Answer:
d) [tex]2\sqrt{26}[/tex] units
The perpendicular distance from the point to the line is
[tex]d = 2 \sqrt{26}[/tex] units
Step-by-step explanation:
Step(i):-
Given point is ( -10 , -5)
Given line is y = -5 x -3
5 x + y + 3 = 0
The perpendicular distance from the point to the line is
[tex]d = \frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2}+b^{2} } }[/tex]
Given line is 5 x + y +3 =0
a = 5 , b = 1 , c = 3
Given point is (x₁ , y₁) = ( -10 , -5)
Step(ii):-
The perpendicular distance from the point to the line is
[tex]d = \frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2}+b^{2} } }[/tex]
[tex]d = \frac{|5(-10) + 1(-5) + 3|}{\sqrt{(5)^{2}+1^{2} } }[/tex]
[tex]d =\frac{-50-5+3}{\sqrt{25+1} } = \frac{|-52|}{\sqrt{26} }[/tex]
[tex]d = \frac{2 X {26} }{\sqrt{26} }[/tex]
[tex]d = \frac{2 X \sqrt{26} X \sqrt{26} }{\sqrt{26} }[/tex] ( ∵ 26 = √26 × √26)
[tex]d = 2 \sqrt{26}[/tex]
Final answer:-
The perpendicular distance from the point to the line is
[tex]d = 2 \sqrt{26}[/tex]
