Point-slope equation

Consider any straight line on the two-dimensional coordinate plane. Let its slope be m and (x₁, y₁) be a fixed point lying on the line. There can be infinite lines with slope m but when we specify that the line passes through the fixed point (x₁, y₁) then we get a unique straight line. Let (x, y) be any arbitrary point on the line. We know the slope of the line is m, i.e. slope between any two points lying on the line will always be m. Using this fact we can derive the equation of any straight line with a given slope and passing through a fixed point. 

We know,

(change in y) / (change in x) = slope,

Since (x, y) and (x₁, y₁) are two points on the line and m is the slope of the straight line, slope between (x, y) and (x₁, y₁) will be equal to m.

Now, slope between (x, y) and (x₁, y₁) = (y – y₁) / (x – x₁)

∴ (y − y₁) / (x − x₁) = m               …(1)

Multiplying both sides of equation (1) by (x − x₁) we get,

⟹ (y − y₁) = m (x − x₁)               …(2)

which is the equation for the considered straight line or the point-slope form.

The reason this equation is called the point-slope form is very obvious. The equation only contains the slope and the fixed point on the straight line as constants, so it is called point-slope form.

Sample Problems on Point-slope Equation

Problem 1: Find the equation of the straight line with slope 3 and 
passing through (-1, 5).

Solution: 

Here slope of the line is 3 or m = 3 and the line passes through (-1, 5),

hence a fixed point on the line is

(-1, 5) or (x₁, y₁) = (-1, 5)

Let (x, y) be any point on the line

∴ Slope between (x, y) and (x₁, y₁) = (y – y₁) / (x – x₁)

Since (x₁, y₁) and (x, y) are two points on the line,

the slope between them will be the slope of the line (m)

∴ (y – y₁) / (x – x₁) = m

Multiplying both sides by (x – x₁) we get,

⇒ y – y₁ = m(x – x₁)

Putting m = 3 and (x₁, y₁) = (-1, 5) we get,

⇒ y – 5 = 3(x – (-1))

⇒ y – 5 = 3(x + 1)

⇒ y – 5 = 3x + 3

⇒ y – 3x – 5 – 3 = 0

⇒ y – 3x – 8 = 0, which is the required equation 

Problem 2: Find the equation of the straight line with slope -2 and passing through (7, -4).

Solution: 

Here slope of the line is -2 or m = -2 and the line passes through (7, -4), hence a fixed point on the

line is (7, -4) or (x₁, y₁) = (7, -4)

Let (x, y) be any point on the line

Since (x₁, y₁) and (x, y) are two points on the line, the

slope between them will be the slope of the line

∴ (y – y₁) / (x – x₁) = m

Putting m = 3 and (x₁, y₁) = (7, -4) we get,

⇒ (y – (-4)) / (x – 7) = -2

Multiplying both sides by (x – 7) we get,

⇒ y + 4 = (-2)(x – 7)

⇒ y + 4 = -2x + 14

⇒ y + 2x + 4 – 14 = 0

⇒ y + 2x – 10 = 0, which is the required equation 

Problem 3: Find the equation of the straight line with slope 1/4 and passing through (2, 3).

Solution: 

Here slope of the line is 1/4 or m = 1/4 and the line passes through (2, 3), 

hence, a fixed point on the line is (2, 3) or (x₁, y₁) = (2, 3)

We know, point-slope form is:  (y – y₁) = m(x – x₁)

Putting the values of m, x₁ and y₁ in the equation we get,

⇒ (y – 3) = (1/4)(x – 2)

Multiplying both sides by 4 we get,

⇒ 4(y – 3) = 1(x – 2)

⇒ 4y – 12 = x – 2

⇒ 4y – x – 12 + 2 = 0

⇒ 4y – x – 10 = 0, which is the required equation

There are several forms to represent the equation of a straight line on the two-dimensional coordinate plane. Three major of them are point-slope form, slope-intercept form, and general or standard form. The point-slope form includes the slope of the straight line and a point on the line as the name suggests. There can be infinite lines with a given slope, but when we specify that the line passes through a given point then we get a unique straight line. Thus, only a point on the line and its slope is required to represent a straight line in the point-slope form.