Step 1:

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The equation of AC is $\"\"$ .

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Rewrite the equation in the slope -intercept form : $\"\"$ .

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$\"\"$

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Subtract x on each side.

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$\"\"$

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Divide each side by 2.

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$\"\"$

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Compare the above equation with standard form.

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Slope $\"\"$ and y - intercept is 8.

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Step 2:

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The perpendicular from point B to AC meets AC at the point X.

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The slope of the perpendicular to line AC is $\"\"$ .

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Point slope form of equation is $\"\"$ .

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Substitute slope $\"\"$ and point $\"\"$ .

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$\"\"$

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A line perpendicular to AC is $\"\"$ .

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Step 3:

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Find the point of intersection of line $\"\"$ and its perpendicular $\"\"$ .

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$\"\"$

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Substitute x = 4 in $\"\"$ .

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$\"\"$

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The point of intersection line AC and the perpendicular from B to AC is X = (4, 6). \ \

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Solution:

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The coordinates of X is (4, 6).

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(2)

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Step 1:

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The equation of AC is $\"\"$ .

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The point A is on the x - axis, so substitute y = 0 in AC to find the point A.

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$\"\"$

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The point A is (16, 0).

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The point C is on the y - axis, so substitute x = 0 in AC to find the point C.

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$\"\"$

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The point C is (0, 8).

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Step 2:

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The point X is

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The line AD is perpendicular to AC. \ \

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The equation of AC is $\"\"$ .

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The slope of the AC is $\"\"$ .

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The slope of the perpendicular to line AC is $\"\"$ .

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The equation of line AD :

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Point slope form of equation is $\"\"$ .

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Substitute slope $\"\"$ and point $\"\"$ .

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Point slope form of the equation is $\"\"$ .

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$\"\"$

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The equation of line CD :

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Two point form of the equation is $\"\"$ .

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Substitute A (0, 8) and D (a, b).

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$\"\"$

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Solve for a and b.

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(2)

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Step 1:

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The equation of AC is $\"\"$ .

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The point A is on the x - axis, so substitute y = 0 in AC to find the point A.

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$\"\"$

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The point A is (16, 0).

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The point C is on the y - axis, so substitute x = 0 in AC to find the point C.

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$\"\"$

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The point C is (0, 8).

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Step 2:

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The line AD is perpendicular to AC. (Since line AC is symmetry) \ \

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The equation of AC is $\"\"$ .

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The slope of the AC is $\"\"$ .

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The slope of the perpendicular to line AC is $\"\"$ .

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The equation of line AD :

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Point slope form of equation is $\"\"$ .

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Substitute slope $\"\"$ and point $\"\"$ .

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Point slope form of the equation is $\"\"$ .

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$\"\"$

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The equation of line CD :

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The line CD is perpendicular to AC. (Since line AC is symmetry) \ \

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The equation of AC is $\"\"$ .

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The slope of the AC is $\"\"$ .

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The slope of the perpendicular to line AC is $\"\"$ .

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Point slope form of equation is $\"\"$ .

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Substitute slope $\"\"$ and point $\"\"$ .

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Point slope form of the equation is $\"\"$ .

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$\"\"$

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The point D is the intersection of line AD and CD.

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$\"\"$

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(2)

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Step 1:

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The point X is the mid point of BD . (Since line AC is symmetry)

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Mid point $\"\"$ .

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Substitute X (4, 6) and $\"\"$ .

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Let the point D (a, b).

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$\"\"$

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So the point D is (6, 10).

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(3) \ \

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Step 1:

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The equation of AC is $\"\"$ .

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The point A is on the x - axis, so substitute y = 0 in AC to find the point A.

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$\"\"$

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The point A is (16, 0).

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The point C is on the y - axis, so substitute x = 0 in AC to find the point C.

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$\"\"$

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The point C is (0, 8).

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Step 2:

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The perimeter of the quadrilateral is P = sum of all sides.

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The given quadrilateral is look like kite.

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It has 2 set of equal length sides.

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BC = CD and AB = AD.

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Perimeter P = BC + CD + AB + AD.

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P = 2BC + 2AB .

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Step 3:

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Length of BC.

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Length of the two points $\"\"$ is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ .

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$\"\"$

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Length of AB.

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Length of the two points $\"\"$ is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ .

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$\"\"$

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Perimeter :

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$\"\"$

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\