Question

How do you find the slope of the line perpendicular to the line $ x = 10 $ ?

Answer 1

Hint: In this question, we should remember that the line $ x = 10 $ , is a vertical line parallel to $ y $ -axis, and perpendicular to it must be a horizontal line which is parallel to $ x $ -axis. To calculate the slope we can use the formula, $ m = \tan \theta $ , where $ m $ is the slope and is the smallest angle which the given line makes with the positive direction of $ x $ -axis.

Complete step-by-step solution:
Perpendicular lines are two or more lines that intersect at an $ {90^\circ} $ angle. These $ {90^\circ} $ angles are also known as right angles.
The slope of a line is the change in a line's y-value with respect to the change in the line's x-value.
The slope or gradient of a line is a number that describes both the direction and the steepness of the line. The steepness, incline, or grade of a line is measured by the absolute value of the slope.
If $ \theta $ $ \left( { \ne {{90}^\circ}} \right) $ is the inclination of a straight line, then $ \tan \theta $ is called its slope or gradient. The slope of any inclined plane is the ratio between the vertical rise of the plane and its horizontal distance where, $ \theta $ is the smallest angle which the given line makes with the positive direction of $ x $ -axis.
The slope of a straight line is the tangent of its inclination and is denoted by letter ‘m’ i.e. if the inclination of a line is $ \theta $ , its slope $ m = \tan \theta $ .
Two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. Example: If the slope of the first equation is 4, then the slope of the second equation will need to be $ \dfrac{1}{4} $ for the lines to be perpendicular.
Given line is $ x = 10 $ ,
It is a vertical line $ x = 10 $ is a line parallel to $ y $ -axis. (Angle between line $ x = 10 $ and $ y $ -axis is $ {90^\circ} $ ),
The Slope ( $ {m_1} $ ) of line $ x = 10 $ is $ \tan {90^\circ} $ ,
 $ {m_1} = \tan {90^\circ} - - - - (1) $ ,
Thus the slope ( $ {m_2} $ ) of the line perpendicular to the line $ x = 10 $ is:-
We know that,
 $ {m_2} = \dfrac{{ - 1}}{{{m_1}}} $ ,
Now substituting the values we get,
 $ \Rightarrow {m_2} = - \dfrac{1}{{\tan {{90}^\circ}}} $ ,
And using the trigonometric identity $ \cot \theta = \dfrac{1}{{\tan \theta }} $ , we get,
 $ \Rightarrow {m_2} = - \cot {90^\circ} $ ,
We know that $ \cot {90^\circ} = 0 $ ,we get,
 $ \Rightarrow {m_2} = 0 $ ,
The required slope is $ 0 $ .

Slope of the line perpendicular to the line $ x = 10 $ is $ 0 $.

Note: The slope of a line is positive if it makes an acute angle in the anti-clockwise direction with $ x $ -axis
The slope of a line is negative, if it makes an obtuse angle in the anti-clockwise direction with the $ x $ -axis or an acute angle in the clockwise direction with the $ x $ -axis.
Since $ \tan \theta $ is not defined when $ \theta = {90^\circ} $ , therefore, the slope of a vertical line is not defined. i.e., slope of $ y $ -axis is $ m = \tan {90^\circ} = \infty $ i.e., not defined.
Slope of $ x $ axis $ m = \tan {0^\circ} = 0 $ .
Since the inclination of every line parallel to the $ x $ axis is $ {0^\circ} $ , so its slope $ m = \tan {0^\circ} = 0 $. Therefore, the slope of every horizontal line is $ 0 $ .