How do you write an equation of a line going through (-1,2), (3,-4)?

1 Answer
smendyka
Dec 23, 2016

y - 2 = -3/2(x + 1)y−2=−32(x+1)

or

y = -3/2x + 1/2y=−32x+12

Explanation:

To find a linear equation for the line going through these two points we can use the point-slope formula.

However, first we need to determine the slope of the line.

The slope can be found by using the formula: color(red)(m = (y_2 - y_1)/(x_2 - x_1)m=y2−y1x2−x1
Where mm is the slope and (color(red)(x_1, y_1)x1,y1) and (color(red)(x_2, y_2)x2,y2) are the two points.

Substituting the two points given in the problem we can solve for mm as:

m = (-4 - 2)/(3 - (-1))m=−4−23−(−1)

m = -6/4 = (2/2)(-3/2) = 1(-3/2)m=−64=(22)(−32)=1(−32)

m = -3/2m=−32

Now that we have the slope we can use the slope-point formula to write the equation for the line.

The point-slope formula states: color(red)((y - y_1) = m(x - x_1))(y−y1)=m(x−x1)
Where color(red)(m)m is the slope and (color(red)((x_1, y_1))) is a point the line passes through.

Substituting the slope of -3/2 and using the point (-1, 2) we can get the equation of the line as:

y - 2 = -3/2(x - (-1))

y - 2 = -3/2(x + 1)

If we want this in the slope-intercept for we can solve for y as follows:

y - 2 = -3/2x - (3/2 * 1)

y - 2 = -3/2x - 3/2

y - 2 + 2 = -3/2x - 3/2 + 2

y - 0 = -3/2x - 3/2 + 2

y = -3/2x - 3/2 + 2

y = -3/2x - 3/2 + 2(2/2)

y = -3/2x - 3/2 + 4/2

y = -3/2x + 1/2

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