Question

3x - 4y + 3 = 0
6x - 8y +7 = 0
a. Find the distance between the two lines.
b. Find the equation of the perpendicular line passing through. (-6,4)
c. Determine the distance of 2 given equations to point (-6.4)

Answer 1

$d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ - formula to find distance between some point $M(x_0,y_0)$ and a line $Ax+By+C=0$

We can use it to solve our first problem.

a)

Let $x=0$, then $3\cdot 0-4y+3=0 \iff y=\frac{3}{4}$

So point $M(0,\frac{3}{4})$ belong to the first line.

$A_2=6, B_2=-8, C_2=7$ - coeffiсients of second line.

$d=\frac{|6\cdot 0+(-8)\cdot\frac{3}{4}+7|}{\sqrt{6^2+(-8)^2}}=\frac{|1|}{\sqrt{100}}=\frac{1}{10}$

(Lines don't intersect because $\frac{A_1}{A_2}=\frac{B_1}{B_2}\neq\frac{С_1}{С_2}$. They are parallel)

b) If two lines are perpendicular, their normal vectors are perpendicular too.

$\vec {n_1}=\{3;-4\}$ normal vector of first line.

$\vec {n_3}=\{a,b\}$ normal vector of perpendicular line

$\vec{n_1}\cdot\vec{n_3}=3\cdot a+(-4)\cdot b=0$ (scalar product )

Let $a=4,$ then $b=3$

So $\vec{n_3}=\{4,3\}$ (what means $A_3=4, B_3=3$ )

$4\cdot(-6)+3\cdot4+C_3=0$ (substitution of (-6, 4))

$C_3=12$

$4x+3y+12=0$ - perpendicular line

c) We can use our first formula

$d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

First line:

$d_1=\frac{|3\cdot(-6)+(-4)\cdot4+3|}{\sqrt{3^2+(-4)^2}}=\frac{|-31|}{\sqrt{25}}=\frac{31}{5}=6.2$

$d_2=\frac{|6\cdot(-6)+(-8)\cdot4+7|}{\sqrt{6^2+(-8)^2}} =\frac{|-61|}{\sqrt{100}}=\frac{61}{10}=6.1$