An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (4 ,9 ) to (8 ,5 ) and the triangle's area is 34 , what are the possible coordinates of the triangle's third corner?

Mar 4, 2017

$\left(\frac{29}{2} , \frac{31}{2}\right) \mathmr{and} , \left(- \frac{5}{2} , - \frac{3}{2}\right) .$

Explanation:

We have $2$ Methods to solve this Problem.

${1}^{s t}$ Method :-

Let $Q \left(4 , 9\right) \mathmr{and} R \left(8 , 5\right) .$

$A = Q R = \sqrt{{\left(4 - 8\right)}^{2} + {\left(9 - 5\right)}^{2}} = 4 \sqrt{2.}$

$\text{Area of "Delta=1/2"(Base)(Altitude)}$

$\therefore 34 = \frac{1}{2} \left(A\right) \text{(Altitude)"rArr"Altitude=} \frac{\left(34\right) \left(2\right)}{4 \sqrt{2}} = \frac{17}{\sqrt{2.}}$

This means the the ${3}^{r d} \text{Vertex}$, call it $P \left(x , y\right) ,$ is situated at a

$\bot - \text{distance}$ of $\frac{17}{\sqrt{2}}$ from the line-sgmt. $Q R = A .$

At the same, $\Delta P Q R \text{ is isisceles with, } P Q \left[= B = C\right] = P R .$

Therefore, $P$ lies on the $\bot -$bisector of $Q R$, at a distance of

$\frac{17}{\sqrt{2.}}$

Let $M$ is the mid-point of $Q R .$

$\therefore M = M \left(6 , 7\right) , P M = \frac{17}{\sqrt{2}} , \mathmr{and} , P M \bot Q R .$

Slope of $Q R = - 1 \Rightarrow \text{Slope of } P M = 1 = \tan \left(\frac{\pi}{4}\right) .$

Now, to find $P ,$we use the following very useful Result:-

Result :- On a line $l ,$ passing through $\left({x}_{0} , {y}_{0}\right)$ & making an

angle $\theta$ with the $+ v e$ X-axis, the point on $l$ at a dist.

$r$ from $\left({x}_{0} , {y}_{0}\right)$ is given by $\left({x}_{0} \pm r \cos \theta , {y}_{0} \pm r \sin \theta\right) .$

Here, we have $l = P M , \left({x}_{0} , {y}_{0}\right) = M \left(6 , 7\right) , \theta = \frac{\pi}{4} , \mathmr{and} , r = P M = \frac{17}{\sqrt{2.}}$

$\therefore , \text{ for } P \left(x , y\right) , x = 6 \pm \left(\frac{17}{\sqrt{2}}\right) \cos \left(\frac{\pi}{4}\right)$

$= 6 \pm \left(\frac{17}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right)$

$\therefore x = \frac{29}{2} , \mathmr{and} , - \frac{5}{2.}$

Similarly, $y = \frac{31}{2} , \mathmr{and} , - \frac{3}{2.}$

Hence, the possible co-ords. of the ${3}^{r d}$ vertex are $\left(\frac{29}{2} , \frac{31}{2}\right)$

or, $\left(- \frac{5}{2} , - \frac{3}{2}\right) .$

Enjoy Maths.!

Mar 4, 2017

$\left(\frac{29}{2} , \frac{31}{2}\right) , \mathmr{and} , \left(- \frac{5}{2} , - \frac{3}{2}\right) .$

Explanation:

${2}^{n d}$ Method :-

Let $Q \left(4 , 9\right) \mathmr{and} R \left(8 , 5\right)$ and let the ${3}^{r d}$ vertex be $P \left(x , y\right) .$

Now, the Area of $\Delta P Q R = 34$

$\Rightarrow \frac{1}{2} | D | = 34 \Rightarrow , w h e r e ,$,

$D = | \left(x , y , 1\right) , \left(4 , 9 , 1\right) , \left(8 , 5 , 1\right) | = | 4 x + 4 y - 52 | = 4 | x + y - 13 | .$

$\therefore 2 | x + y - 13 | = 34$

$\therefore x + y - 13 = \pm 17.$

Let, $x + y = 30. \ldots \ldots \ldots . . \left(1 '\right) , \mathmr{and} , x + y = - 4. \ldots \ldots \ldots \ldots . \left(1 ' '\right) .$

Next, we have, $P Q = P R$

$\therefore {\left(x - 4\right)}^{2} + {\left(y - 9\right)}^{2} = {\left(x - 8\right)}^{2} + {\left(y - 5\right)}^{2.}$

$\therefore 8 x - 8 y = - 8 , \mathmr{and} , x - y = - 1. \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$.

Solving (1') and (2), x=29/2, y=31/2; and,

$\left(1 ' '\right) \mathmr{and} \left(2\right) \Rightarrow x = - \frac{5}{2} , y = - \frac{3}{2.}$

$\therefore P \left(x , y\right) = \left(\frac{29}{2} , \frac{31}{2}\right) , \mathmr{and} , \left(- \frac{5}{2} , - \frac{3}{2}\right) ,$ as we had in the

${1}^{s t}$ Method.

Enjoy Maths.!