Question

What is the equation of the line that passes through (6, 1) and is perpendicular to the line y 3x+2? CI) At what angle do the lines y= 6x +4 and y-2x+3 cross?

Answer 1

Answer with explanation:

The equation of any line with slope 'm' and passing through any point $(x_1,y_1)$ is given by

$y=mx+(y_1-mx_1)$

As we know that the general equation of a line with slope 'm' is $y=mx+c$

Comparing with the given equation $y=3x+2$ we can conclude slope of the given line is $m_1=3$

Now we know that the product of slopes of perpendicular lines is -1

Mathematically we can write for perpendicular lines

$m_1\times m_2=-1$

Thus the slope of the required line is obtained from the above relation since it is given that they are perpendicular

$3\times m_{2}=-1\\\\\therefore m_{2}=\frac{-1}{3}$

Hence using the given and the obtained values the equation of the required line is

$y=-\frac{1}{3}x+(1-\frac{-1}{3}\times 6)}\\\\y=\frac{-x}{3}+3$

Part b)

The angle of intersection between 2 lines with slopes $m_{1},m_2$ is given by

$\theta =tan^{-1}(\frac{m_2-m_1}{1+m_1m_2})$

Comparing the equations of given lines

$y=6x+3\\\\y=2x+3$

with the standard equation we get

$m_{1}=6\\\\m_{2}=2$

Thus the angle of intersection becomes

$\theta =tan^{-1}(\frac{2-6}{1+2\times6})=17.10$