# Urgent Finding Coordinates

• Oct 7th 2008, 03:26 PM
kaley015
Urgent Finding Coordinates
Three points have coordinates A(2,5), B(10,9) and C(6,2). Line L1 passes through A and B. Line L2 passes through C and is perpendicular to L1.
Find the coordinates of the point of intersection of L1 and L2? With working.
• Oct 7th 2008, 03:32 PM
icemanfan
The slope of AB = L1 is $\displaystyle \frac{9-5}{10-2} = \frac{4}{8} = \frac{1}{2}$. So the slope of the line through C (L2) will have opposite reciprocal slope, which is -2. Now use point-slope form ($\displaystyle y - y_0 = m(x - x_0)$) to find the equations for both lines and then solve the system of equations given by the lines.
• Oct 7th 2008, 03:42 PM
masters
Quote:

Originally Posted by kaley015
Three points have coordinates A(2,5), B(10,9) and C(6,2). Line L1 passes through A and B. Line L2 passes through C and is perpendicular to L1.
Find the coordinates of the point of intersection of L1 and L2? With working.

First, find the slope of the L1. Then define the equation.

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{4}{8}=\frac{1}{2}$

Use y=mx+b with (2, 5) and m = 1/2

$\displaystyle y=mx+b$

$\displaystyle 5=\frac{1}{2}(2)+b$

$\displaystyle 5=1+b$

$\displaystyle 4=b$

Equation of L1 = $\displaystyle \boxed{y=\frac{1}{2}x+4}$

Now use the negative reciprocal of the slope of L1 for your perpendicular.

That would be -2. Use point C(6,2) and define the equation of your perpendicular as above.

Use y=mx+b, m = -2, passing through C(6, 2)

$\displaystyle 2=-2(6)+b$

$\displaystyle 2=-12+b$

$\displaystyle 14=b$

Equation of L2 = $\displaystyle \boxed{y=-2x+14}$

Now, you can solve the system of those two linear equations and find where they intersect. Have fun.