amaths

1. y = 2x+1
2x - y + 1 = 0
Distance required = [2(5) - (-3) + 1]/sqrt(2^2 + 1^2)
= 14/sqrt(5)
= 14sqrt(5)/5
2(a) Let m be the slope of the angle bisectors.
Slope of the two given lines are -3/4 and 5/12
Hence, for angle bisector,
|(m+3/4)/(1-3m/4)| = |(m-5/12)/(1+5m/12)|
|(4m+3)/(4-3m)| = |(12m - 5) / (12+5m)|
(4m+3)(12+5m) = (4-3m)(12m-5) or (4m+3)(12+5m) = -(4-3m)(12m-5)
20m^2 + 63m+ 36 = -36m^2 + 63m - 20 or 20m^2 + 63m +36 = 36m^2 - 63m + 20
56m^2 +56 = 0 (no solution) or 16m^2 - 126m - 16 = 0
8m^2 - 63m - 8 = 0
(m-8)(8m+1) = 0
m = 8 or -1/8
The intersection of the two given lines can be computed as:
3x+4y-1 = 0 ...(1)
5x- 12y + 3 = 0 ...(2)
(1)x3: 9x+12y - 3 = 0 ...(3)
(2)+(3): 14x = 0
x = 0
y = 1/4
The intersection of the two lines is (0,1/4)
The two angle bisectors are y-1/4 = 8x or y - 1/4 = (-1/8)x
32x-4y+1 = 0 or x + 8y - 2 = 0
(b) The slope of the two angle bisectors found in (a) are 8 and -1/8
Since (8)(-1/8) = -1, the two bisectors are perpendicular to each other