Question

Find the equation of the straight line which cuts off intercepts on x-axis twice as that of Y-axis and is at a unit distance from origin.

Answer 1

Hint: In order to answer this question, to find the equation of the given straight line, we will first assume the variable of both the axes. And then we will write the equation of the straight line on the basis of the assumed variables. And then we will first find the value of the assumed variable to find the final equation of the line.

Complete step by step solution:
Let the straight line be $ ax + by + c = 0 $ and the intercepting point be $ ({x_1},{y_1}) $ .
And the distance of the line from the point $ ({x_1},{y_1}) $ is $ d $ .
So, the value of $ d $ $ = |\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}| $
Now, as we know that,
Intercept form of line:
 $ \because \dfrac{x}{a} + \dfrac{y}{b} = 1 $
here, $ a $ is the $ x - \operatorname{int} ercept $
and $ b $ is the $ y - \operatorname{int} ercept $
Now, according to the question: $ x - \operatorname{int} ercept $ twice as that of $ y - \operatorname{int} ercept $ , i.e.. $ a = 2b $ .
 $
   \Rightarrow \dfrac{x}{{2b}} + \dfrac{y}{b} = 1\,\,\,\,\,\,(replacing\,a\,by\,2b) \\
   \Rightarrow x + 2y = 2b \\
   \Rightarrow x + 2b - 2b = 0 \;
  $
Now, as per the question, the distance of the straight line from the origin is a unit distance:
 $
  \therefore Dis\tan ce\,from\,({x_1},{y_1}) = Dis\tan ce\,from\,origin(0,0) \\
   \Rightarrow |\dfrac{{1 \times 0 + 2 \times 0 - 2b}}{{\sqrt {{1^2} + {2^2}} }}| = 1 \\
   \Rightarrow |\dfrac{{ - 2b}}{{\sqrt {1 + 4} }}| = 1 \\
   \Rightarrow |\dfrac{{ - 2b}}{{\sqrt 5 }}| = 1 \\
   \Rightarrow | - 2b| = \sqrt 5 \\
   \Rightarrow - 2b = \pm \sqrt 5 \\
   \Rightarrow 2b = \mp \sqrt 5 \;
  $
After intercepting, the equation of the line is $ x + 2y - 2b = 0 $ .
We have a value of $ 2b $ .
 $
  \because x + 2y - 2b = 0 \\
   \Rightarrow x + 2y - ( \mp \sqrt 5 ) = 0 \\
   \Rightarrow x + 2y \pm \sqrt 5 = 0 \;
  $
Hence, the required equation of the straight line is $ x + 2y \pm \sqrt 5 = 0 $ .
So, the correct answer is “ $ x + 2y \pm \sqrt 5 = 0 $ ”.

Note: A straight line equation can be calculated in a variety of ways depending on the circumstances. This note discusses the straight line, horizontal line, slope of a straight line, collinear line, equation of a straight line and other topics.