Question

The inclination of the line joining the points $A(-2,3)$ and $B(-1,4)$ with the positive direction of the $X$-axis is

• A. $45^{\circ}$
• B. $60^{\circ}$
• C. $90^{\circ}$
• D. $120^{\circ}$

Hint:

Whenever the change in $x$ and the change in $y$ are equal, the slope becomes \[ m=\frac{\Delta y}{\Delta x}=1. \] Since \[ \tan 45^{\circ}=1, \] the inclination is immediately \[ 45^{\circ}. \] For the points $(-2,3)$ and $(-1,4)$, \[ \Delta x=1,\qquad \Delta y=1, \] so the answer can be obtained within seconds.

Answer 1

The Correct Option is A

Concept: The inclination of a straight line is the angle $\theta$ made by the line with the positive direction of the $X$-axis, measured in the anticlockwise direction. The slope $m$ of a line joining two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by \[ m=\frac{y_2-y_1}{x_2-x_1}. \] The slope and inclination are related by \[ m=\tan\theta. \] Therefore, once the slope is known, the inclination can be obtained using the corresponding trigonometric value.

Step 1: Identify the coordinates of the given points. The coordinates of the two points are \[ A(-2,3) \] and \[ B(-1,4). \] Comparing with the standard notation, \[ x_1=-2,\qquad y_1=3, \] \[ x_2=-1,\qquad y_2=4. \]

Step 2: Calculate the slope of the line joining the two points. Using the slope formula, \[ m=\frac{y_2-y_1}{x_2-x_1}. \] Substituting the given values, \[ m=\frac{4-3}{-1-(-2)}. \] Simplifying the numerator, \[ 4-3=1. \] Simplifying the denominator, \[ -1+2=1. \] Therefore, \[ m=\frac{1}{1}=1. \] Hence, the slope of the line is \[ m=1. \]

Step 3: Relate the slope to the angle of inclination. We know that \[ m=\tan\theta. \] Since $m=1$, \[ \tan\theta=1. \] From standard trigonometric values, \[ \tan 45^{\circ}=1. \] Therefore, \[ \theta=45^{\circ}. \]

Step 4: Verify the result geometrically. Observe that moving from point $A(-2,3)$ to point $B(-1,4)$, \[ \Delta x=(-1)-(-2)=1, \] and \[ \Delta y=4-3=1. \] Since the horizontal change and vertical change are equal, the line rises one unit for every one unit moved to the right. Such a line always makes an angle of \[ 45^{\circ} \] with the positive $X$-axis. Hence, the inclination of the line is \[ \boxed{45^{\circ}}. \] Therefore, the correct answer is \[ \boxed{(A)\ 45^{\circ}}. \]