Determine the nature of the quadrilateral formed by four lines $3 x + 4 y - 5 = 0; 4 x - 3 y - 5 = 0; 3 x + 4 y = 5 = 0$ and $4 x - 3 y + 5 = 0$ . Find the equation of the circle inscribed and circumscribing this quadrilateral.

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Solution

$L_{1} = 3 x + 4 y = 5 = m = - 3 / 4$

$L_{2} = 4 x - 3 y = 5 = m = 4 / 3$

$L_{3} = 3 x + 4 y = - 5 = m = - 3 / 4$

$L_{4} = 4 x - 3 y = - 5 = m = 4 / 3$

Diagonal Ac = $\frac{1 / 5 + 1 / 5}{7 / 5 + 7 / 5} = \frac{2 / 5}{14 / 5} = \frac{1}{7}$

DiagonalBD= $\frac{7 / 5 + 7 / 5}{- 1 / 5 - 1 / 5} = \frac{14 / 5}{- 2 / 5} = - 7$

$M_{A C} \times M_{B D} = \frac{1}{7} \times - 7$

Diagonals are at right angle

ABCD is a square

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Determine the nature of the quadrilateral formed by four lines 3x+4y−5=0 ; 4x−3y−5=0 ; 3x+4y=5=0 and 4x−3y+5=0. Find the equation of the circle inscribed and circumscribing this quadrilateral.

L1=3x+4y=5=m=−3/4 L2=4x−3y=5=m=4/3 L3=3x+4y=−5=m=−3/4 L4=4x−3y=−5=m=4/3 Diagonal Ac = 1/5+1/57/5+7/5=2/514/5=17 DiagonalBD=7/5+7/5−1/5−1/5=14/5−2/5=−7 MAC×MBD=17×−7 Diagonals are at right angle ABCD is a square

Determine the nature of the quadrilateral formed by four lines $3 x + 4 y - 5 = 0; 4 x - 3 y - 5 = 0; 3 x + 4 y = 5 = 0$ and $4 x - 3 y + 5 = 0$ . Find the equation of the circle inscribed and circumscribing this quadrilateral.