Let ( , ) be the centroid of the triangle formed by the lines 15x-y=82,6x-5y=-4 and 9x+4y=…

Question

Let $(\alpha ,\beta )$ be the centroid of the triangle formed by the lines $15x-y=82,6x-5y=-4$ and $9x+4y=17$. Then $\alpha +2\beta $ and $2\alpha -\beta $ are the roots of the equation :

Accepted Answer

Solve the equations $15x-y=82$ and $6x-5y=-4$

Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:

$75x-5y=410$

$6x-5y=-4$

Subtracting these gives $69x=414$ which leads to $x=6$

Substitute $x=6$ into the first equation to get $y=8$

So, the first vertex is $(6,8)$.

Solve the equations $15x-y=82$ and $9x+4y=17$

Multiply the first equation by 4 and the second by 1 and then add:

$60x-4y=328$

$9x+4y=17$

Adding these gives $69x=345$ which leads to $x=5$

Substitute $x=5$ into the second equation to get $y=-7$

So, the second vertex is $(5,-7)$.

Solve the equations $6x-5y=-4$ and $9x+4y=17$

Multiply the first equation by 4 and the second by 5 and then add:

$24x-20y=-16$

$45x+20y=85$

Adding these gives $69x=69$ which leads to $x=1$

Substitute $x=1$ into the first equation to get $y=2$

So, the third vertex is $(1,2)$.

So, the vertices of the triangle are $(6,8)$, $(5,-7)$, and $(1,2)$.

Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\alpha ,\beta )$ of a triangle with vertices $({x}_{1},{y}_{1})$, $({x}_{2},{y}_{2})$, and $({x}_{3},{y}_{3})$ is given by :

$\alpha =\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3}$,

$\beta =\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3}$

Substituting the coordinates of the vertices into these equations, we get:

$\alpha =\frac{6+5+1}{3}=4$,

$\beta =\frac{8-7+2}{3}=1$

Then, we find $\alpha +2\beta $ and $2\alpha -\beta $:

$\alpha +2\beta =4+2(1)=62\alpha -\beta =2(4)-1=7$

So, $\alpha +2\beta =6$ and $2\alpha -\beta =7$ are the roots of the quadratic equation.

We can write the quadratic equation as

${x}^{2}-(\alpha +2\beta +2\alpha -\beta )x+(\alpha +2\beta )(2\alpha -\beta )=0$

Substituting $\alpha =4$, $\beta =1$ gives us:

${x}^{2}-13x+42=0$

Solutions · 1