The line $l$ has equation $4x+2y+7=0$. Find the gradient of line $l$. Answer ... Find the coordinates of the point where $l$ cuts the y-axis. Answer ... Another line $k$ is parallel to $y = \frac{1}{2}x + 5$ and it passes through the point $(8, 3).$
Find the equation of line $k$. Answer ...

1. Rewrite the equation of the line in gradient-intercept form and identify the gradient.

$4x+2y+7=0$

$2y=-4x-7$

$y=-2x-\dfrac{7}{2}$

Comparing with $y=mx+c$, the gradient is $m=-2$.

1. Use the fact that the y-intercept has x-coordinate 0 and solve for y.

At the y-axis, $x=0$.

Substitute into $4x+2y+7=0$:
$4(0)+2y+7=0$
$2y=-7$
$y=-\frac{7}{2}$

So the point is $(0,-\frac{7}{2})$.

1. Identify the gradient of line $k$ from the parallel line.

Since $k$ is parallel to $y=\frac{1}{2}x+5$, it has the same gradient.

So, for line $k$, $m=\frac{1}{2}$.

2. Substitute the point $(8,3)$ into $y=mx+c$ to find $c$, then write the equation.

Line $k$ has equation

$y=\frac{1}{2}x+c.$

Substitute $(8,3)$:

$3=\frac{1}{2}(8)+c$

$3=4+c$

$c=-1$

Therefore,

$y=\frac{1}{2}x-1.$