pertanyaan no.23 gradien garis QT 1/2 gimana cara cari titik T?

a) misal koordinat S = (a, b)
y = -1/2 x + 5 => b = -1/2 a + 5
Luas OPSQ = 12
=> OP . OQ = 12
=> b . a = 12
=> (-1/2 a + 5) a = 12
=> -1/2 a^2 + 5a - 12 = 0 =====> kali -2
=> a^2 - 10a + 24 = 0
=> (a - 6)(a - 4) = 0
=> a = 6 atau a = 4
Jika a = 6 => b = -1/2 (6) + 5 = 2 => (6, 2)
Jika a = 4 => b = -1/2 (4) + 5 = 3 => (4, 3)
Karena nilai x lebih besar 1 dari y maka
koordinat S yang memenuhi (4,3)

2) misal koordinat T = (x, y) => y = -1/2 x + 5
Koordinat Q = (4, 0)
Gradien QT = 1/2 =======> rumus gradien : (y2 - y1)/(x2 - x1)
=> (0 - y)/(4 - x) = 1/2
=> -2y = 4 - x
=> -2(-1/2 x + 5) = 4 - x
=> x - 10 = 4 - x
=> 2x = 14
=> x = 7
y = -1/2 (7) + 5 = -7/2 + 10/2 = 3/2
Koordinat T = (7, 3/2)

c) PSTU adalah jajar genjang maka PS sejajar UT dan PU sejajar ST
P = (0, 3), S(4, 3), T(7, 3/2), U(X, Y)
1) Gradien PS = Gradien UT
=> (3 - 3)/(4 - 0) = (3/2 - Y)/(7 - X)
=> 0/4 = (3/2 - Y)/(7 - X)
=> 4(3/2 - Y) = 0(7 - X)
=> 6 - 4Y = 0
=> -4Y = -6
=> Y = 3/2
2) Gradien PU = gradien ST
=> (Y - 3)/(X - 0) = (3/2 - 3)/(7 - 4)
=> (3/2 - 3)/X = (3/2 - 6/2)/3
=> (-3/2)/X = (-3/2)/3
=> 1/X = 1/3
=> X = 3
Jadi koordinat U = (3, 3/2)