A line has slope -2 and contains P(3,4) and Q(-4, a). How do you find the value of a?

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Answer 1 · 2017-01-09T14:36:46

See full solution process below.

First, because there is a slope and point given we can use the point-slope formula to obtain the equation for this line:

The point-slope formula states: $(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))$

Where $color(blue)(m)$ is the slope and $color(red)(((x_1, y_1)))$ is a point the line passes through.

Substituting the slope and point from the problem results in:

$(y - color(red)(4)) = color(blue)(-2)(x - color(red)(3))$

We can now convert this to more familiar slope-intercept form by solving the equation for $y$ :

$y - color(red)(4) = color(blue)(-2)x + (color(blue)(2) xx color(red)(3))$

$y - color(red)(4) = color(blue)(-2)x + 6$

$y - color(red)(4) + 4 = color(blue)(-2)x + 6 + 4$

$y - 0 = color(blue)(-2)x + 10$

$y = -2x + 10$

To find $a$ we can substitute $-4$ for $x$ and $a$ for $y$ and calculate $a$ :

$a = (-2 xx -4) + 10$

$a = 8 + 10$

$a = 18$