How do you find the slope-intercept form of the equation of the line passing through the points $(\frac{3}{4}, \frac{3}{2}) and (- \frac{4}{3}, \frac{7}{4})$ $(3/4, 3/2) and (-4/3, 7/4)$ ?

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Answer 1 · 2017-02-16T14:48:09

$y = - \frac{3}{25} x + 1 \frac{59}{100}$ $y = -3/25x +1 59/100$

$m = - \frac{3}{25} and c = 1.59$ $m = -3/25 and c = 1.59$

Lets find the slope first: $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m = (y_2-y_1)/(x_2-x_1)$

$\to m = \frac{\frac{7}{4} - \frac{3}{2}}{- \frac{4}{3} - \frac{3}{4}}$ $rarr" " m = (7/4-3/2)/(-4/3-3/4)$

$m = \frac{\frac{7 - 6}{4}}{\frac{- 16 - 9}{12}} = \frac{\frac{1}{4}}{\frac{- 25}{12}}$ $m= ((7-6)/4)/((-16-9)/12) = (1/4)/((-25)/12)$

$m = (1xxcancel12^3)/(cancel4xx-25) = -3/25$

Substitute $m and$ one point into $y-y_1 =m(x-x_1)$

$m = -3/25 " and the point, " (3/4, 3/2)$

$y-3/2 = -3/25(x-3/4)$

$y = -3/25x+9/100+3/2$

$y = -3/25x +1 59/100$

How do you find the slope-intercept form of the equation of the line passing through the points (3/4, 3/2) and (-4/3, 7/4)(34,32)and(−43,74)(3/4, 3/2) and (-4/3, 7/4)?