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In geometry, to describe straight objects with negligible width and depth, the notion of line or straight line was introduced by ancient mathematicians. Lines are an idealization of certain objects that are often represented or referred to with a single letter in terms of two points.

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A point typically refers to a part called space in a certain set. More precisely, a point is a primitive notion in Euclidean geometry on which the geometry is constructed, meaning that a point can not be described in terms of objects previously defined.

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The slope or gradient of a line in mathematics is a number that defines both the direction and the steepness of the line Slope is often referred to by the letter m; there is no straightforward answer to the question of why the letter m is used for slope, but in O'Brien who wrote the straight-line equation as

y = mx + b, its earliest use in English appears.

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Select x and solve the equation for y, or the equation for y.

Select y and solve for x.

The ratio of the "vertical change" to the "horizontal change" between (any) two distinct points on a line is determined by finding the slope. The ratio is often represented as a quotient ("rise over run"), giving the same number on the same line for any two distinct points. A line that decreases has a "rise" that is negative.

Equations of horizontal and vertical lines

Point-slope form equation of a line

Two-point form equation of a line

Slope-intercept form equation of a line

Intercept form

Normal form

We can learn how to find the equation of a line given one point and slope that is inclined at a given angle to the positive direction of the x-axis in the anticlockwise sense and passes through a given point by the equation of a line in point-slope form.Â

Let the MN line form an angle with the positive x-axis direction in the anticlockwise sense and pass through the Q point (x1, y1). We need to find an equation for the MN line. Let any point on the line MN be P(x, y). But Q (x1, y1) is a point on the same side as well.Â

Therefore, the slope of the MN line = (y-y1) / (x-x1)

Again, the MN line produces an angle with the positive direction of the x-axis; thus, the line slope = tan = m (say).

Therefore, (y-y1) / (x-x1) = mÂ

â‡’ y-y1 = m (x-x1)

The equation y-y1 = m (x-x1) above is fulfilled by the coordinates of any point P on the line MN.Â

Therefore, y-y1 = m (x-x1) represents the AB straight line equation.

Write the line's point-slope form with the value of slope being 3 that goes through the point (2,5).

Solution: The slope is given as m=3, and there are coordinates of x1 = 2 andÂ y1 = 5 for the point (2,5). To get the final answer, plug the known values into the slope-intercept form now.

y-y1 = m (x-x1)

Y-5 = 3(x-2)

A straight line passes through the point (2, -3) and the positive orientation of the x-axis gives an angle of 135 Â°. Find the Straight Line Equation.

Solution: An angle of 135 Â° with the positive direction of the axis of x renders the appropriate line.Â

The slope of the appropriate line, therefore,Â Â

m= tan 135 Â° = tan (90 Â° + 45 Â°) = - cot 45 Â° = -1.Â

Again, the line that is needed passes through the point (2, -3).Â

We know that a straight line formula passes through a given point (x1, y1) and that the slope of 'm' is y-y1 = m (x-x1).

Therefore, the necessary straight line formula is y - (-3) = -1 (x -2)Â

y + 3 = -x + 2

x + y + 1 = 0

FAQ (Frequently Asked Questions)

1. Find the equation of a line with point and slope.

To find an equation with slope and point, a line's equation is usually written as y=mx+b, where the slope is m and the y-intercept is b. The line equation in the point-slope form will teach us how to find a straight line equation that is inclined at a given angle to the positive direction of the x-axis in an anticlockwise direction and passes through a given point.Â

The equation of a line with point and slope is,Â

The slope of point (m) = (y-y_{1}) / (x-x_{1})

2. Find a line equation passing through a point (âˆ’4,3) with a slope Â½

Given, slope (m) = Â½Â

Value of x_{1}= -4

Value of y_{1}=3

Therefore, the equation of the line becomes,

=> Â½ = (y-3) / (x-(-4))

=> Â½ = (y-3) / (x+4)

=> x+4 = 2(y-3)

Answer => x+4 = 2y-6