The-hhpotenuse-of-a-right-angled-triangle-has-its-ends-at-the-points-1-3-and-4-1-Find-an-equation-of-the-legs-perpendicar-sides-of-the-triangle-

Question Number 74948 by vishalbhardwaj last updated on 04/Dec/19

The hhpotenuse of a right angled triangle

has its ends at the points (1, 3) and (- 4, 1)

. Find an equation of the legs (perpendicar

sides) of the triangle .

Answered by MJS last updated on 04/Dec/19

given hypothenuse \Rightarrow the 3^{rd} point is located

on a circle with center in the center of the

hyp . and radius = \frac{hyp .}{2}

center = \frac{(\begin{matrix} 1 \\ 3 \end{matrix}) + (\begin{matrix} - 4 \\ 1 \end{matrix})}{2} = (\begin{matrix} - \frac{3}{2} \\ 2 \end{matrix})

radius = \frac{1}{2} abs ((\begin{matrix} 1 \\ 3 \end{matrix}) - (\begin{matrix} - 4 \\ 1 \end{matrix})) = \frac{1}{2} \sqrt{29}

equation of circle is

{(x + \frac{3}{2})}^{2} + {(y - 2)}^{2} = \frac{29}{4}

y = 2 \pm \sqrt{- x^{2} - 3 x + 5} \Rightarrow - \frac{3}{2} - \frac{\sqrt{29}}{2} ⩽ x ⩽ - \frac{3}{2} + \frac{\sqrt{29}}{2}

so {there}^{'} s no unique triangle

triangles {A B C}_{1}, {A B C}_{2}

let A = (\begin{matrix} - 4 \\ 1 \end{matrix}) B = (\begin{matrix} 1 \\ 3 \end{matrix}) C_{1, 2} = (\begin{matrix} p \\ 2 \pm \sqrt{- p^{2} - 3 p + 5} \end{matrix})

equations of lines

A B : y = \frac{2}{5} x + \frac{13}{5}

{A C}_{1} : y = \frac{1 - \sqrt{- p^{2} - 3 p + 5}}{p + 4} x + \frac{p + 8 - 4 \sqrt{- p^{2} - 3 p + 5}}{p + 4}

{B C}_{1} : y = - \frac{1 + \sqrt{- p^{2} - 3 p + 5}}{p - 1} x + \frac{3 p - 2 + \sqrt{- p^{2} - 3 p + 5}}{p - 1}

{A C}_{2} : y = \frac{1 + \sqrt{- p^{2} - 3 p + 5}}{p + 4} x + \frac{p + 8 + 4 \sqrt{- p^{2} - 3 p + 5}}{p + 4}

{B C}_{2} : y = - \frac{1 - \sqrt{- p^{2} - 3 p + 5}}{p - 1} x + \frac{3 p - 2 - \sqrt{- p^{2} - 3 p + 5}}{p - 1}

\forall p \in R ∣ - \frac{3}{2} - \frac{\sqrt{29}}{2} ⩽ x ⩽ - \frac{3}{2} + \frac{\sqrt{29}}{2} \land p \neq 1 \land p \neq - 4

Commented by vishalbhardwaj last updated on 04/Dec/19

sir equation of perpendicular sides ? ?

Commented by MJS last updated on 04/Dec/19

ran out of time, will finish in about an hour

or two

Commented by MJS last updated on 04/Dec/19

is this what you need ?

Commented by vishalbhardwaj last updated on 04/Dec/19

yes sir

Commented by vishalbhardwaj last updated on 05/Dec/19

but sir the equations of perpendicular

lines are given x = 1 and y = 1, why ? ?

Commented by MJS last updated on 05/Dec/19

I {don}^{'} t know . the triangle is not unique, we

would need more information

anyway the point (\begin{matrix} 1 \\ 1 \end{matrix}) lies on the circle, p = 1

\Rightarrow we cannot use my formulas but we can

calculate the lines A C and B C and yes, they

are x = 1 and y = 1

with the given information this is only one

solution

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