Perpendicular lines

So that two lines are perpendicular, the slopes must be negative reciprocals. We compute the slope of the line $x+23y=138$ by transforming the equation into slope-intercept form. We have

$x+23y=138$

$23y=-x+138$

$y=\dfrac{-1}{23}x+6$

The slope is $\dfrac{-1}{23}$ . That means that the slope of the other line must be $23$ . We have

$y-ax=2$

$y=ax+2$

$\boxed{a=23}$

The slopes of the two lines are:

$m_1 = a$ and $m_2 = \frac{-1}{23}$

because the lines are perpendicular, we know that $m_1 \times m_2 = -1$

therefore, $a = 23$

solve for the slope of the line in the equation $x + 3y = 138$ by transforming it into the slope-intercept form, $y=$ $\frac{-1}{23}x+6$ . We can see that slope is $\frac{-1}{23}$ . We know that slopes of two perpendicular lines are $negative$ $reciprocals$ . So the slope of the other line must be $23$ . Transforming the first equation of the line to the slope-intercept form, we have $y=ax+2$ . We can see that $a=23$ .