# Math Help - Challenging coordinate geometry?

1. ## Challenging coordinate geometry?

Find the correct statement(s)?

A) the lines x + log b(base a).y + log c(base a) = 0, log a(base b).x + y + log c(base b)= 0 and log a(base c).x + log b(base c).y + 1 = 0 are concurrent

B)Equation of a straight line passing through origin and making with x axis an angle twice the size of the angle made by the line 5y = x with the x-axis is 5y = 2x

C) The triangle formed by 2x - y + 5 = 0, x + y - 5 = 0 and x - 2y - 5 = 0 is isoceles

D) None of these.

2. Originally Posted by fardeen_gen
Find the correct statement(s)?

A) the lines x + log b(base a).y + log c(base a) = 0, log a(base b).x + y + log c(base b)= 0 and log a(base c).x + log b(base c).y + 1 = 0 are concurrent

B)Equation of a straight line passing through origin and making with x axis an angle twice the size of the angle made by the line 5y = x with the x-axis is 5y = 2x

C) The triangle formed by 2x - y + 5 = 0, x + y - 5 = 0 and x - 2y - 5 = 0 is isoceles

D) None of these.
B) False. Check the arctan of their slopes. Almost, but not twice the size

$\tan^{-1}\left(\frac{1}{5}\right)\approx 11.3^{\circ}$

$\tan^{-1}\left(\frac{2}{5}\right)\approx 21.8^{\circ}$

C) True. The triangle is isosceles

$2x-y+5=0 \ \ and \ \ x+y-5=0$ intersect at $A(0, 5)$

$2x-y+5=0 \ \ and \ \ x-2y-5=0$ intersect at $B(-5, -5)$

$x+y-5=0 \ \ and \ \ x-2y-5=0$ intersect at $C(5, 0)$

Using the distance formula, it can be shown that $AB=BC=\sqrt{125}$.

3. Thank you! I figured out C just now. But what about A? I am not familiar with logarithmic quantities to a great extent.

4. Originally Posted by fardeen_gen
Find the correct statement(s)?

A) the lines x + log b(base a).y + log c(base a) = 0, log a(base b).x + y + log c(base b)= 0 and log a(base c).x + log b(base c).y + 1 = 0 are concurrent
$x+(\log_a b)y + \log_a c =0$

$(\log_b a)x+y+\log_b c=0$

$(\log_c a)x+(\log_c b)y+1=0$

I'm not sure about this one, either. I suggest you post it as a separate thread and see if any of our other astute tutors can tackle it. I hope I translated your code correctly.

5. Hello, fardeen_gen!

Ah, masters explained (B) and (C) . . . good!

Find the correct statement(s).
A very challenging problem!

A) The lines: . $\begin{Bmatrix}x + (\log_ab)y &=& -\log_ac & [1] \\ (\log_ba)x + y &=& -\log_bc & [2]\\
(\log_ca)x + (\log_cb)y &=& -1 & [3] \end{Bmatrix}\quad\text{ are concurrent.}$
. . . . True
Using the Base-Change Formula:

$[1]\;\;x + \left(\frac{\log b}{\log a}\right)y \:=\:-\frac{\log c}{\log a} \quad\Rightarrow\quad (\log a)x + (\log b)y \:=\:-\log c$

$[2]\;\;\left(\frac{\log a}{\log b} \right)x + y \:=\:-\frac{\log c}{\log b} \quad\Rightarrow\quad (\log a)x + (\log b)y \:=\:-\log c$

$[3]\;\;\left(\frac{\log a}{\log c}\right)x + \left(\frac{\log b}{\log c}\right)y \:=\:-1 \quad\Rightarrow\quad (\log a)x + (\log b)y \:=\:-\log c$

The lines are identical . . . They are concurrent.
. . (They share a common point.)