Question

Prove that the following sets of three lines are concurrent:

$\left(i\right)15x-18y+1=0,12x+10y-3=0$ and $6x+66y-11=0$

$\left(ii\right)3x-5y-11=0,5x+3y-7=0$ and $x+2y=0$

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

Answer 1

(i) Given

$15x-18y+1=0...\left(i\right)$

$12x+10y-3=0...\left(ii\right)$

$6x+66y-11=0...\left(iii\right)$

Now, consider the following determinant:

$\left|\begin{array}{ccc}15& -18& 1\\ 12& 10& -3\\ 6& 66& -11\end{array}\right|$

$=15(-110+198)+18(-132+18)+1(792-60)$

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

Hence, the given lines are concurrent.

$\left(ii\right)3x-5y-11=0...\left(i\right)$

$5x+3y-7=0...\left(ii\right)$

$x+2y=0...\left(iii\right)$

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:

$bx+ay-ab=0...\left(i\right)$

$ax+by-ab=0...\left(ii\right)$

$x-y=0...\left(iii\right)$

Now, consider the following determinant:

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

$=-b\times ab-a\times ab-ab\times (-a-b)=0$

Hence, the given lines are concurrent.