Question

Let $L$ be the line $2x + y = 2$. If the axes are rotated anticlockwise by ${45^\circ }$, then the intercepts are made by the line $L$ on the new axes respectively.
A) $\sqrt 2 $ and $1$
B) $1$ and $\sqrt 2 $
C) $2\sqrt 2 $ and $2\dfrac{{\sqrt 2 }}{3}$
D) $\dfrac{{2\sqrt 2 }}{3}$ and $2\sqrt 2 $

Answer 1

Hint:We will be using the concept of the intercept form of the equation of the line to solve the given equation.
If we have an equation of line $ax + by + c = 0$ then its intercept form will be $\dfrac{x}{a} + \dfrac{y}{b} = 1$ . Equations of lines can be written in three forms:
The Two-Intercept form, The Point-Slope form, and The Slope-Intercept form.

Complete step by step Solution:
We have the provided equation of the line,
$2x + y = 2$
if we have to express the equation in intercept form i.e.,
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
Then we divide the given equation into both sides by $2$
On simplifying we get,
$\dfrac{x}{1} + \dfrac{y}{2} = 1$
On comparing with the general equation of the line in intercept form we get, intercepts $(1,2)$
Now, we rotate the axis to $\alpha = - {45^\circ }$ i.e., in anti-clockwise direction, then we will have
$x = X\cos \alpha - Y\sin \alpha $
$y = X\sin \alpha + Y\cos \alpha $
Put $\alpha = - {45^\circ }$
$x = X\cos ( - {45^\circ }) - Y\sin ( - {45^\circ })$
$y = X\sin ( - {45^\circ }) + Y\cos ( - {45^\circ })$
Solving these two we get
$x = \dfrac{{X + Y}}{{\sqrt 2 }}$ and $y = \dfrac{{X - Y}}{{\sqrt 2 }}$
Substituting these values in the intercept form of the equation of the line, we get
$\dfrac{{X + Y}}{{\sqrt 2 }} + \dfrac{{X - Y}}{{2\sqrt 2 }} = 1$
Solving further, we have
$\dfrac{X}{{\dfrac{{2\sqrt 2 }}{3}}} + \dfrac{Y}{{2\sqrt 2 }} = 1$
Hence, the intercepts of this equation are $\dfrac{{2\sqrt 2 }}{3}$ and $2\sqrt 2 $ .

Therefore, the correct option is (D).

Note: In calculus, the most common form of linear equation you will see is $y = ax + b$ where, $a$ and $b$ are constants. An equation of such form is called a linear equation in slope-intercept form. An equation of the form, $y - {y_0} = a(x - {x_0})$ is called the point-slope form of a linear equation.