# Math Help - coordinate geometry

1. ## coordinate geometry

Hi all,

The points A(-1,-2), B(7,2) and C(k,4), where K is a constant, are the vertices of $\triangle$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 $p\sqrt{5}$, where p is an integer to be found.

d) Find the exact value of the area $\triangle$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

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

so

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

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

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

and

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

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

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

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

so solve

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

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

so

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

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

then

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

then solve

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

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

thanks sammy

2. 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 $\triangle$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 $p\sqrt{5}$, where p is an integer to be found.

d) Find the exact value of the area $\triangle$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

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

so

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

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

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

and

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

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

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

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

so solve

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

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

so

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

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

then

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

then solve

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

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

thanks sammy
That's what I've done, but I'm not sure if you are allowed to use this method:

$\overline{AB} = (7,2) - (-1,-2) = (8,4)~\implies~|\overline{AB}|^2=80$

$\overline{AC} = (k,4) - (-1,-2) = (k+1,6)~\implies~|\overline{AC}|^2=(k+1)^2+36$

$\overline{BC} = (k,4) - (7,2) = ((k-7)^2,2)~\implies~|\overline{BC}|^2=(k-7)^2+4$

Since ABC is a right triangle with the hypotenuse AC you have to solve the equation

$80+(k+1)^2+36=(k-7)^2+4$

which will yield k = -4

3. thanks earboth . maybe using pythagoras thm is just as quick as both come to the same quadratic.

i checked both pythagoras and gradient method and both resolve to (sorry mind was scrambled by the time i wrote above quadratic)

$k^2-6k+5=0$

I checked the book and it gives the answer as 6, but

$(k-1)(k-5)$

tells me that the answer is 1 or 5.

thanks

4. This was troubling me, because the book is a very good book, and i double checked by plotting the equations on a graphing calc and there are two solutions if ACB is a right angle.

I should read the question more carefully