Solve the following quadratic equation: y=5x2+10x Solutions: x= and x=
1 answer:
You might be interested in
Answer:
Point of intersection (-11/3 , 1/3)
All distances from the vertices are
Step-by-step explanation:
The vertices of the triangle is A(-3,5) , B (1,1) and C (-7,-3)
We need to find the perpendicular bisector of the triangle first
let's take one side connecting A and B
mid point of A and B = (-1,3)
Slope of the line joining A and B = -1
slope of perpendicular to line joining A and B = 1
equation of line passing through (-1,3) with slope 1
y - 3 = 1(x-(-1))
y -3 = x+1
x-y = -4 ............(1)
similarly
mid point joining B and C = (-3,-1)
slope perpendicular to line joining B and C = -2
Equation of perpendicular bisector of line joining B and C =
y +1 = -2(x +3 )
y+1 = -2x -6
2x+y = -7 ..........(2)
On solving 1 and 2
x= -11/3 , y= 1/3
Distances
From A =
From B =
From C =