equation of line perp. to 3x + 2y = 12, thru (0, 2)
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3x+2y = 0
2y = -3x
y = (-3/2)x
Now the line perpendicular to y = (-3/2)x will have a slope which is a multiplicative inverse with sign change from the original line's slope.
so y = (2/3)x is perpendicular to y = (-3/2)x
Now your slope is (2/3) so we can use the point-slope form for the equation of a line.
y-y1=m(x-x1)
y-2=(2/3)(x-0)
y-2=(2/3)x
y=(2/3)x + 2
which is perpendicular to y = (-3/2)x and passes through (0,2)
2y = -3x
y = (-3/2)x
Now the line perpendicular to y = (-3/2)x will have a slope which is a multiplicative inverse with sign change from the original line's slope.
so y = (2/3)x is perpendicular to y = (-3/2)x
Now your slope is (2/3) so we can use the point-slope form for the equation of a line.
y-y1=m(x-x1)
y-2=(2/3)(x-0)
y-2=(2/3)x
y=(2/3)x + 2
which is perpendicular to y = (-3/2)x and passes through (0,2)
First think of your slope from.
3x+2y=0
so y = (-3/2)x
how what would the slope be for your line "perpendicular" to y = (-3/2)x ?
It would be (2/3) (switch the numerator and denominator, and change the sign)
so now you have y = (2/3)x which is perpendicular to y = (-3/2)x
you can then use the point-slope form for the equation of a line to find the equation for the line which is perpendicular to y = (-3/2)x and passes through (0,2)
Cheers,
John.
3x+2y=0
so y = (-3/2)x
how what would the slope be for your line "perpendicular" to y = (-3/2)x ?
It would be (2/3) (switch the numerator and denominator, and change the sign)
so now you have y = (2/3)x which is perpendicular to y = (-3/2)x
you can then use the point-slope form for the equation of a line to find the equation for the line which is perpendicular to y = (-3/2)x and passes through (0,2)
Cheers,
John.
pka
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Given two lines \(\displaystyle \left\{ \begin{array}{l}
l_1 :\;y = m_1 x + b_1 \\
l_2 :\;y = m_2 x + b_2 \\
\end{array} \right\) these two lines are perpendicular if \(\displaystyle m_1 m_2 = - 1\) or one is vertical and the other is horizontal.
Now line \(\displaystyle 3x + 2y = 0 \to y = \frac{{ - 3}}{2}x\) any line perpendicular must have slope \(\displaystyle m = \frac{2}{3}\) because \(\displaystyle \left( {\frac{2}{3}} \right)\left( {\frac{{ - 3}}{2}} \right) = - 1.\)
l_1 :\;y = m_1 x + b_1 \\
l_2 :\;y = m_2 x + b_2 \\
\end{array} \right\) these two lines are perpendicular if \(\displaystyle m_1 m_2 = - 1\) or one is vertical and the other is horizontal.
Now line \(\displaystyle 3x + 2y = 0 \to y = \frac{{ - 3}}{2}x\) any line perpendicular must have slope \(\displaystyle m = \frac{2}{3}\) because \(\displaystyle \left( {\frac{2}{3}} \right)\left( {\frac{{ - 3}}{2}} \right) = - 1.\)
