Question

Find the distance between the following perpendicular lines x+3y=7 and (sqr(1+ sqr64)x +5y +5=0

Answer 1

The lines $x+3y=7$ and $(\sqrt{1+\sqrt{64}})x+5y+5=0$ are neither perpendicular nor parallel lines

$$y=-\dfrac{1}{3}x+\dfrac{7}{3}, slope_1=-\dfrac{1}{3}$$

$$y=-\dfrac{\sqrt{1+\sqrt{64}}}{5}x-1, slope_2=-\dfrac{9}{5}$$

$$slope_1\not=slope_2$$

$$slope_1(slope_2)=-\dfrac{1}{3}(-\dfrac{9}{5})=\dfrac{3}{5}\not=-1$$

The lines $x+3y=7$ and $(\sqrt{1+\sqrt{64}})x+5y+5=0$ intersect.

We the shortest distance between the given lines is zero.

When two parallel lines ($slope_1=slope_2$ ) are given by

$$ax+by+c_1=0$$

$$ax+by+c_2=0$$

the distance between them can be expressed as

$$d=\dfrac{|c_2-c_1|}{\sqrt{a^2+b^2}}$$