Question

Find the value of k for which the lines $3x+y=2,k+2y=3$ and $2x-y=3$ may intersect at a point.

Answer 1

The given lines are :

$3x+y-2=0$......(i)

$kx+2y-3=0$......(ii)

$2x-y-3=0$......(iii)

On solving (i) and (iii)by cross multiplication, we get

$\frac{x}{(-3-2)}=\frac{y}{(-4+9)}=\frac{1}{(-3-2)}\Rightarrow \frac{x}{-5}=\frac{y}{5}=\frac{1}{-5}$

$\Rightarrow x=\left(\frac{-5}{-5}\right)=1$ and $y=\left(\frac{5}{-5}\right)=-1$

Thus, the point of intersection of (i) and (iii) is P(1, -1)

For the given lines to intersect at a point, $x=1$ and $y=-1$ must satisfy (ii) also.

$\therefore (k\times 1)+2\times (-1)-3=0\Rightarrow k-5=0\Rightarrow k=5$

Hence, k = 5