Find the equation of the line that passes through (–5, 2) and the intersection of the lines x +3y =0 and 4x −4y −13=0.

Question

Find the equation of the line that passes through (–5, 2) and the intersection of the lines x +3y =0 and 4x −4y −13=0.

Answer 1

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Answer 2

We need to find the equation of a line, but we are given only one explicit point, which is (–5,2). If we had another point, that would define the line in question. That second point is the intersection point of the two lines specified by their equations.

That intersection point is given by the x and y that satisfy each of the equations. Thus, we need to solve the two equations, considering them as a system of equations.

x + 3y = 0

4x — 4y — 13 = 0

I will use the substitution method, solving the first equation for x and then substitution the expression of x into the second equation where I see an x. Then, I will solve the second equation for y.

x + 3y = 0

—3y —3y

x = —3y

4x — 4y — 13 = 0

4 (—3y) —4y — 13 = 0

—12y —4y — 13 = 0

—16y — 13 = 0

+16y +16y

—13 = 16y

____ ____

16 16

—13

y = ________

16

Now we solve for x, by substituting the fraction above for y in the first equation of a line, since it is easier.

x + 3y = 0

x + 3 (—13/16) = 0

x + (–39/16) = 0

x = (39/16)

Now we know that the line for which you need an equation contains the two points (—5,2) and

(39/16 , —13/16).

Take a deep breath. This is a long problem, but we are almost there. We are going to use the slope-intercept template of an equation of a line: y = mx + b. To get the equation, we need to know m and b. The slope is m. The y-intercept is b.

Since we have two points on the line, we can calculate the slope, which is rise over run or, in other words, the change (difference) in y (vertical = rise) over the change (difference) in x (horizonta l= run).

2 — (—13/16) 2 + (13/16) (32/16) + (13/16) 45/16

m (slope) = _______________ = ________________ = _________________ = ____________

—5 — (39/16) —5 —(39/16) (—80/16) — (39/16) —119/16

45 —119 45 16 45

____ ÷ _______ = ______ x ______ = _________

16 16 16 –119 –119

Now we can fill in the slope (m) into our template equation to get y = (45/–119)x + b. We only need to find b. The way we do this is to substitute in for x and y the coordinates a point that we know is on that line and must satisfy the equation or, in other words, make it true. Let's use the "easier" point, (–5,2).

y = (45/–119)x + b

2 = (45/–119)(—5) + b

2 = (–225/–119) + b

2 = (225/119) + b

2 — (225/119) = b

(238/119) — (225/119) = b

13/119 = b

Tada! We have the makings of an equation, even if it's not pretty, by substituting this value of b into our template, y = (45/–119)x + b

y = (45/–119)x + 13/119

Most algebra problems do not use such "ugly" fractions, so I wonder whether you or I made a mistake. However, the important thing is that YOU UNDERSTAND and LEARN the METHOD!!!! Good luck, Sada!!!

Answer 3

Lets first find the intersection of x +3y =0 and 4x −4y −13=0

Multiplying both sides of the first equation by -4 we get:

-4x - 12y = 0

Adding 13 to both sides of the second equation we get:

4x - 4y = 13

Adding the two equations together we get:

-4x - 12y = 0

4x - 4y = 13

----------------

-16y = 13

or y = -13/16

Using the first equation x +3y =0 we get:

x = -3y

x = -3(-13/16)

x = 39/16

So the intersection is at the point (39/16, -13/16)

Then the slope of the line we are looking for is found using the slope equation:

m = (y₂ - y₁)/(x₂ - x₁)

m = (2 - -13/16)/(-5 - 39/16)

m = (32/16 + 13/16)/(-80/16 - 39/16)

m = (45/16)/(-119/16)

m = -45/119

Using the point-slope form of a line (y - y1) = m(x - x1) we get:

y - 2 = -45/119(x - -5)

y - 2 = -45/119(x + 5)

You can switch this to slope-intercept form if you want but the instructions don't specify so I'd leave it like this.