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The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.

$a.)x+3y-1=0\\ b.)x-3y+1=0\\ c.)2x-y-5=0\\ \color{green}{d.)x+2y-5=0}\\$

$\quad\\~\\~\\$

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Let the slope of the required line be $m$. I used the slope formula between two lines $\angle B=\angle C\\~\\\left|{\dfrac{(-1-m)}{(1-m)}}\right|=\left|{\dfrac{(7-m)}{(1+7m)}}\right|\\ \implies m=-3,\dfrac{1}{3}$

But the book is giving right answer as option $d.)$.

By finding $m$ i only found slope not the whole equation.

Also i would like to know if their is clean simple short way , and also using $Area=5$ .

I have studied maths upto $12th$ grade.

Update : by using Geogebra I found that $x-3y+1=0$ fits perfectly

enter image description here

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    $\begingroup$ If side AB goes with equation $x+y=5$ then it seems the point labeled B should be the one your diagram has labeled C. (and vice-versa for the point labeled C). I don't know if this affects the calculation... $\endgroup$ – coffeemath May 25 '15 at 21:26
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Let the lines: $x+y=5$ & $7x-y=3$ represent AC & AB. Then the acute angle $\theta$ between the equal sides AB & AC is given as $$\tan \theta=\left|\frac{7-(-1)}{1+7(-1)}\right|=\left|\frac{8}{-6}\right|=\frac{4}{3} \implies \sin \theta=\frac{4}{5}$$ The length of two sides AB & AC of isosceles triangle are equal hence, the area of triangle is given as $$\frac{1}{2}(AB)(AC)\sin\theta=5 \implies \frac{1}{2}(AB)^2\left(\frac{4}{5}\right)=5 \implies AB=\frac{5}{\sqrt{2}}$$ The equations of lines bisecting the angle between the equal sides: $x+y=5$ & $7x-y=3$ are given as $$\frac{x+y-5}{\sqrt{1^2+1^2}}=\pm \frac{7x-y-3}{\sqrt{7^2+(-1)^2}}\implies 5x+5y-25=\pm(7x-y-3)$$$$ \implies x-3y+11=0 \quad \text{&} \quad 3x+y-7=0 $$ From above, equations it's clear that $3x+y-7=0$ is acute angle bisector having negative slope & passing through vertex A. We can also check it out by a rough sketch.
Let the equation of third (unknown) line i.e. side BC be $x-3y+c=0$ normal to the angle bisector: $3x+y-7=0$. Now, solving: $x-3y+c=0$ & any given line say $7x-y=3$, we get the intersection point $\left(\frac{c+9}{20}, \frac{7c+3}{20} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(1, 4 \right)$. Now, calculating the length $AB=\frac{5}{\sqrt{2}}$ of equal sides of isosceles triangle using point-to-point distance-formula as $$AB=\sqrt{\left(1-\frac{c+9}{20}\right)^2+\left(4-\frac{7c+3}{20}\right)^2}=\frac{5}{\sqrt{2}}$$$$ (c-11)^2=100 \implies c-11=\pm 10 \implies c=21 \quad \text{&} \quad c=1$$ Thus by setting the values of $c$, we get two equations of parallel lines representing the third unknown side BC as: $\color{#0ae}{x-3y+1=0}$ & $ \color{#0ae} {x-3y+21=0}$ lying on either side of vertex A of given isosceles triangle ABC. Thus, we see that the option (b) $\color{#0b4}{x-3y+1=0}$ is correct.

According to the option (d) in your book the unknown side is: $x+2y-5=0$. Note that this line has slope $\frac{-1}{2}$ i.e. negative but see in the figure you drew the slope of unknown side must be only positive. Thus option provided in your book is absolutely wrong (may be due to some printing mistake).

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    $\begingroup$ I might be wrong but by checking ur equations $Area(\triangle ABC)$ is coming $20$ units $\endgroup$ – R K May 25 '15 at 22:32
  • $\begingroup$ No! you are absolutely right, actually I got wrong in calculations. But I have pointed it out & corrected. Now, you may check it out. $\endgroup$ – Harish Chandra Rajpoot May 25 '15 at 22:45
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the two two lines that bisects the $\angle BAC$ are given by $$\frac{7x-y-3}{\sqrt{50}} = \pm\frac{x+y-5}{\sqrt2}.$$ they are $$x-3y+11 = 0, \quad 3x + y - 7=0 \text{ and } A = (1, 4).$$
we will pick a point $$D = (1+2k, 4-6k), k \text{ to be fixed later} $$ on the angle bisector $3x+y-7 = 0.$ the line through $D$ parallel to the other bisector is $$x - 3y = 1+2k -3(4-6k) = 20k-11.\tag 1$$ the point $C$ is the intersection of $(1)$ and $$7x - y = 3 \tag 2 $$ therefore $$C = (1-k, 4-7k). $$ we can now compute $$AD^2 = 4k^2 + 36k^2 = 40k^2\\ CD^2 = 9k^2 + k^2 =10k^2$$ the constraint that the area of the triangle $ABC = 5$ translates to $$AD^2 \times CD^2 = 25\implies 400k^2 = 25, k = \pm \frac14 $$ the line $BC$ are $$x-3y = -6, x-3y = -16. $$

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This question is not that difficult to solve.

(1) A = … = (1, 4), which will be used as the center of a circle.

(2) Let $\theta$ be the angle between the 2 given lines. Then, $\theta$ or $\tan \theta$ can be obtained by "the angle between 2 lines formula".

(3) From (2), the value of $\sin \theta$ is found.

(4) Putting the 5 in the area formula "$Area = 0.5ab \sin C$", we get $\frac {25}{2} = AB^2$, which will be used as $(radius)^2$.

(5) Form the circle equation $K: (x – 1)^2 + (y – 4)^2 = \frac {25}{2}$.

(6) The co-ordinates of B and C can be found by solving ($K$ and e) and also ($K$ and f).

(7) From the co-ordinates found, form the equations back to see which one meets the options.


However, I am sorry to say that “This question is badly put”. See the following for the reasons:-

  1. The answer supplied was incorrect in the first place.

  2. There are 4 possible answers (instead of one), namely lines d, n, p, q depending on which points are chosen as B and C.

For example, if we choose B on line e: y = -x – 5 and C’ as C on f: y = 7x – 3, then the green-shaded triangle ABC’ will also be isosceles and its area is also 5 square units (meeting the description completely). The same is true for the other 3 triangles. This is exactly what you have found ($m=-3,\dfrac{1}{3}$, the slopes of the 2 sets of possible lines).

Think of it this way: What would happen if I chose B as B and C' as C. Then, I get $p: 3x + y = 12$. It matches none of the given options. But did I do any thing wrong?

enter image description here

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So, the coordinate of $A(1,4)$

We can set the coordinates of $B(a,5-a);C(b,7b-3)$

We have $|AB|^2=|AC|^2\implies(a-1)^2+(a-1)^2=(b-1)^2+\{7(b-1)\}^2$

$\implies a-1=\pm(5b-5)=\pm5(b-1)$

$$\triangle ABC=\dfrac12\begin{vmatrix}1 & 4 & 1 \\ a & 5-a & 1 \\ b & 7b-3 &1\end{vmatrix}=4|(a-1)(b-1)|$$

$\implies\triangle ABC=4\cdot5|(b-1)^2|$

$\implies20(b-1)^2=5\iff b-1=\pm\dfrac12$

Can you take it home from here?

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