this question is related to coordinate geometry(maths).i need to know how to solve this one.
Responses (3)
Consider the three given points
A( -2,1) B(2, -2) C(5,2)
as the vertices of a triangle, and
a = BC, b = AC, c = AB
as the sides of this triangle.
Two lines being at a right angle to each other is just a different way of saying that they are perpendicular to each other.
If g is a line with slope m and and g' is a line with slope m', these lines are perpendicular, if and only if m is the negative inverse of m' :
m = -m'^(-1)
The slope m of a line g which is given by two points P1(x1,y1) P2(x1,y1) is
mg = (y2 - y1) / (x2 - x1).
The sum of the interior angles of a triangle equals 180. So, only one of them can be a right angle. Make a plot of the given triangle. Only B looks pretty much to be the vertex of a right angle. But "pretty much" isn't good enough.
We, therefore, have to figure out the slopes ma of a=BC and mc of c=AB and see wether the condition for being perpendicular is met.
ma = (2 - (-2)) / (5 - 2) = 4/3
mc = (-2 - 1) / (2 - (-2)) = -3/4
Although you see at a glance that the condition is met, we'll write it out properly:
4/3 = - ( -3/4)^(-1)
4/3 = - ( -4/3)
4/3 = 4/3.
So, there is one vertex, B, in which the two lines a and c include a right angle.
A right triangle has two perpendicular lines. Perpendicular lines have negative reciprocal slopes.
Slope is rise over run. Run is horizontal distance, left to right. Run is always positive because we always go left to right. Rise is the vertical change in that same distance. A negative rise means it drops.
Line from (-2, 1) to (2, -2) rises -3 units from 1 to -2 and runs 4 units from -2 to 2 so slope is -3/4.
Line from (2, -2) to (5, 2) rises 4 units and runs 3 units so slope is 4/3.
Line from (-2, 1) to (5, 2) rises 7 units and runs 1 unit so slope is 7.
-3/4 is the negative reciprocal of 4/3 so the lines are perpendicular and the figure is a right triangle.