Question

A is the point of intersection of the lines 3x − y = 7 and x + 4y + 2 = 0. Find
(a) in normal form, the equation of the line which passes through A and is parallel
to the line 2x + 3y − 40 = 0,
(b) the perpendicular distance of A from the line 2x + 3y − 40 = 0

Answer 1

Find the point of intersection A by solving together two equations:$\left\{\begin{matrix} 3x-y=7 \\ x+4y=-2 \end{matrix}\right.$ . As a result obtain: $A=(2,-1)$ .

Part (a):

1. Find the equation of of the line which passes through A and is parallel to the line $2x + 3y − 40 = 0$. This two lines have common normal vector $b=(2,3)$. The equation of the line passes through the point $(2,-1)$ and has normal vector (2,3) is: $2(x-2) + 3(y+1) = 0 \Rightarrow 2x+3y-1=0$.
2. To transform this equation into the normal form one should divide both sides by the following expression: $\sqrt{2^2 + 3^2}$. Thus obtain: $\dfrac{1}{\sqrt{13}}(2x+3y-1)=0 \Rightarrow \dfrac{2}{\sqrt{13}}x+\dfrac{3}{\sqrt{13}}y-\dfrac{1}{\sqrt{13}}=0$.

Part (b):

1. The distance between the point $(x_0,y_0)$ and a line is given by the formula: $d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2 + b^2}}$ , where $a,b,c$ - coefficients in the line equation.
2. With given data: $d=\dfrac{|2\cdot 2+3\cdot(-1)-40|}{\sqrt{13}} = \dfrac{39}{\sqrt{13}}$.