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

`6`.

Explanation:

A line that goes through
`(x_(0),\, y_(0))` and
`(x_(1),\, y_(1))` where
`x_(0) \\e 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) &= (1 - 4)/((-1) - 8) \\ &= ((-3))/((-9)) \\ &= (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) &= ((-1))/(m_(2)) \\ &= ((-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* (-1) + 3 = 6`.