Question

What is the equation of the line parallel to 3x-y-7=0 that passes through point (-5, -3)

Answer 1

y = 3x+12 is the line parallel to 3x-y-7=0 that passes through point (-5, -3).

Explanation:

We have given an equation:

3x-y-7=0

above equation in standard form:

y = 3x-7 (eq I)

We have to the line parallel to to 3x-y-7=0 that passes through point (-5, -3).

The slope of (eq I) is m = 3.

The parallel lines have same slopes.

So, the line parallel to given lines has slope m = 3.

The general form of slope-point form of equation is:

y = m(x) +c (eg II)

putting y= -3 , x= -5 and m = 3 in above equation we get,

-3 = 3(-5)+c

-3=-15+c

c = -3+15

c =12

Putting the value of c and m=3 in ( eq II) weget,

y = 3x+12 is the line parallel to 3x-y-7=0 that passes through point (-5, -3).

Answer 2

The equation of the line in slope-intercept form is y =3x+12

Step-by-step explanation:

The first step is to rewrite the equation of the line in the slope-intercept form in order to identify the slope of the line. The re-written equation is;

y = 3x-7.

This implies that the slope of the line is 3. Since the two lines are parallel they they will have equal slope of 3. The slope-intercept form of the equation of this line will be;

y =3x+c.

Since the line passes through (-5, -3), substitute x with -5 and y with -3 to solve for c;

-3 =3(-5)+c.

We find c =12 and the equation of the line becomes;

y =3x+12.