Question

What is the slope of the line that passes through the points (-4, -3) and (14, 0)?

Answer 1

Answer:

m is equal to y2-y1/x2-x1

Step-by-step explanation:

1/6 is the answer

Answer 2

We know that :-

$\rm\large Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: (x_2 , y_2) :$

$\boxed{\rm\large y - y_1 = x-x_1\bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg )}$

$\rm \large\rightarrow \: here \: \bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg ) is \: slope \: of \: line.$

Now , we have to find out the equation of the line that passes through the points (-4, -3) and (14, 0).

$\tt \large\implies y - 3 = (x + 4)\bigg( \dfrac{0-3}{ 14 + 4} \bigg ) \\ \\ \\ \tt \large\implies y - 3 = (x + 4)\bigg( \dfrac{-3}{ 18} \bigg )\\ \\ \\ \tt\large \implies y - 3 = (x + 4)\bigg( \dfrac{-1}{ 6} \bigg )\\ \\ \\ \tt\large \implies6 (y - 3) = (x + 4) ( - 1)\\ \\ \\ \tt\large \implies6y - 18 = - x - 4\\ \\ \\ \tt \large\implies6y - 18 + x + 4 = 0\\ \\ \\ \large\tt \implies6y + x - 14 = 0$

The equation of the line that passes through the points (-4, -3) and (14, 0). :-

6y+x-14=0