Question

What is y in the first line if the line passing though (-1,y) and (1,0) perpendicular to the line passing (8,4) and (-1,1) ?

Answer 1

Answer:

$6$.

Step-by-step explanation:

A line that goes through $(x_{0},\, y_{0})$ and $(x_{1},\, y_{1})$ where $x_{0} \ne x_{1}$ would have a slope of $m = (x_{1} - x_{0}) / (y_{1} - y_{0})$.

The slope of the line that goes through $(8,\, 4)$ and $(-1,\, 1)$ would thus be:

$\begin{aligned}m_{2} &= \frac{1 - 4}{(-1) - 8} \\ &= \frac{(-3)}{(-9)} \\ &= \frac{1}{3}\end{aligned}$.

Two lines in a cartesian plane are perpendicular to one another if and only if the product of their slopes is $(-1)$.

Thus, if $m_{1}$ and $m_{2}$ denote the slope of the first and second lines in this question, $m_{1}\, m_{2} = (-1)$ since the two lines are perpendicular to one another. Since $m_{2} = (1/3)$, the slope of the first line would be:

$\begin{aligned} m_{1} &= \frac{(-1)}{m_{2}} \\ &= \frac{(-1)}{(1/3)} \\ &= (-3)\end{aligned}$.

Given that the first line goes through the point $(1,\, 0)$, the point-slope equation of that line would be:

$(y - 0) = (-3)\, (x - 1)$.

$y = -3\, x + 3$.

Substitute in $x = (-1)$ to find the $y$-coordinate of the point in question:

$y = -3\times (-1) + 3 = 6$.