Determine, algebraically, the vertices of the triangle formed by the lines
3 x – y = 3
2x – 3y = 2
and x + 2y = 8
Given equation of lines are
3 x – y = 3 …………..(i)
2x – 3y = 2 ………….(ii)
and x + 2y = 8 ………….(iii)
Let lines (i), (ii) and (iii) represent the sides of a ΔABC i.e., AB, BC and CA, respectively. On solving lines (i) and (ii), we will get the intersecting
point B.
On multiplying Eq. (i) by 3 in Eq. (i) and then subtracting, we get
On putting the value of $x$ in Eq. (i), we get
$3 \times 1-y=3$
$\Rightarrow$ $y=0$
So, the coordinate of point or vertex $B$ is $(1,0)$.
Or solving lines (ii) and (iii), we will get the intersecting point $C$.
On multiplying Eq. (iii) by 2 and then subtracting, we get
$\Rightarrow \quad y=2$
On putting the value of $y$ in Eq. (iii), we gat
$x+2 \times 2=8$
$\Rightarrow \quad x=8-4$
$\Rightarrow \quad x=4$
Hence, the coordinate of point or vertex $C$ is $(4,2)$.
On solving lines (iii) and (i), we will get the intersecting point $A$.
On multiplying in Eq. (i) by 2 and then adding Eq. (iii), we get
$\Rightarrow \quad x=2$
On putting the value of $x$ in Eq. (i), we get
$3 \times 2-y=3$
$\Rightarrow \quad y=6-3$
$\Rightarrow \quad y=3$
So, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the ΔABC formed by the given lines are A (2, 3), B (1, 0) and C(4,2).