Answer with explanation:
The equation of any line with slope 'm' and passing through any point [tex](x_1,y_1)[/tex] is given by
[tex]y=mx+(y_1-mx_1)[/tex]
As we know that the general equation of a line with slope 'm' is [tex]y=mx+c[/tex]
Comparing with the given equation [tex]y=3x+2[/tex] we can conclude slope of the given line is [tex]m_1=3[/tex]
Now we know that the product of slopes of perpendicular lines is -1
Mathematically we can write for perpendicular lines
[tex]m_1\times m_2=-1[/tex]
Thus the slope of the required line is obtained from the above relation since it is given that they are perpendicular
[tex]3\times m_{2}=-1\\\\\therefore m_{2}=\frac{-1}{3}[/tex]
Hence using the given and the obtained values the equation of the required line is
[tex]y=-\frac{1}{3}x+(1-\frac{-1}{3}\times 6)}\\\\y=\frac{-x}{3}+3[/tex]
Part b)
The angle of intersection between 2 lines with slopes [tex]m_{1},m_2[/tex] is given by
[tex]\theta =tan^{-1}(\frac{m_2-m_1}{1+m_1m_2})[/tex]
Comparing the equations of given lines
[tex]y=6x+3\\\\y=2x+3[/tex]
with the standard equation we get
[tex]m_{1}=6\\\\m_{2}=2[/tex]
Thus the angle of intersection becomes
[tex]\theta =tan^{-1}(\frac{2-6}{1+2\times6})=17.10[/tex]
