a) BD and AC meet at a point which is the midpoint of AC,
the mid point of AC is ((-1+5)/2, (5+1)/2) = (2,3)
Line BD is also parallel to a line with the equation y= -5x +2
But parallel lines have equal slope, thus the slope of BD is -5
Therefore, since BD passes through (2,3) and has a slope of -5 , its equation will be;
(y-3)/(x-2)=-5
y-3 = -5(x-2)
y-3 = -5x + 10
y= -5x+13
b) BC is perpendicular to AC,
the product of slopes of two perpendicular lines is -1
The slope of AC is (1-5)/(5--1)= -4/6 that is -2/3
Therefore the gradient of BC will be 3/2
Equation of BC;
(y-1)/(x-5) = 3/2
2(y-1) = 3(x-5)
2y - 2 = 3x - 15
2y= 3x -13
ii) the coordinates of B
Line BC and line BD meet at point B,
Therefore, to get the coordinate of B, we solve the equation of BC and BD simultaneously
y + 5x =13
2y - 3x = -13 ( by eliminating y)
= 2y + 10x = 26
2y - 3x = -13 (subtracting the two equations)
13 x = 39
x = 3 and y = -5(3) +13
y= -2
thus the coordinates of B is (3,-2)
iii) the coordinates of D,
The mid point of line BD is (2,3) and the coordinate of B is (3,-2)
thus (x+3)/2 =2 and (y +-2)/2 = 3, solving for x and y;
therefore x = 1 and y= 8
hence the coordinate of D is (1,8)
