Question

Write the equation of the line passing through (−3, 4) and perpendicular to −2xx + 7yy = −3

Answer 1

Answer:

$y=-3.5x-6.5$

Step-by-step explanation:

Fist thing we need to do is find the slope of the original line, we have:

$-2x+7y=-3$

we want to clear for $y$ to get a slope-intercept equation ($y=mx+b$, where $m$ is the slope and $b$ is the interception of the line with the y axis)

$7y=2x-3\\y=\frac{2}{7}x-\frac{3}{7}$

this way we can see that the number that represents the slope is $\frac{2}{7}$. I will call this $m_{1}=\frac{2}{7}$ because it is the slope of the fisrt line.

Now to find the equation of the second line (the line perpendicular to $-2x+7y=-3$), We need to apply the condition so that two lines are perpendicular:

$m_{1}m_{2}=-1$

we have $m_{1}=\frac{2}{7}$, so the slope of the perpedicular line $m_{2}$ is:

$\frac{2}{7}m_{2}=-1\\ m_{2}=-\frac{7}{2}$

we already have the slope, and the problem mentions that the new line passes through the point (-3, 4), so we use the point-slope equation

$y=m(x-x_{0})+y_{0}$

where $m$ is the slope, and $(x_{0},y_{0})$ is the point, so $x_{0}=-3$, and $y_{0}=4$

thus:

$y=-3.5(x-(-3))+4 \\y=-3.5(x+3)+4\\y=-3.5x-10.5+4\\y=-3.5x-6.5$

the equation is $y=-3.5x-6.5$