Question

Determine the slope-intercept form of a linear equation, given the listed attributes: a) Slope = -2 and y-intercept = (0,10) b) Slope = -3 and (4, -2) lies on line c) Slope = 0 and (2,4) lies on line d) (3, -2) and (-12,1) lies on line e) (20, 240) and (15,450) lies on line

Answer 1

a) If slope is $-2$ and y-intercept is $(0,10)$ then the slope-intercept form of a linear equation is $y=-2x+10.$

b) If slope $-3$ and $(4, -2)$ lies on line then the slope-intercept form of a linear equation is $y=-3x+b,$ where $-2=-3\cdot 4+b,$ that is $b=10.$ Therefore, $y=-3x+10.$

c) If slope is and $(2,4)$ lies on line then the slope-intercept form of a linear equation is $y=4.$

d) The points (3, -2) and (-12,1) lies on line given by the following equation:

$\frac{x-3}{-12-3}=\frac{y-(-2)}{1-(-2)},$ which is equivalent to the following equations

$\frac{x-3}{-15}=\frac{y+2}{3}$

$y+2=\frac{x-3}{-5}$

$y=-2-\frac{x}{5}+\frac{3}{5}$

Therefore, the slope-intercept form of a linear equation is

$y=-\frac{1}{5}x-\frac{7}{5}.$

e) The points $(20, 240)$ and $(15,450)$ lies on the line given by the following equation:

$\frac{x-15}{20-15}=\frac{y-450}{240-450},$ which is equivalent to the following equations

$\frac{x-15}{5}=\frac{y-450}{-210}$

$y-450=-42(x-15)$

Therefore, the slope-intercept form of a linear equation is

$y=-42x+1080$