Question

plz answer this from chapter equation of line step by step plz olz ​

Answer 1

Question:

The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slope of its sides.

Answer:

Slope of AB = 0

Slope of BC = -√3

Slope of CA = √3

Note:

• The slope of a straight line is given by the tangent of the angle which is made between the straight line and the +ve x-axis and measured in anti-clockwise direction.

• The slope is generally denoted by m , thus;

m = tan@ , where @ is the angle made between the straight line and the +ve x-axis measured in anti-clockwise direction.

• For a straight line, the slopes at every points are always equal.

• Slope of x-axis = tan0° = 0

• Slope of y-axis = tan90° = ∞

• The slope of two or more parallel lines are equal.

• The product of slopes of two mutually perpendicular straight lines is equal to -1.

• The slopes of all the lines parallel to x-axis is equal to 0.

• The slopes of all the straight lines parallel to y-axis is equal to ∞.

• Each interior angles of a equilateral triangle is equal to 60°.

• Each exterior angles of a equilateral triangle is equal to (180°-60°) ie; 120°.

Solution:

• Since AB is parallel to the x-axis, thus the slope of AB will be zero.

ie;

=> m(AB) = 0.

• Since, BC makes 120° with the +ve x-axis measured in anti-clockwise direction, thus the slope of BC will be tan120°.

ie;

=> m(BC) = tan120°

=> m(BC) = tan(180°-60°)

=> m(BC) = -tan60°

=> m(BC) = -√3

• Since, CA makes 60° with the +ve x-axis measured in anti-clockwise direction, thus the slope of CA will be tan60°.

ie;

=> m(CA) = tan60°

=> m(CA) = √3

Hence,

The slopes of AB , BC and CA are 0 , -√3 and √3 respectively.

Answer 2

Question :-------

• The side AB of an equilateral triangle ∆ABC is parallel to X-axis. Find the slopes of all sides.

Concept used :------

• The slope of a line parallel to the x -axis is 0..
• The slope of a straight line is the tangent of its inclination and is denoted by letter 'm' .... if the inclination of a line is θ, its slope m = tan θ.
• All angles of Equaliteral Triangle is 60° .
• Exterior angle of Equaliteral ∆ is 120° .

Solution :--------

1) Side AB is parallel to x- axis , and we know that , slope of x - axis = 0.

also , slope of parallel lines are Equal .

Hence,

→ slope of AB = 0 .

___________________

2) Angle BAC = 60° ( in anti-clockwise direction)

slope of line = tanθ.

→ tan60° = √3 .

Hence,

→ Slope of line CA = √3 ....

_______________________

3) Now, angle made by line BC with positive x axis = 120° .

→ slope of line BC = tan120°

→ tan(120°) = tan(90°+30°)

[ tan(90+θ) = -cotθ ]

→ tan120° = (-cot30°)

→ tan120° = (-√3)

Hence,

→ Slope of line BC = (-√3)

________________________

Extra knowledge :---

→ if given any two points (x1,x2) and (y1,y2) in the line ,

its slope is given by :---

(y2-y1)/(x2-x1)

→ if two lines are perpendicular slope of their product is Equal to (-1) ...