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.
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.
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.
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.