Question

line m contains the points -3,4 and 1,0. write the equation of a line that would be perpendicular to this one and pass through the point -2,6 answer for algebra 1 need help

Answer 1

The equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6) in point slope form is y = x + 8

Solution:

Given that line m contains points (-3, 4) and (1, 0)

We are asked to find the equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6)

Let us first find slope of the line "m"

Given two points are (-3, 4) and (1, 0)

`m = (y_2 - y_1)/(x_2 - x_1)`

`\left(x_(1), y_(1)\right)=(-3,4) \text { and }\left(x_(2), y_(2)\right)=(1,0)`

`m=(0-4)/(1-(-3))=-1`

Thus slope of line m is -1

We know that product of slope of given line and perpendicular line are always -1

So, we get

`\begin{array}{l}{\text { slope of line } m * \text { slope of perpendicular line }=-1} \\\\ {-1 * \text { slope of perpendicular line }=-1} \\\\ {\text { slope of perpendicular line }=1}\end{array}`

So we have got the slope of perpendicular line is 1 and it passes through (-2, 6)

Let us use the point slope form to find the required equation

The point slope form is given as:

`y - y_1 = m(x - x_1)`

`(x_1, y_1) = (-2, 6)` and m = 1

y - 6 = 1(x - (-2))

y - 6 = x + 2

y = x + 8

Thus equation of required line in point slope form is y = x + 8