Prove that the following sets of three lines are concurrent:
(i) 15x − 18y + 1 = 0, 12x + 10y − 3 = 0 and 6x + 66y − 11 = 0

(ii) 3x − 5y − 11 = 0, 5x + 3y − 7 = 0 and x + 2y = 0

(iii) $\frac{x}{a} + \frac{y}{b} = 1, \frac{x}{b} + \frac{y}{a} = 1 and y = x .$

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Solution

(i) Given:
15x − 18y + 1 = 0 ... (1)

12x + 10y − 3 = 0 ... (2)

6x + 66y − 11 = 0 ... (3)

Now, consider the following determinant:

$|\begin{matrix} 15 & - 18 & 1 \\ 12 & 10 & - 3 \\ 6 & 66 & - 11 \end{matrix}| = 15 (- 110 + 198) + 18 (- 132 + 18) + 1 (792 - 60)$

$\Rightarrow |\begin{matrix} 15 & - 18 & 1 \\ 12 & 10 & - 3 \\ 6 & 66 & - 11 \end{matrix}| = 1320 - 2052 + 732 = 0$

Hence, the given lines are concurrent.

(ii)

Given:
3x − 5y − 11 = 0 ... (1)

5x + 3y − 7 = 0 ... (2)

x + 2y = 0 ... (3)

Now, consider the following determinant:

$|\begin{matrix} 3 & - 5 & - 11 \\ 5 & 3 & - 7 \\ 1 & 2 & 0 \end{matrix}| = 3 \times 14 + 5 \times 7 - 11 \times 7 = 0$

Hence, the given lines are concurrent.

(iii)

Given:
$b x + a y - a b = 0$ ... (1)

$a x + b y - a b = 0$ ... (2)

x − y = 0 ... (3)

Now, consider the following determinant:

$|\begin{matrix} b & a & - a b \\ a & b & - a b \\ 1 & - 1 & 0 \end{matrix}| = - b \times a b - a \times a b - a b \times (- a - b) = 0$

Hence, the given lines are concurrent.

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Similar questions

15 x - 18 y + 1 = 0, 12 x + 10 y - 3 = 0

6 x + 66 y - 11 = 0

Find the angle between each of the following pairs of straight lines:

(i) $3 x + y + 12 = 0$ and $x + 2 y - 1 = 0$

(ii) $3 x - y + 5 = 0$ and $x - 3 y + 1 = 0$

(iii) $3 x + 4 y - 7 = 0$ and $4 x - 3 y + 5 = 0$

(iv) $x - 4 y = 3$ and $6 x - y = 11$

(v) $(m^{2} - m n) y = (m n + n^{2}) x + n^{2}$ and $(m n + m^{2}) y = (m n + n^{2}) x + m^{3}$ .

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