Question

What is an equation of the line that passes through the point (1,-7) and is parallel to the line 3x+y=3?

Answer 1

Answer

the required equation is  3x+y+4=0

Step-by-step explanation:

given point is (1,-7)  // x₁=1 ,y₁= -7

and the given line is 3x+y=3

in the given line a=3 ,b=1

slope of given line=m=-a/b

• m=-3/1
• m=-3 is the slope

by using point slope form

• (y-y₁)=m(x-x₁)

by putting the value

• [y-(-7)]=-3(x-1)
• y+7=-3x+3
• 3x+y+7-3=0  // by changing the position

• 3x+y+4=0 is the required equation

• and the slope of this line is=m=-3
• so prove these lines are parallel

Answer 2

Answer:

Required equation of line is y+3x = -4

Step-by-step explanation:

Given point is (1,-7) and line is 3x+y=3.

We want to find an equation of the line that passes through the point (1,-7) and is parallel to the line 3x+y=3.

We know,If a line passes through the point (a,b) and is parallel to the line y = mx + c then equation of the line will be (y-b) = m (x-a)

Here,

$3x + y = 3 \\ y = - 3x + 3....(1)$

We are comparing equation (1) with equation y = mx+c and get m = -3 where m is slope of line.

According to rule ,

Equation of line which passes through (1,-7) and 3x+y=3 is

$(y - ( - 7)) = - 3(x - 1) \\ y + 7 = - 3(x - 1) \\ y + 7 = - 3x + 3 \\ y +3x = 3 - 7 \\ y + 3x = - 4$

So, required equation of line is y+3x = -4

This is a problem of line related.

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