Please help me!

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.

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- Oct 7th 2008, 03:26 PMkaley015Urgent Finding Coordinates
Please help me!

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 PMicemanfan
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 PMmasters
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.