Question

Find the linear approximation of the data that passes through the points 5,3 and 20,6

Answer 1

namely, the equation of that line with those points

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 5}}\quad ,&{{ 3}})\quad % (c,d) &({{ 20}}\quad ,&{{ 6}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{6-3}{20-5}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \begin{array}{llll} \textit{plug in the values for } \begin{cases} y_1=3\\ x_1=5\\ m=\boxed{?} \end{cases}\\ \textit{and solve for "y"} \end{array}\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}$

Answer 2

Answer:

$y=\dfrac{1}{5}(x)+2$

Step-by-step explanation:

The linear approximation of the data that passes through the points (5,3) and (20,6).

We need to find the equation of line.

If a line passes through two points, then the equation of line is

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

The line passes through (5,3) and (20,6). So, the equation of line is

$y-3=\dfrac{6-3}{20-5}(x-5)$

$y-3=\dfrac{3}{15}(x-5)$

$y-3=\dfrac{1}{5}(x-5)$

$y-3=\dfrac{1}{5}(x)-\dfrac{1}{5}(5)$

Add 3 on both sides.

$y=\dfrac{1}{5}(x)-1+3$

$y=\dfrac{1}{5}(x)+2$

Therefore, the equation of linear approximation is $y=\dfrac{1}{5}(x)+2$.