Question: Find the equation of a straight line parallel to the line y=8x+3 and passing through the point (2,-8).

Solution:

Here, equation of line 1 is (y=8x+3)
Comparing given equation with y =mx+c, we get,
m = 8. So, the slope of line 1 ($m_1$) = m = 8.

We know, if two lines are parallel then their slopes are equal. Let $m_2$ be the slope of the other line then,
$m_1 = m_2$
$or, m_2 = m_1$
$\therefore m_2 = 8$

For line 2, we have coordinate of a single point (2,-8). Let (2,-8) be ($x_1,y_1$).
We know, slope of the line ($m_2$) is -8. Using formula, equation of line 2 is,

$or, y - y_1 = m_2 (x -x_1)$

$or, y -(-8) = 8 (x -2)$

$or, y +8 = 8x -16$

$or, 8x -y -16 -8 = 0$

$or, 8x -y -24 = 0$ is the required equation.


Hence, the required equation of the line passing through (2,-8) and parallel to the line represented by (y=8x +3) is (8x -y -24 =0).