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(i) \(5x - 4y + 8 = 0, 7x + 6y – 9 = 0\)

(ii)\(9x + 3y + 12 = 0, 18x + 6y + 24 = 0\)

(iii)\( 6x - 3y + 10 = 0, 2x – y + 9 = 0\)

Answer :

Ans.(i) \(5x - 4y + 8 = 0, 7x + 6y – 9 = 0\)

On comparing these equations with the general form: \(ax^2 + bx + c\)

\(a_1 = 5, b_1 = -4, c_1 = 8\),

\(a_2 = 7, b_2 = 6, c_2 = -9 \)

Here \({{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}\) as,

\({{5}\over {7}} \ne {{-4}\over {6}}\)

So, these lines have a unique solution which means they intersect at one point.

(ii) \(9x + 3y + 12 = 0,18x + 6y + 24 = 0\)

On comparing these equations with the general form: \(ax^2 + bx + c\)

\(a_1 = 9, b_1 = 3, c_1 = 12\)

\(a_2 = 18, b_2 = 6, c_2 = 24 \)

Here \({{a_1}\over {a_2}} = {{b_1}\over {b_2}} = {{c_1}\over {c_2}}\) as,

\({{9}\over {18}} = {{3}\over {6}} = {{12}\over {24}}\)

So, these lines are coincident.

(iii) \(6x - 3y + 10 = 0,2x – y + 9 = 0\)

On comparing these equations with the general form: \(ax^2 + bx + c\)

\(a_1 = 6, b_1 = -3, c_1 = 10\)

\(a_2 = 2, b_2 = -1, c_2 = 9 \)

Here \({{a_1}\over {a_2}} = {{b_1}\over {b_2}} \ne {{c_1}\over {c_2}}\) as,

\({{6}\over {2}} = {{-3}\over {-1}} \ne {{10}\over {9}}\)

So, these lines are parallel to each other.

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