急F4 Equations of straight line

24.
(a)
AB: 3x - y - 1 = 0 ...... (1)
AC: x + 2y - 5 = 0 ...... (2)

(2)´3 - (1) :
7y - 14 = 0
y = 2

(1)´2 + (2) :
7x - 7 = 0
x = 1

Hence, the coordinates of A = (1, 2)

(b)
Put x = 0 into the equation of AB :
0 - y - 1 = 0
y = -1
Hence, the coordinates of B = (0, -1)

Put y = 0 into the equation of AC:
x + 0 - 5 = 0
x = 5
Hence, the coordinates of C = (5, 0)

(c)
The equation of BC (intercept form) :
(x/5) + [y/(-1)] = 1
x - 5y - 5 = 0

(d)
Slope of BC = (0 + 1)/(5 - 0) = 1/5
Slope of the required altitude = -1/(1/5) = -5
The required altitude passes through A(1, 2).
The equation of the required altitude (point-slope form) :
(y - 2)/(x - 1) = -5
y - 2 = -5x + 5
5x + y - 7 = 0


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30.
(a)
Slope of BC = (0 - 3)/(2 - 0) = -3/2

(b)
AP is perpendicular to BC.
Slope of AP = -1/(-3/2) = 2/3
The equation of AP (point-slope form) :
(y - 0)/(x + 1) = 2/3
2x + 2 = 3y
AP : 2x - 3y + 2 = 0

(c)(i)
Put x = 0 into the equation of AP:
0 - 3y + 2 = 0
y = 2/3
Hence, the coordinates of H = (0,2/3)

(c)(ii)
(Slope of BH) ´ (Slopeof AC)
= {[0 -(2/3)]/(2 - 0)} ´ (0 - 3)/(-1 - 0)
= (-1/3) ´ 3
= -1
Hence, the altitude from B to passes through H.
Both the altitude AP and the altitude from C to AB pass through H.
Hence, the three altitudes of ΔABC pass through the same point, H.