Question

Find the equation of a line parallel to y=x−1 that contains the point (−3,−2). Write the equation in slope-intercept form.

Answer 1

The equation is :

↬ y = x + 1

Solution:

We Know

If two lines are parallel to each other, then their slopes are equal. The slope of y = x - 1 is 1. Hence, the slope of the line that is parallel to that line is 1.

We shouldn't forget about a point on the line : (-3, -2).

I plug that into a point-slope which is :

`\sf{y-y_1=m(x-x_1)}`

Slope is 1 so

`\sf{y-y_1=1(x-x_1)}`

Simplify

`\sf{y-y_1=x-x_1}`

Now I plug in the other numbers.

-3 and -2 are x and y, respectively.

`\sf{y-(-2)=x-(-3)}`

Simplify

`\sf{y+2=x+3}`

We're almost there, the objective is to have an equation in y = mx + b form.

So now I subtract 2 from each side

`\sf{y=x+1}`

Hence, the equation is y = x + 1

Answer 2

y = x + 1

Explanation:

Parallel lines have same slope.

y = x - 1

Compare with the equation of line in slope y-intercept form: y = mx +b

Here, m is the slope and b is the y-intercept.

m =1

Now, the equation is,

y = x + b

The required line passes through (-3 ,-2). Substitute in the above equation and find y-intercept,

-2 = -3 + b

-2 + 3 = b

`\boxed{b= 1}`

Equation of line in slope-intercept form:

`\boxed{\bf y = x + 1}`