Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a fixed angle keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$, then
A) ${a^2} + {b^2} = {p^2} + {q^2}$
B) $\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} = \dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}$
C) ${a^2} + {p^2} = {b^2} + {q^2}$
D) $\dfrac{1}{{{a^2}}} + \dfrac{1}{{{p^2}}} = \dfrac{1}{{{b^2}}} + \dfrac{1}{{{q^2}}}$
Answer
303k+ views
Hint: When two objects are measured along a line that is perpendicular to one or both then, the distance between them is known as the perpendicular distance. It contains the perpendicular distance between a point and a line, the perpendicular distance between skew lines, and many more. Perpendicular distance is basically the shortest distance between two objects.
Formula Used:
The perpendicular distance $d$ of a point $({x_1},{y_1})$ from the line $ax + by + c = 0$ is given by $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ .
Complete step by step Solution:
According to the question, line $L$ has intercepts $a$ and $b$.
So, the equation of the line with these intercepts will be,
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
On rotating the axes, the intercepts become $p$ and $q$ which gives the equation of line $L$ as
$\dfrac{x}{p} + \dfrac{y}{q} = 1$
The perpendicular distance from the origin of the lines (1.1) and (1.2) remaining the same are given by,
$\left| {\dfrac{1}{{\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} }}} \right| = \left| {\dfrac{1}{{\sqrt {\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}} }}} \right|$
$\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} = \sqrt {\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}} $
Taking the square of both the sides, we get the equation
$\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} = \dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}$
Therefore, the correct option is (B).
Note: The condition for which the two lines ${a_1}x + {b_1}y + c = 0$ and ${a_2}x + {b_2}y + c = 0$ are perpendicular is given by ${a_1}{a_2} + {b_1}{b_2} = 0$ .
If the axes of the equation of line are rotated in a clockwise direction, then the angle will be $\alpha $ and if they are rotated in an anticlockwise direction, then the angle will become $( - \alpha )$.
Formula Used:
The perpendicular distance $d$ of a point $({x_1},{y_1})$ from the line $ax + by + c = 0$ is given by $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$ .
Complete step by step Solution:
According to the question, line $L$ has intercepts $a$ and $b$.
So, the equation of the line with these intercepts will be,
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
On rotating the axes, the intercepts become $p$ and $q$ which gives the equation of line $L$ as
$\dfrac{x}{p} + \dfrac{y}{q} = 1$
The perpendicular distance from the origin of the lines (1.1) and (1.2) remaining the same are given by,
$\left| {\dfrac{1}{{\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} }}} \right| = \left| {\dfrac{1}{{\sqrt {\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}} }}} \right|$
$\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} = \sqrt {\dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}} $
Taking the square of both the sides, we get the equation
$\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} = \dfrac{1}{{{p^2}}} + \dfrac{1}{{{q^2}}}$
Therefore, the correct option is (B).
Note: The condition for which the two lines ${a_1}x + {b_1}y + c = 0$ and ${a_2}x + {b_2}y + c = 0$ are perpendicular is given by ${a_1}{a_2} + {b_1}{b_2} = 0$ .
If the axes of the equation of line are rotated in a clockwise direction, then the angle will be $\alpha $ and if they are rotated in an anticlockwise direction, then the angle will become $( - \alpha )$.
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