Question

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

Answer 1

(i) Given:
15x − 18y + 1 = 0 ... (1)

12x + 10y − 3 = 0 ... (2)

6x + 66y − 11 = 0 ... (3)

Now, consider the following determinant:

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

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

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:

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

Hence, the given lines are concurrent.

(iii)

Given:
$bx+ay-ab=0$ ... (1)

$ax+by-ab=0$ ... (2)

x − y = 0 ... (3)

Now, consider the following determinant:

$\left|\begin{array}{ccc}b& a& -ab\\ a& b& -ab\\ 1& -1& 0\end{array}\right|=-b\times ab-a\times ab-ab\times \left(-a-b\right)=0$

Hence, the given lines are concurrent.