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WRITE an equation of the line that passes through the given point and is PARALLEL to the given line.

(10, -12), 3x + 4y = 1

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Answer:


y=-(3)/(4)x-(9)/(2)

Explanation:

Hi there!

What we need to know:

  • Linear equations are typically organized in slope-intercept form:
    y=mx+b where m is the slope and b is the y-intercept (the value of y when x is 0)
  • Parallel lines always have the same slope

1) Determine the slope (m)


3x + 4y = 1

Rearrange the given equation into slope-intercept form (this will help us determine m)

Subtract 3x from both sides to isolate 4y


3x + 4y-3x = -3x+1\\4y= -3x+1

Divide both sides by 4 to isolate y


y= -(3)/(4) x+(1)/(4)

Now, we can identify that
-(3)/(4) is the slope of the line. Parallel lines have the same slope, so this would also be the slope of the line we're solving for. Plug this into
y=mx+b:


y=-(3)/(4)x+b

2) Determine the y-intercept (b)


y=-(3)/(4)x+b

Plug in the given point (10,-12) and solve for b


-12=-(3)/(4)(10)+b\\-12=-(3)/(2)(5)+b\\-12=-(15)/(2)+b

Add
(15)/(2) to both sides to isolate b


-12+(15)/(2)=-(15)/(2)+b+(15)/(2)\\-12+(15)/(2)=b\\-(9)/(2) =b

Therefore, the y-intercept is
-(9)/(2). Plug this back into
y=-(3)/(4)x+b:


y=-(3)/(4)x-(9)/(2)

I hope this helps!

User Shahjapan
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