On comparing the ratio coefficients, find out whether the lines representing the following linear equations intersect at a point ace parallel or coincide.
(a) 3x - 5y + 6 = 0
7x + 6y-9= 0
(b) 9x – 3y +12 = 0
18 + 6y +2 = 0​

Answer 1

Parallel lines -

a1/a2 = b1/b2 ≠ c1/c2

Intersecting lines -

a1/a2 ≠ b1/b2

Coincident lines -

a1/a2 = b1/b2 = c1/c2

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Given equations :-

(a)

3x - 5y = -6

7x + 6y = 9

Comparing the ratios,

3/7 ≠ -5/6

a1/a2 ≠ b1/b2

•°• The lines intersect.

(b)

9x - 3y = -12

18 + 6y = -2

Comparing the ratios,

9/18 = 1/2

-3/6 = 1/2

-12/-2 = 6/1

a1/a2 = b1/b2 ≠ c1/c2

•°• The lines are parallel.

Answer 2

To find :-

Whether the lines representing the following linear equations intersect at a point, are parallel or coincide.

Acknowledgement :-

◘ Unique solution (Lines intersect at a single point) :-

$\displaystyle\sf{\frac{a_1}{a_2}\neq \frac{b_1}{b_2} }$

◘ No solutions (Parallel lines) :-

$\displaystyle\sf{\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2} }$

◘ Infinitely many solutions (Lines coincide each other) :-

$\displaystyle\sf{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} }$

Given equations :-

(a) 3x - 5y + 6 = 0

7x + 6y-9= 0

(b) 9x – 3y +12 = 0

18 + 6y +2 = 0​

Solution :-

(a) a₁ = 3, a₂ = 7, b₁ = -5, b₂ = 6, c₁ = 6 & c₂ = -9.

Checking :-

$\displaystyle{\sf{\frac{a_1}{a_2}=\frac{3}{7} }}$

$\displaystyle{\sf{\frac{b_1}{b_2}=\frac{-5}{6} }}$

So, here we are already getting that, $\displaystyle\sf{\frac{a_1}{a_2}\neq \frac{b_1}{b_2} }$. So, the lines intersect at a point.

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(b) a₁ = 9, a₂ = 18, b₁ = -3, b₂ = 6, c₁ = 12 & c₂ = 2.

Checking :-

$\displaystyle{\sf{\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2} }}$

$\displaystyle{\sf{\frac{b_1}{b_2}=\frac{-3}{6}=\frac{-1}{2} }}$

So, here we are already getting that,$\displaystyle\sf{\frac{a_1}{a_2}\neq \frac{b_1}{b_2} }$.So, the lines intersect at a point.