A triangle abc with vertices a(-1,0), b(-2,3/4), c(-3,7/6) has its orthocentre h.then the orthocentre of triangle bch will be.?
Answers
Answer:
A(-1,0)
Step-by-step explanation:
Given triangle is obtused angled triangle, being obtused at vertice B.
Orthocenter is point of intersection of altitudes , and for an obtuse angled triangle orthocenter lies outside the triangle.
Now, Given that H is the Orthocenter for triangle ABC,
from attached diagram it is evident that
A will be the orthocenter for triangle BHC
B will be orthocenter for triangle AHC
C will be the orthocenter for triangle AHB.
Answer:A(-1,0).
Answer:
Step-by-step explanation:
The orthocenter is the intersecting point or all the altitude of the triangle. The point where the altitude of a triangle meet is known as the orthocenter.
Given the points,
A(−1,0)≡(x
1
,y
1
),B(−2,
4
3
)≡(x
2
,y
2
),C(−3,
6
−7
)
To find the orthocenter, H
SlopeofAB=
x
2
−x
1
y
2
−y
1
=
−2+1
4
3
−0
=
4
−3
Slope of CF = perpendicular slope of AB
=
SlopeofAB
−1
=
3
4
∴ Equation of CF⇒y−y
1
=m(x−x
1
)
y
4
−3
=
3
4
(x+3)⇒16x−12y+45=0→(1)
SlopeofBC=
x
2
−x
1
y
2
−y
1
=
−3+2
6
−7
4
−3
=
12
23
∴ Slope of AD=perpendicular slope of BC=
23
−12
∴ Equation of AD=y−y
1
=m(x−x
1
)
y
12
−23
=
23
−12
(x+1)⇒144x+276y=385⟶(2)
Obtaining the value of x and y by solving can (1) and (2)
16x−12y+45=0⟶(1)×144
144x+276y=385⟶(2)×16
6144y=12640
2304x+4416y=6160−2304x±1728y=±6480
⇒y=2.057,x=4.36
To find the orthocenter of △BCH
i.e, Let the orthocenter be denoted as 0
SlopeofBH=
x
2
−x
1
y
2
−y
1
=
−2−4.36
0.75−2.06
=0.206
∴ Slope of CR = perpendicular slope of BH
=
SlopeofBH
−1
=
0.206
−1
=−4.85
Equation of CR⇒y−y
1
=m(x−x
1
)
y−0.206=−4.85(x+3)
⇒4.85x+y=−14.36⟶(1)
SlopeofBC=
x
2
−x
1
y
2
−y
1
=
−3+2
6
−7
4
−3
=
12
23
=1.92
Slope of HP = perpendicular slope of HP
slopeofHP
−1
=−0.52
Equation of HP=y−1.92=−0.52(x−4.36)
⇒0.52x+y=4.19⟶(2)
Obtaining the value of x and y by solving equation (1) and (2)
∴4.33x=−18.55
4.85x+y=−14.36−0.52x∓y=−4.19
⇒x=−4.28 and y=6.42
∴ The orthocenter of △BCH=(−4.28,6.42)The orthocenter is the intersecting point or all the altitude of the triangle. The point where the altitude of a triangle meet is known as the orthocenter.
Given the points,
A(−1,0)≡(x
1
,y
1
),B(−2,
4
3
)≡(x
2
,y
2
),C(−3,
6
−7
)
To find the orthocenter, H
SlopeofAB=
x
2
−x
1
y
2
−y
1
=
−2+1
4
3
−0
=
4
−3
Slope of CF = perpendicular slope of AB
=
SlopeofAB
−1
=
3
4
∴ Equation of CF⇒y−y
1
=m(x−x
1
)
y
4
−3
=
3
4
(x+3)⇒16x−12y+45=0→(1)
SlopeofBC=
x
2
−x
1
y
2
−y
1
=
−3+2
6
−7
4
−3
=
12
23
∴ Slope of AD=perpendicular slope of BC=
23
−12
∴ Equation of AD=y−y
1
=m(x−x
1
)
y
12
−23
=
23
−12
(x+1)⇒144x+276y=385⟶(2)
Obtaining the value of x and y by solving can (1) and (2)
16x−12y+45=0⟶(1)×144
144x+276y=385⟶(2)×16
6144y=12640
2304x+4416y=6160−2304x±1728y=±6480
⇒y=2.057,x=4.36
To find the orthocenter of △BCH
i.e, Let the orthocenter be denoted as 0
SlopeofBH=
x
2
−x
1
y
2
−y
1
=
−2−4.36
0.75−2.06
=0.206
∴ Slope of CR = perpendicular slope of BH
=
SlopeofBH
−1
=
0.206
−1
=−4.85
Equation of CR⇒y−y
1
=m(x−x
1
)
y−0.206=−4.85(x+3)
⇒4.85x+y=−14.36⟶(1)
SlopeofBC=
x
2
−x
1
y
2
−y
1
=
−3+2
6
−7
4
−3
=
12
23
=1.92
Slope of HP = perpendicular slope of HP
slopeofHP
−1
=−0.52
Equation of HP=y−1.92=−0.52(x−4.36)
⇒0.52x+y=4.19⟶(2)
Obtaining the value of x and y by solving equation (1) and (2)
∴4.33x=−18.55
4.85x+y=−14.36−0.52x∓y=−4.19
⇒x=−4.28 and y=6.42
∴ The orthocenter of △BCH=(−4.28,6.42)
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