# Thread: Find cordinates of the point of intersection.

1. ## Find cordinates of the point of intersection.

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Length A right triangle is formed in the first quadrant by the x- and y-axes and the line through the point (3,2). write the length L of the hypotenuse as a function of x.

2. Originally Posted by bilalownsu

one median is the line segment joining (b,c) and the midpt of (-a,0) and (a,0) which is (0,0).. so, get the equation on the line joining (b,c) and (0,0)..

another median is the line segment joining (a,0) and the midpt of (-a,0) and (b,c).. get the midpt first then get the equation of the line joining that midpt and (a,0)..

so you have two linear equations, just get their intersection..

Originally Posted by bilalownsu
Length A right triangle is formed in the first quadrant by the x- and y-axes and the line through the point (3,2). write the length L of the hypotenuse as a function of x.

so, $L=\sqrt{x^2+y^2}$

get the equation of the line joining (x,0) and (3,2): $Y = \frac{2}{3-x}(X-x)$.. since (0,y) is in this line, it must be satisfied.. thus $y = \frac{2}{3-x}(-x) = \frac{2x}{x-3}$ (we just sabstitute 0 to X and y to Y)

therefore, the length of the hypotenuse as a function of L is $L(x) = \sqrt{x^2+\left(\frac{2x}{x-3}\right)^2}$

3. Hello, bilalownsu!

A right triangle is formed in the first quadrant by the x- and y-axes
and the line through the point $P(3,2).$
Write the length $L$ of the hypotenuse as a function of its $x$-intercept.
Code:
      |
|B
(0,b) o
|   * (3,2)
|       o
|       P   *
|               *   A
- - + - - - - - - - - - o - -
|                 (a,0)

The line has intercepts: $A(a,0)\,\text{ and }\,B(0,b)$
. . Note that: $a > 3$

The slope of the line through $P$ and $A$ is: . $m \:=\:\frac{-2}{a-3}$

The equation of the line through $A$ with slope $m$ is:

. . $y \:=\:\frac{-2}{a-3}(x-a) \quad\Rightarrow\quad y \:=\:\frac{-2}{a-3}x + \frac{2a}{a-3}$

Hence: line $L$ has intercepts: . $(a, 0)\,\text{ and }\,\left(0,\frac{2a}{a-3}\right)$

Therefore, the length of $L$ is: . $\sqrt{a^2 + \left(\frac{2a}{a-3}\right)^2} \;=\;\boxed{\frac{a\sqrt{a^2-6a+13}}{a-3}}$