A-triangle-has-an-area-of-20-square-units-and-two-vertices-are-3-4-and-2-7-What-is-the-position-of-the-third-vertex-

Question Number 76252 by necxxx last updated on 25/Dec/19

A t r i a n g l e h a s a n a r e a o f 20 s q u a r e

u n i t s a n d t w o v e r t i c e s a r e (3, 4) a n d (2, 7) .

W h a t i s t h e p o s i t i o n o f t h e t h i r d v e r t e x ?

Commented by MJS last updated on 25/Dec/19

in words : we know the area = \frac{side \times height}{2}

and we know the side = \sqrt{{(3 - 2)}^{2} + {(4 - 7)}^{2}} =

= \sqrt{10} \Rightarrow height = \frac{2 \times area}{side} = \frac{40}{\sqrt{10}} = 4 \sqrt{10}

all triangles with s = \sqrt{10} \land h = 4 \sqrt{10} have the

same area \Rightarrow 3^{rd} vertex is located on two

lines parallel to the line through the given

vertices and with distance 4 \sqrt{10} from this line

let A = (\begin{matrix} 3 \\ 4 \end{matrix}) B = (\begin{matrix} 2 \\ 7 \end{matrix})

the distance of two parallel lines

y = a_{0} x + b_{0}, y = a_{0} x + b_{1}

is d = \frac{∣ b_{1} - b_{0} ∣}{\sqrt{a^{2} + 1}} \Leftrightarrow b_{1} = b_{0} \pm d \sqrt{a^{2} + 1}

\Rightarrow

line A B : y = 13 - 3 x

line C_{1} : y = - 27 - 3 x

line C_{2} : y = 53 - 3 x

\Rightarrow

C = (\begin{matrix} x \\ - 27 - 3 x \end{matrix}) \lor C = (\begin{matrix} x \\ 53 - 3 x \end{matrix})

Answered by benjo last updated on 25/Dec/19

Commented by necxxx last updated on 30/Dec/19

t h a n k y o u s o m u c h . I^{'} m m o s t g r a t e f u l .

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