# The base of an isosceles triangle lies on the line x-2y=6, the opposite vertex is (1,5), and the slope of one side is 3. How do you find the coordinates of the other vertices?

Feb 21, 2016

Two vertices are $\left(- 2 , - 4\right)$ and $\left(10 , 2\right)$

#### Explanation:

First let us find the midpoint of the base. As base is on $x - 2 y = 6$, perpendicular from vertex $\left(1 , 5\right)$ will have equation $2 x + y = k$ and as it passes through $\left(1 , 5\right)$, $k = 2 \cdot 1 + 5 = 7$. Hence equation of perpendicular from vertex to base is $2 x + y = 7$.

Intersection of $x - 2 y = 6$ and $2 x + y = 7$ will give us midpoint of base. For this, solving these equations (by putting value of $x = 2 y + 6$ in second equation $2 x + y = 7$) gives us

$2 \left(2 y + 6\right) + y = 7$
or $4 y + 12 + y = 7$
or $5 y = - 5$.

Hence, $y = - 1$ and putting this in $x = 2 y + 6$, we get $x = 4$, i.e. mid point of base is $\left(4 , - 1\right)$.

Now, equation of a line having a slope of $3$ is $y = 3 x + c$ and as it passes through $\left(1 , 5\right)$, $c = y - 3 x = 5 - 1 \cdot 3 = 2$ i.e. equation of line is $y = 3 x + 2$

Intersection of $x - 2 y = 6$ and $y = 3 x + 2$, should there give us one of the vertices. Solving them, we get $y = 3 \left(2 y + 6\right) + 2$ or $y = 6 y + 20$ or $y = - 4$. Then $x = 2 \cdot \left(- 4\right) + 6 = - 2$ and hence one vertex is at $\left(- 2 , - 4\right)$.

We know that one of the vertices on base is $\left(- 2 , - 4\right)$, let other vertex be $\left(a , b\right)$ and hence midpoint will be given by $\left(\frac{a - 2}{2} , \frac{b - 4}{2}\right)$. But we have midpoint as $\left(4 , - 1\right)$.

Hence $\frac{a - 2}{2} = 4$ and $\frac{b - 4}{2} = - 1$ or $a = 10$ and $b = 2$.

Hence two vertices are $\left(- 2 , - 4\right)$ and $\left(10 , 2\right)$