The vertex A(-4,2) is a vertex of the triangle ABC. The edge BC lies on the line y=3x-7, and the height from the vertex C to the edge AB lies on the line y=2x+8. What are the coordinates of B and C of the triangle ABC?
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The vertex A(-4,2) is a vertex of the triangle ABC. The edge BC lies on the line y=3x-7, and the height from the vertex C to the edge AB lies on the line y=2x+8. What are the coordinates of B and C of the triangle ABC?
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The coordinates of the vertices of a triangle are (0,1),(2,3) and (3,5):
(a) Find centroid of the triangle.
(b) Find circumcentre and the circumradius.
(c) Find Orthocentre of the triangle.
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Lets coordinate of a triangle are A(x1,y1),B(x2,y2),C(x3,y3)
Here A(0,1),B(2,3),C(3,5)
(a) Centroid of the triangle=(xc,yc)
xc=3x1+x2+x3,yc=3y1+y2+y3
xc=30+2+3=35,yc=31+3+5=3
So the centroid is (35,3).
(b) To find out the circumcenter we have to solve any two bisector equations and find out the intersection points.
So, mid point of AB =(20+2,21+3)=F(1,2)
Slope of AB=3−12−0=1
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector =−1
Equation of line with slope −1 and the coordinates F(1,2) is,
(y−2)=−1(x−1)
x+y=3………………(1)
Similarly, for AC
Mid point of AC =(20+3,21+5)=E(23,3)
Slope of AC=3−05−1=34
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector =−43
Equation of line with slope −43 and the coordinates E(23,3) is,
(y−3)=−43(x−23)
y−3=−43x+89
43x+y=833………………(2)
By solving equation (1) and (2),
(1)−(2)⇒4x=3−833;x=−29
Substitute the value of x in to (1)
−29−y=3
y=−215
So the circumcenter is (−29,−215)
(c) Let perpendicular bisectors from A,B,C of triangle be F (on side AB), E (on side AC),D(on side BC)
We know, slope of AB=1
Slope of CF = Perpendicular slope of AB
=−Slope of AB1
=−1
The equation of CF is given as,
y−5=−1(x−3)
x+y=8 ………………(1)
Slope of AC=34
Slope of AD = Perpendicular slope of BC
=−Slope of BC1
=−43
The equation of BE is given as,
y−3=−43(x−2)
y−3=−43x+23
43x+y=29 ………………(2)
By solving equation (1) and (2),
(1)−(2)⇒4x=8−29;x=14
Substitute the value of x in to (1)
14+y=8
y=−6
So the Orthocenter is (14,−6)