Question

Find the equation to the straight line drawn at right angles to the straight line $\dfrac{{\text{x}}}{{\text{a}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{b}}}$ =1 through the point where it meets the axis a of x.

Answer 1

Hint- this question is based on the theory of straight line. So we need to first recall the basics of straight line. As we know any straight line with slope m is written in the form of y = mx +c. and the product of slope of two perpendicular line is -1 i.e. $ m \times {{m}_{1}} = -1 $ in this case as discussed below.

Complete step-by-step solution -
The diagram for the question as follow:

Given line is
$\dfrac{{\text{x}}}{{\text{a}}}{\text{ - }}\dfrac{{\text{y}}}{{\text{b}}}$ =1
Let, Slope of the line is m and its value is
m = $\dfrac{{\text{b}}}{{\text{a}}}$
Then perpendicular slope is ${{\text{m}}_{\text{1}}}$ and its value is
${{\text{m}}_{\text{1}}}$= $\dfrac{{{\text{ - a}}}}{{\text{b}}}$
As we know the equation of straight line with slope ${{\text{m}}_{\text{1}}}$ is y = mx +c.
Then the equation of the line with this slope is y= $\dfrac{ - a}{b}$ x +c ………………… (1)
As (a,0) passes through the above line (Given line meets at x−axis i.e., put y=0)
Then, 0= $\dfrac{{{\text{ - a}}}}{{\text{b}}}$$ \times $a +c
$ \Rightarrow $${\text{c = }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{b}}}$
Put, value of c in equation (1), we get
$\Rightarrow y = \dfrac{-a}{b}\times x + \dfrac{{a}^{2}}{b} $
On taking L.C.M we get the required line $by + ax $ = ${{\text{a}}^{\text{2}}}$.

Note- The important thing is that you should know that the general equation of a straight is given as ax+by = c, where a,b $ \ne $0. You should know that when two lines are perpendicular then the product of their slope is -1 and also if two lines are parallel then their slopes are equal.