Find the equation of side of isoceles triangle for the given conditions.

Question

Two equal sides of an isosceles triangle are given by the equations $7x-y+3 = 0$ and $x + y -3 = 0$ & its third side passes through the point $(1, -10)$. Determine the equation of the third side.

My attempt is as follows:-

Attempt $1$:

Let $AB=AC=r$

$$B\equiv\left(r\cos\theta,3+r\sin\theta\right)$$ $$\tan\theta=7$$ $$B\equiv\left(\dfrac{r}{\sqrt{50}},3+\dfrac{7r}{\sqrt{50}}\right)$$

In the same way

$$C\equiv\left(r\cos\alpha,3+r\sin\alpha\right)$$ $$\tan\alpha=-1$$ $$C\equiv\left(-\dfrac{r}{\sqrt{2}},3+\dfrac{r}{\sqrt{2}}\right)$$

As point $(1,-10)$ lies on $BC$

$$\dfrac{3+r\sin\alpha+10}{r\cos\alpha-1}=\dfrac{3+r\sin\theta+10}{r\cos\theta-1}$$

$$13r\cos\theta-13+r^2\sin\alpha\cos\theta-r\sin\alpha=13r\cos\alpha-13+r^2\sin\theta\cos\alpha-r\sin\theta$$

$$13\cos\theta-13\cos\alpha+\sin\theta-\sin\alpha=r(\sin\theta\cos\alpha-\sin\alpha\cos\theta)$$

$$\dfrac{13}{5\sqrt{2}}+\dfrac{13}{\sqrt{2}}+\dfrac{7}{5\sqrt{2}}-\dfrac{1}{\sqrt{2}}=r\left(-\dfrac{7}{5\sqrt{2}}\cdot\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{5\sqrt{2}}\right)$$

$$\dfrac{1}{\sqrt{2}}\left(\dfrac{13}{5}+13+\dfrac{7}{5}-1\right)=-\dfrac{4r}{5}$$

$$r=-10\sqrt{2}$$

Hence $C\equiv\left(10,-7\right)$

So equation of third side will be $$y+10=\dfrac{-7+10}{10-1}(x-1)$$

$$3y+30=x-1$$ $$3y-x+31=0$$

But there were two answers, other answer is $3x+y+7=0$. Am I ignoring some other value of $r$ in my calculations? Please help me in this.

Attempt $2$:

Let's try find out $\angle BAC$

Case $1$: Assuming $\angle BAC=\theta$ as acute

$$\tan\theta=\left|\dfrac{7-(-1)}{1-7}\right|$$ $$\tan\theta=\dfrac{4}{3}$$

Let's find $\angle ABC=\alpha$

$$\alpha=\dfrac{\pi-\tan^{-1}\dfrac{4}{3}}{2}$$

Let's find slope of $BC$

$$\tan\alpha=\left|\dfrac{7-m}{1+7m}\right|$$

$$\tan\alpha=\dfrac{7-m}{1+7m} \text { or } \tan\alpha=\dfrac{m-7}{1+7m}\tag{1}$$

$$\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$$ $$-\dfrac{4}{3}=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$$ $$2\tan^2\alpha-3\tan\alpha-2=0$$ $$\tan\alpha=2 \text{ or } -\dfrac{1}{2}$$

We will only consider $\tan\alpha=2$ as $\alpha$ cannot be obtuse

Putting the value of $\tan\alpha$ in equation $(1)$

$$2(1+7m)=7-m \text { or } 2(1+7m)=m-7$$ $$15m=5 \text { or } 13m=-9$$ $$m=\dfrac{1}{3} \text { or } m=-\dfrac{9}{13}$$

Using $m=\dfrac{1}{3}$, we get equation of BC as $3y-x+31=0$

Using $m=-\dfrac{9}{13}$, we get equation of BC as $13y+9x+121=0$

Case $2$: Assuming $\angle BAC=\theta$ as obtuse

$$\tan\theta=-\dfrac{4}{3}$$ $$\theta=\tan^{-1}\left(-\dfrac{4}{3}\right)$$

$$\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$$ $$\dfrac{4}{3}=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$$ $$2\tan^2\alpha+3\tan\alpha-2=0$$ $$\tan\alpha=-2 \text{ or } \dfrac{1}{2}$$

