Could u please explain this geometry question?

Question

Find the value of y so that the line through ( -2, -1) and (10,y) is perpendicular to the line through ( 6, -2) and ( 5, 7)

Answer 1

You first need to solve for the slope of (6,-2) and (5,7).

When you already have it, you have to apply the negative reciprocal of that slope to the second set of points.

Answer 2

In order for two lines to be perpendicular to each other, their slopes must be the negative reciprocals of each other....

So..... let's find the slope of the line that runs through points (6, -2) and (5,7)

You find the slope by using the following formula....

Slope of Line1 = m1 = (y2-y1) / (x2-x1)

..... so let (x1,y1) = (6,-2).... and let (x2,y2) = (5,7)

Plugging in the values for x1, y1, x2, and y2.... you get....

m = [ 7 - (-2) ] / (5 - 6) = (7+2) / (5 - 6) = 9 / -1 = - 9/1 = -9

What is the negative reciprocal of -9?  The negative reciprocal of m = -9 is - 1/m = -1 / (-9) = +1/9 = 1/9

So the slope of Line 2 to be perpendicular to Line 1.... must have a slope of 1/9

Slope of Line 2 = m2 = [ y - (-1) ] / [ 10 - (-2) ]

m2 = (y+1) / (10+2) = (y+1) / 12

.......... (y+1) ......... 1

m2 = -------------- = -------- <== because m2 = -1 / m1 = -1 / (-9) = +1/9

............. 12 .......... 9

Cross multiplying.... you get.....

(y+1)(9) = (12)(1)

Distribute the "9" to both the "y" and the "+1".... like this...

(y)(9) + (+1)(9) = 12

9y + 9 = 12

Subtract "9" from both sides like this...

9y + 9 = 12

..... - 9 ... -9

---------------------

9y + 0 = 3

9y = 3

Divide both sides by "9"... like this...

9y ... 3

---- = ----

9 ..... 9

The 9's on the left side cancel each other out... leaving you with.... y = 3/9 = 1/3.....

So when " y = 1/3 ".... the line that runs through the first set of points.... will be perpendicular to the line that runs through the 2nd set of points.....