How do you determine the value of 4 so that (6,4), (9,2) has slope 1/3?

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How do you determine the value of 4 so that (6,4), (9,2) has slope 1/3?

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Answer 1 · 2017-12-27T05:31:20

For the line to have a slope of $\frac{1}{3}$ $1/3$ , $4$ $4$ should take the value $1$ $1$

Explanation:

The slope of a straight line is $m = \frac{1}{3}$ $m=1/3$

The line connecting points $(6, 4); (9, 2)$ $(6,4);(9,2)$ cannot have the slope $\frac{1}{3}$ $1/3$

The option open to us is to change the value of $4$ $4$ in the point $(6, 4)$ $(6,4)$

Let us assume instead of $4$ $4$ , we supply value $n$ $n$

Then that point is $(6, n)$ $(6,n)$ .

The line connecting points $(6, n); (9, 2)$ $(6,n); (9,2)$ has a slope of $\frac{1}{3}$ $1/3$

We have to find the value of $n$ $n$

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m=(y_2-y_1)/(x_2-x_1)$
$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1}{3}$ $(y_2-y_1)/(x_2-x_1)=1/3$

$x_{1} = 6$ $x_1=6$
$y_{1} = n$ $y_1=n$
$x_{2} = 9$ $x_2=9$
$y_{2} = 2$ $y_2=2$

$\frac{2 - n}{9 - 6} = \frac{1}{3}$ $(2-n)/(9-6)=1/3$
$\frac{2 - n}{3} = \frac{1}{3}$ $(2-n)/3=1/3$
$2 - n = \frac{1}{3} \times 3 = 1$ $2-n=1/3xx3=1$
$- n = 1 - 2 = - 1$ $-n=1-2=-1$
$n = 1$ $n=1$

For the line to have a slope of $\frac{1}{3}$ $1/3$ , $4$ $4$ should take the value $1$ $1$

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How do you determine the value of 4 so that (6,4), (9,2) has slope 1/3?

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