We will only consider $\tan\alpha=\dfrac{1}{2}$ as $\alpha$ cannot be obtuse

Putting the value of $\tan\alpha$ in equation $(1)$

$$\dfrac{1}{2}=\dfrac{7-m}{1+7m} \text { or } \dfrac{1}{2}=\dfrac{m-7}{1+7m}$$

$$1+7m=14-2m \text { or } 1+7m=2m-14$$ $$m=\dfrac{13}{9} \text { or } m=-3$$

Using $m=-3$, we get equation of BC as $y+3x+7=0$

Using $m=\dfrac{13}{9}$, we get equation of BC as $9y-13x+103=0$

Its crazy, now by this method I am getting four equations

$$3y-x+31=0$$ $$13y+9x+121=0$$ $$y+3x+7=0$$ $$9y-13x+103=0$$

Now what mistake I am doing here, I checked it multiple times but everything seems correct? What am I violating here?

Answer 1

Hint:

Use

The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

So, if $D$ is the midpoint of $BC, AD\perp BC$

$AD$ will bisect $\angle A$

So, the equation of $AD$ will be $$\dfrac{7x-y+4}{\sqrt{7^2+1^2}}=\pm\dfrac{x+y-3}{\sqrt{1^2+1^2}}$$

So, there will be possible equations of $AD$ each one will correspond to one $BC$

Answer 2

Could not identify the mistake in your steps.

WLOG $$B(h,7h+3);C(k,3-k)$$

So, the gradient of $BC, m_{BC}=\dfrac{7h+3-(3-k)}{h-k}=\dfrac{7h+k}{h-k}$

We need $\dfrac{7h+k}{h-k}=\dfrac{k+10}{3-k-1}\ \ \ \ (1)$

As $|AB=|CA|,|$

$$h^2+(7h+3-3)^2=k^2+(3-k-3)^2\implies k=\pm5h$$

Case $\#1:$ if $k=-5h,$ $$\dfrac{7h+k}{h-k}=?$$ Now use $(1)$

Case $\#2:$ if $k=5h$ $$\dfrac{7h+k}{h-k}=?$$ Now use $(1)$ again

Answer 3

Use the concept that In isosceles triangle third side is equally inclined to both equal sides. $$ $$ Line equally inclined to given two intersecting lines is parallel to the bisectors. $$. $$ Required line is parallel to $$(\frac{7x-y+3}{\sqrt{50}})=\pm(\frac{x+y-3}{\sqrt{2}})$$ Hence slope of two required lines are $m=\frac{ 1}{3} , -3$ $$ $$ Hence 3rd side can be $$(y+10)=\frac{1}{3}(x-1)$$ Or. $$(y+10)=-3(x-1)$$

Answer 4

Finally got my mistakes, it was hard to realize it by looking at written calculations and formulae, so only way was to draw the picture.

Mistake in Attempt $1$:

I assumed that $B$ and $C$ are below $A$ and that's why for both $B$ and $C$, I took same sign of $r$. I wrote $B\equiv\left(r\cos\theta,3+r\sin\theta\right)$, $C\equiv\left(r\cos\alpha,3+r\sin\alpha\right)$

So we also have to consider the case when one of the point is above $A$ and the other point is below $A$. So if we consider $B\equiv\left(r\cos\theta,3+r\sin\theta\right)$, $C\equiv\left(-r\cos\alpha,3-r\sin\alpha\right)$, then we will also get the second possible equation of $BC$

Mistake in Attempt $2$:

Mistake in Case $1$: Assuming $\angle BAC=\theta$ as acute

Here I got two values for slope of $BC$ as $\dfrac{1}{3},-\dfrac{9}{13}$. But I should have ensured that $BC$ still makes angle $\alpha$ with side $AC$ for both values of $m$

By further checking, I got to know that for $m=-\dfrac{9}{13}$, $BC$ doesn't make angle $\alpha$ with side $AC$ which makes triangle non-isoceles. Hence $m=-\dfrac{9}{13}$ is not valid value.

Same mistake I did in Case $2$

So really these mistakes were eye opener for me. But this upsets me also as in time pressure these mistakes are very common, so in such situations we are bound to do such types of problems in the standard way as given by @lab bhattacharjee.