2) the line AB has equation 3x - 4y + 5 = 0
a) the point with coordinates (p, p+2) lies on the line AB. Find the value of the constant P
b) find the gradient of AB
c) the point A has coordinates (1,2). The point C (-5, k) is such that AC is perpendicular to AB. Find the value of K.
D) the line AB intersects the line with equation 2x - 5y = 6 at the point D. Find the coordinates of D
+6248 a) \(x=p,~y=p+2\\ 3(p) - 4(p+2) + 5 = 0\\ 3p-4p+8 = 0\\ -p+8=0 \\ p=8\)
b) There are a few ways to do this. I'd rewrite the equation for AB as
\(3x-4y+5=0 \\ 4y = 3x+5 \\ y = \dfrac 3 4 s + \dfrac 5 4 \\ \text{and you can read the gradient right off as }m=\dfrac 3 4\)
c) The key here is that the product of the slopes over perpendicular lines is -1.
\(m_{AC} = \dfrac{k-2}{-5-1} = -\dfrac{k-2}{6} \\ \text{we know from (a) that }m_{AB}=\dfrac 3 4 \\ m_{AC} \cdot m_{AB}=-1\\ -\dfrac{k-2}{6} \cdot \dfrac 3 4 = -1 \\ \dfrac{3k-6}{24}=1 \\ 3k=30\\ k=10\)
d)
\(AB: y=\dfrac 3 4 x+\dfrac 5 4\\ 2x-5y=6 : y = \dfrac 2 5 x - \dfrac 6 5 \\ \dfrac 3 4 x+\dfrac 5 4 = \dfrac 2 5 x - \dfrac 6 5 \\ \dfrac{15-8}{20}x = -\dfrac{24+25}{20}\\ \dfrac{7}{20}x=-\dfrac{49}{20} \\ x=-7 \\ y = \dfrac{2}{5}\cdot (-7)-\dfrac 6 5 = -\dfrac{20} 5 = -4\\ D=(x,y) =(-7,-4)\)
