Question

Line JK passes through points J(-3, 11) and K(1, -3). What is the equation of line JK in standard form? The answers are
7x + 2y = -1
7x + 2y = 1
14x + 4y = -1
14x + 4y = 1

Answer 1

To find the standard form of line JK, we have to find the slope, so we can put these information to y-y1 = m (x-x1)

To find the slope, we can use the equation of m = y2 - y1 / x2 - x1

So the slow is -3 - 11 / 1 - (-3) = -14 / 4 = -7/2

To put this on that equation,

y - 11 = -7/2 ( x + 3)

y - 11 = -7/2 x - 21/2

y + 7/2 x = 11 - 21/2

To multiply this result and 2,

2y + 7x = 22 - 21 = 1

So the answer is 7x + 2y = 1

Answer 2

Linear equation is the equation of a straight line. A linear equation in standard form is: $ax + by = c$

The equation of JK in standard form is: $7x + 2y = 1$

Given that:

$(x_1,y_1) = (-3,11)$ --- point J

$(x_2,y_2) = (1,-3)$ --- point K

First, we calculate the slope (m) of JK using:

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

So, we have:

$m = \frac{-3-11}{1--3}$

$m = \frac{-14}{4}$

$m = \frac{-7}{2}$

The equation is then calculated using:

$y = m(x - x_1) + y_1$

This gives:

$y = \frac{-7}{2}(x - -3) + 11$

$y = \frac{-7}{2}(x +3) + 11$

Multiply through by 2

$2y = -7(x +3) + 22$

Open bracket

$2y = -7x -21 + 22$

$2y = -7x +1$

Rewrite in standard form as

$7x + 2y = 1$

Hence, the equation of JK in standard form is:

$7x + 2y = 1$

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