How do you find the equation of a line passing through (4, 1) with the slope m = -1/2?

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Answer 1 · 2018-04-29T14:03:22

See below:

Point $(x_{1}, y_{1}) = (4, 1)$ $(x_1, y_1)=(4, 1)$

Slope ( $m$ $m$ ) $= - \frac{1}{2}$ $=-1/2$

Let the equation of the line be

$y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$

$y - 1 = - \frac{1}{2} (x - 4)$ $y-1=-1/2(x-4)$

$y - 1 = - \frac{1}{2} x + 2$ $y-1=-1/2x+2$

$\frac{x}{2} + y = 3$ $x/2+y=3$

which is the required equation of the line.

Answer 2 · 2018-04-29T14:08:55

$y = - \frac{1}{2} x + 3$ $y=-1/2x+3$

Slope-intercept form of a line: $y = m x + b$ $y=mx+b$ , with $m$ $m$ as the slope and $b$ $b$ as the y-intercept

We know that the slope is $- \frac{1}{2}$ $-1/2$ , so we can put that in for $m$ $m$ :

$y = - \frac{1}{2} x + b \to$ $y=-1/2x+b rarr$ To find the y-intercept, plug in the point $(4, 1)$ $(4, 1)$ and solve

$1 = - \frac{1}{2} \cdot 4 + b$ $1=-1/2*4+b$

$1 = - 2 + b$ $1=-2+b$

$b = 3$ $b=3$

The equation in slope-intercept form is $y = - \frac{1}{2} x + 3$ $y=-1/2x+3$