F.4 Equations of straight line

(30) (a) Slope of BC = (3 - 0) / (0 - 2) = -1.5
(b) Since AP is perpendicular to BC, slope of AP = -1/-1.5 = 2/3
(y - 0) / (x + 1) = 2/3
3y = 2x + 2
(c) (i) x = 0 => y = 2/3
H is (0, 2/3)
(ii) OC is another altitude, so OC passes through this point.
Consider the third altitude, we can follow the same step as above.
Slope of AC = (3 - 0) / (0 + 1) = 3
Slope of altitude = -1/3
Equation of altitude is (y - 0) / (x - 2) = -1/3
3y = -x + 2
The x-ordinate of the point the altitude cuts the y-axis is given by,
3y = 2
y = 2/3
The point is (0, 2/3) the same point H.
(31) (a) For A, y = 0, 2x + 4 = 0 => x = -2. So A is (-2,0)
For B, x = 0, -y + 4 = 0 => y = 4. So B is (0,4)
(b) Slope of L1 = 2
Slope of L2 = -1/2
(y - 4) / (x - 0) = -1/2
2y - 8 = -x
2y + x - 8 = 0
(c) The x-ordinate of C is given by 2(0) + x - 8 = 0 => x = 8
Triangle CAB is similar to triangle COD
AC = 8 + 2 = 10
OC = 8
Area ODC : Area ABC = 8^2 : 10^2 = 16 : 25
Area ODC : Area OABD = 16 : 25 - 16 = 16 : 9