0
$\begingroup$

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?

  • 0
    So the slop of $BC$ is $45$. – 2017-01-29
  • 0
    How is that? @MyGlasses – 2017-01-29
  • 0
    $m_{AB}.m_{AC}=-1$ then $AB\perp AC$ that is $A=90$. – 2017-01-29
  • 0
    Hint: the two equal sides are symmetric with respect to the $y$ axis, so the base will be parallel to the $x$ axis. – 2017-01-29
  • 0
    @dxiv, what is meant by 'symmetric with respect to the y-axis'? – 2017-01-29
  • 0
    @NeWtoN Draw the two lines. The intersection is at $(0, 3)\,$ (i.e. on the $y$ axis), and the slopes are opposite (as you noticed already). – 2017-01-29
  • 0
    @dxiv, yeah that I know? then... – 2017-01-29
  • 0
    @NeWtoN Then the bisector of the top angle is the $y$ axis, and the base is orthogonal to that, so it must be parallel to the $x$ axis. Which means its equation is something like $y = \text{constant}\,$. – 2017-01-29
  • 0
    @dxiv, How do you know that, the bisector of $\angle A$ is the $Y-$ axis and is $\perp $ to $BC$? – 2017-01-29
  • 0
    @NeWtoN (a) by symmetry with respect to the $y$ axis, and (b) since the triangle is isosceles (so that the top vertex bisector is also a perpendicular bisector, altitude etc). A [simple drawing](http://www.wolframalpha.com/input/?i=plot+x%2By%3D3,+x-y%2B3%3D0,+y%3D-10) is worth many words. – 2017-01-29

1 Answers 1

3

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.

enter image description here

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?