Question

What is the equation of the line that passes through the points (-2, -3) and (4, -5)

Answer 1

Answer:

x + 3y + 11 = 0  is the  required equation.

Step-by-step explanation:

Here the given points are A( -2,-3) and B(4,-5)

Now, slope of the equation = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 -(-3))}{4-(-2)} = \frac{-2}{6} =-\frac{1}{3}$

Hence, the slope of the equation is m = (-1/3)

Now, by POINT SLOPE FORMULA:

The equation of a line with slope m and point (x0, y0) is given as

(y - y0) = m (x -x0)

So, here the equation of line with (4, -5) is

$y - (-5) = -\frac{1}{3} (x - 4) \implies 3( y +5) = -x + 4$

or, 3y + 15 + x-4 = 0

or, x + 3y + 11 = 0

⇒ x + 3y + 11 = 0 is the  required equation.