501 Geometry Questions

469. No solution: parallel lines.Sub “(^23 )x– 2” in for yin the other

equation and solve for x:

y= (^23 )x–2

–4x+ 6((^23 )x– 2) = 20

–4x+ 4x– 12 = 20

–12 ≠20. These two lines must be parallel since 12 does not equal 20,

which indicates that there is no solution to this system of equations.

470. (12,–2) Set the two equations equal to one another and then solve

for x:

(^34 )x– 11 = –(^12 )x+ 4

(^34 )x+ (^12 )x= 4 + 11

(^54 )x= 15

x= 12

Sub x= 12 in for xand solve for y:

y= –(^12 )(12) + 4, y= –2

Set 95

471. •A (0,0), •B (3,0), and •C (0,-3).Usually, in pairs, you would
     solve for each point of interception; however, x= 0 (the y-axis) and
     y= 0 (the x-axis) meet at the origin; therefore the origin is the first
     point of interception. One at a time, plug x= 0 and y= 0 into the
     equation y= x– 3 to find the two other points of interception: y= 0

- 3. y= –3; and 0 = x– 3. –3 = x. The vertices of ΔABC are A (0,0),

B (3,0), and C (0,–3).

472. ΔABC is an isosceles right triangle.ABhas zero slope; CAhas
     no slope, or undefined slope. They are perpendicular, and they
     both measure 3 lengths. ΔABC is an isosceles right triangle.


473. Perimeter= 6 units + 3 2 units.ABand CAare three units
     long. Using the Pythagorean theorem or the distance formula, find
     the length of BC. d=  3 ^2 + 3^2. d=  18 . d= 3 2 . The
     perimeter of ΔABC is the sum of the lengths of its sides: 3 + 3 +
     3  2 = 6 + 3 2 .

501 Geometry Questions