Question

Determine, algebraically, the vertices of the triangle formed by the lines
3x – y = 3
2x – 3y = 2
x + 2y = 8

Answer 1

Given equation of lines are
3x – y = 3...(i)
2x – 3y = 2...(ii)
and x + 2y = 8...(iii)

Let lines (i) and (ii) and (iii) represent the sides of a $\Delta$ ABC i.e., AB, BC and CA respectively.
On solving equation of lines (i) and (ii), we will get the intersecting point B.
On multiplying Eq. (i) by 3 and then subtracting, we get
(9x – 3y = 9)
-$(2x-3y=2)\underset{–}{}=7x=7\Rightarrow x=1$
On putting the value of x in Eq. (i) we get
3 × 1 – y = 3
$\Rightarrow$ y = 0
So, the coordinate of point of vertex B is (1, 0)

On solving equation of lines (ii) and (iii), we will get the intersecting point C.
On multiplying Eq. (iii) by 2 and then subtracting we get
$(2x+4y=16)-(2x-3y=2)\underset{–}{}$
$7y=14$
$\Rightarrow y=2$
On putting the value of y in Eq. (iii) we get
x + 2 × 2 = 8
x = 8 – 4
$\Rightarrow$ 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 Eq. (i) by 2 and then adding Eq. (iii), we get
$(6x-2y=6)+(x+2y=8)\underset{–}{}$
$7x=14$
$\Rightarrow x=2$
On putting the value of $x$ in Eq. (i), we get
$3\times 2-y=3$
$6-y=3$
$\Rightarrow y=6-3\Rightarrow y=3$
So, the coordinate of point or vertex A is (2, 3)
Hence, the vertices of the $\Delta$ ABC formed by the given lines are A (2, 3), B (1, 0) and C(4, 2).