Question

Graph a line with a slope of 1/4 that contains the point (6, 3).

Answer 1

Answer:

The graph in the attached figure

Step-by-step explanation:

we know that

The equation of a line in point slope form is

$y-y1=m(x-x1)$

we have

$(x_1,y_1)=(6,3)$

$m=\frac{1}{4}$

substitute

$y-3=\frac{1}{4}(x-6)$

The easiest way to graph a line is to calculate the intercepts

The x-intercept is the value of x when the value of y is equal to zero

For y=0

$0-3=\frac{1}{4}(x-6)$

$-12=(x-6)$

$x=-12+6=-6$

The x-intercept is the point (-6,0)

The y-intercept is the value of y when the value of x is equal to zero

For x=0

$y-3=\frac{1}{4}(0-6)$

$y-3=-\frac{3}{2}$

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

$y=1.5$

The y-intercept is the point (0,1.5)

Plot the intercepts and join the points to graph the line

(-6,0) and (0,1.5)

The graph in the attached figure