Originally Posted by

**sammy28** Hi all,

The points A(-1,-2), B(7,2) and C(k,4), where K is a constant, are the vertices of $\displaystyle \triangle$ABC. Angle ABC is a right angle.

a) Find the gradient of AB

b) Calculate the value of K.

c) Show that the length of AB may be written in the form $\displaystyle p\sqrt{5}$, where p is an integer to be found.

d) Find the exact value of the area $\displaystyle \triangle$ABC.

e) Find an equation for the straight line l passing through B and C. Give your answer in the form ax + by +c = 0, where a,b, and c are integers.

Can someone please indicate the easiest (less time consuming way of answering part b)

I worked out the equations for AC and BC using

$\displaystyle \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}$

so

$\displaystyle \frac{y+2}{4+2}=\frac{x+1}{k+1}$

$\displaystyle (y+2)(k+1)=6(x+1) \equiv ky+y+2k+2=6x+6$

$\displaystyle k=\frac{6x-y+4}{y+2}$

and

$\displaystyle \frac{y-2}{4-2}=\frac{x-7}{k-7}$

$\displaystyle (y-2)(k-7)=2(x-7) \equiv ky-2y-2k+14=2x-14$

$\displaystyle k(y-2)=2x-14-7y-14$

$\displaystyle k=\frac{2x+7y-28}{y-2}$

so solve

$\displaystyle \frac{6x-y+4}{y+2}=\frac{2x+7y-28}{y-2}$

Whilst writing this out perhaps i should be using $\displaystyle y-y_{1}=m(x-x{1})$ where the equations of line AC and line BC are perpendicular.

so

gradient AC = $\displaystyle \frac{4+2}{k+1}$

gradient BC = $\displaystyle \frac{4-2}{k-7}$

then

$\displaystyle \left(\frac{4+2}{k+1}\right) \left(\frac{4-2}{k-7}\right)= -1$

then solve

$\displaystyle 12k^2-8k-6=0$

maybe the second way is the quickest in the exam or are there shortcuts?

thanks sammy