Question

Determine the vertices of the triangle formed by the lines
$y=x,3y=x$ and $x+y=8$.

• A. $(0,0),(0,4),(6,2)$
• B. $(0,0),(4,4),(0,2)$
• C. $(0,0),(7,7),(5,2)$
• D. $(0,0),(4,4),(6,2)$

Answer 1

The correct option is D $(0,0),(4,4),(6,2)$

The vertices of the triangle will be the point of intersection of given lines taken two at a time. So, we need to find the points of intersection of the lines:

A. Taking

$y=x$ $\dots \left(1\right)$

$3y=x$ $\dots \left(2\right)$

Let, the point of intersection of above two lines is $A$.

Subtract $\left(1\right)$ from $\left(2\right)$:

$2y=0$

$y=0$

Put this value of $y$ in equation $\left(1\right)$:

$y=x=0$

Hence, the coordinates of A are $(0,0)$.

B. Taking

$y=x$ $\dots \left(2\right)$

$x+y=8$ $\dots \left(3\right)$

Let, the point of intersection of above two lines is $B$.

Subtract $\left(2\right)$ from $\left(3\right)$:

$x+y-y=8-x$

$2x=8$

$x=4$

Put this value of $x$ in equation $\left(3\right)$:

$y=x=4$

Hence, the coordinates of B are $(4,4)$.

C. Taking

$x+y=8$ $\dots \left(3\right)$

$x=3y$ $\dots \left(1\right)$

Let, the point of intersection of above two lines is $C$.

Subtract equation $\left(1\right)$ from $\left(3\right)$:

$x+y-x=8-3y$

$y=8-3y$

$4y=8$

$y=2$

Put this value of $y$ in equation $\left(1\right)$:

$x=3\left(2\right)=6$

Hence, the coordinates of C are $(6,2)$.

The coordinates of the triangle are $(0,0),(4,4)$ and $(6,2).$