To find the equation of the line with given properties, let's start with the general form of the line equation in normal form. The normal form of the equation of a line is given by:
x‌cos‌θ+ysin‌θ=pwhere:
p is the perpendicular distance from the origin to the line, which is given as 5 units.
θ is the angle that the normal (perpendicular) makes with the positive
x-axis.
Given that the slope of the line is -1 , we can determine the angle
α the line makes with the positive
x-axis, where:
tan‌α=−1⟹α=135∘‌ or ‌315∘ Since the line is perpendicular to the direction of the normal vector, the angle
θ the normal makes with the positive
x-axis will be:
θ=α−90∘=45∘‌ or ‌−45∘ For
θ=45∘, the normal form equation becomes:
x‌cos‌45∘+ysin‌45∘=5Since
cos‌45∘=sin‌45∘=‌, the equation simplifies to:
x‌+y‌=5or
‌(x+y)=5 Multiplying both sides by 2 , we get:
√2(x+y)=10or
x+y=‌=5√2 Therefore, the equation in the standard form is:
x+y−5√2=0‌‌ or
‌‌x+y+5√2=0 Hence, the correct option is: Option A:
x+y±5√2=0