Question

Find the equation of the line which passes through (−2, 3) and the point of intersection of the lines x + 2y=0 and 2x − y − 12=0.

Answer 1

Answer:

First, let's find where the lines:

x + 2y = 0

2x - y - 12 = 0

intersect.

First, let's isolate the variable y in one side of the equalities.

y = (-x/2)

y = 2x - 12

Now we have:

-(x/2) = 2x - 12

We can solve this for x.

12 = 2*x + x/2 = 2.5*x

12/2.5 = x = 4.8

Then, replacing that value in any of the lines, we can find the value of y.

y = 2*8 - 12 = 2*4.8 - 12 = -2.4

Then those lines intersect in the point (4.8, -2.4)

Now we want to find the line that passes through the points (-2, 3) and (4.8, -2.4)

A linear relationship can be written as:

y = a*x + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

Then, the slope for our line is:

a = (3 - (-2.4))/(-2 - 4.8) = -0.79

And the

y-intercept will be:

y = -0.79*x + b

when x  = -2, y = 3.

3 = -2*-0.79 + b

3 - 2*0.79 = b = 1.42

The line is:

y = -0.79*x + 1.42