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

Given:

$3 x-y=3$

$2 x-3 y=2$

$x+2 y=8$

To do:

We have to find the vertices of the triangle formed by the given lines.

Solution:

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

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

$x+2 y-8=0$.............(iii)

Let the lines given by (i), (ii) and (iii) represent the sides of a $\triangle ABC$ respectively.

Solving (i) and (ii), we get the intersection point $B$.

Multiplying (i) by 3 and subtracting (ii) from it, we get,

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

$9x-2x-3y+3y-9+2=$

$7x=7$

$x=1$

Substituting $x=1$ in (i), we get,

$3(1)-y-3=0$

$y=0$

This implies, 

The coordinates of the vertex $B$ is $(1,0)$

Similarly,

$2\times$ (iii) $-$ (ii), we get,

$2x+4y-16-(2x-3y-2)=0$

$7y=14$

$y=2$

Substituting $y=2$ in (ii), we get,

$2x-3(2)-2=0$

$2x=8$

$x=4$

The coordinates of the vertex $C$ is $(4,2)$

$2\times$ (i) $+$ (iii), we get,

$6x-2y-6+x+2y-8=0$

$7x=14$

$x=2$

Substituting $x=2$ in (iii), we get,

$2+2y-8=0$

$2y=6$

$y=3$

The coordinates of the vertex $A$ is $(2,3)$

Hence, the vertices of the triangle formed by the given lines are $(2, 3), (1, 0)$ and $(4,2)$.

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Updated on: 10-Oct-2022

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