Question

The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). What is the standard form of the equation for this line?
x – 4y = 8
x – 4y = 2
4x – y = 8
4x – y = 2

Answer 1

(-4,-3)(12,1)
slope(m) = (1 - (-3) / (12 - (-4)
slope(m) = (1 + 3) / (12 + 4)
slope(m) = 4/16 = 1/4

y - y1 = m(x - x1)...using (12,1)
y - 1 = 1/4(x - 12)
y - 1 = 1/4x - 3
y = 1/4x - 3 + 1
y = 1/4x - 2
-1/4x + y = -2...multiply by -4
x - 4y = 8 <=== here it is

Answer 2

Answer:

Option  A is correct.

x – 4y = 8

Step-by-step explanation:

Using point slope form:

The equation of line is given by:

$y-y_1 = m(x-x_1)$                  ....[1]

where, m is the slope.

As per the statement:

The point passes through the line are:

(-4, -3) and (12, 1)

Using Slope(m) formula:

$m = \frac{y_2-y_1}{x_2-x_1}$

Substitute the given values we have;

$m = \frac{1+3}{12+4}=\frac{4}{16}=\frac{1}{4}$

Substitute the value of m and (12, 1) in [1] we have;

$y-1 = \frac{1}{4}(x-12)$

Multiply both sides by 4 we have;

$4y-4 = x-12$

Add 4 to both sides we have;

$4y= x-8$

⇒$-x+4y = -8$

Multiply both sides by -1 we have;

⇒$x-4y =8$

Therefore, the standard form of the equation for this line is, x – 4y = 8