Question

find the equation of the straight line passing through the point (-2,3)with slope 1/3​

Answer 1

Given,

The slope of the line = 1/3

A point on the line = (-2,3)

To Find,

The equation of the straight line.

Solution,

The equation of the straight line when the slope of the line and one point are given is

(y-y₁) = m(x-x₁), where m is the slope of the line and x₁,y₁ is the given point.

So, the equation of the line will be

(y-3) = 1/3(x+2)

(y-3)3 = x+2

x+2 = 3y-9

x-3y+11 = 0

Hence, the equation of the line is x-3y+11 = 0.

Answer 2

Answer:

The equation of the straight line passing through the point (-2,3) with slope $\frac{1}{3}$ is  x - 3y + 11 = 0

Step-by-step explanation:

The equation of a straight line with slope 'm' passing through the point $(x_{1},y_{1})$  is given by the formula

$y-y_1 = m(x-x_1)$ --------------(1)

Given,

Slope = m = $\frac{1}{3}$ and the point $(x_{1},y_{1})$ = (-2,3)

Substitute the above values in equation (1) we get,

y- 3 = $\frac{1}{3}$ (x-(-2))

3(y-3) = x+2

3y - 9 = x+2

Rearranging the terms we get,

x - 3y + 11 = 0

Hence the equation of the straight line passing through the point (-2,3) with slope $\frac{1}{3}$ is  x - 3y + 11 = 0