Question

find area of parallelogram with 3 vertices are (1,-2) , (2,3) & (-3,2) in order and also find the 4th vertex​

Answer 1

Answer:

$D_{1}$ (- 4, - 3) ; $D_{2}$ (- 4, - 3) ; $D_{3}$ (- 2, 7) .

Step-by-step explanation:

Slope of line m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

Slopes of ║ lines are equal.

y - $y_{1}$ = m(x - $x_{1}$ )

d = $\sqrt{(x_{2} -x_{1} )^2 + (y_{2} -y_{1} )^2}$

~~~~~~~~~~~~~~

A( - 3, 2), B(2, 3), C(1, - 2)

$m_{AB}$ = $\frac{2-3}{-3-2}$ = $\frac{1}{5}$

$m_{BC}$ = $\frac{-2-3}{1-2}$ = 5

$m_{AC}$ = $\frac{-2-2}{1+3}$ = - 1

Location of $4^{th}$ vertex D of parallelogram may vary. It dependent from what side of given triangle will be a diagonal of parallelogram.

Scenario 1. Let AC be that diagonal, then find equation the line DC║AB and passing through point (1, - 2), and equation of the line DA║BC and passing through (- 3, 2)

y - (- 2) = $\frac{1}{5}$ ( x - 1 ) ⇒ y = (1)

y - 2 = 5(x - (- 3)) ⇒ y = 5x + 17 ..... (2)

Line (1) and line (2) intersect at point $D_{1}$ (- 4, - 3).

Scenario 2. Let BC be that diagonal, then find equation the line DC║AB and passing through point (1, - 2), and equation of the line DB║AC and passing through (2, 3).

y - (- 2) = $\frac{1}{5}$ (x - 1) ⇒ y = (1)

y - 3 = (- 1)(x - 2) ⇒ y = - x + 5 ..... (3)

Line (1) and line (3) intersect at point $D_{2}$ (- 4, - 3).

Scenario 3. Line (2) and line (3) intersect at $D_{3}$ (- 2, 7)