The equation of $AB$ and $AC$ of an isosceles $\triangle ABC$ $(AB=AC)$ are $x+y=3$ and $x-y+3=0$. Find the equation of the remaining side $BC$, which passes through the point $P(1,-10)$
My Attempt:
$AB :: x+y=3\tag1.$
$AC :: x-y+3=0\tag2$
Slope of $AB=m_1= -1$.
Slope of $AC=m_2=1$.
What should I do to get it solved?
Solving the two equations, we have A = (0, 3). By considering the slopes, we get $BA \bot CA$. This, together with the fact that AB = AC, means BC must be the hypotenuse.
As indicated, a picture worth many words.
In the picture (or by calculation), we observe that AB and AC are symmetric to each other about the y-axis.
AT this point, there are many possible candidates for the hypotenuse. See those dotted lines.
The requirement that BC must pass through P=(1, -10) forces the hypotenuse BC must lie on the red line. Since that line is horizontal and it passes through P=(1, -10), can you find its equation?