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Mila [183]
3 years ago
6

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) ?

Mathematics
1 answer:
malfutka [58]3 years ago
4 0

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.

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