A,B and C have co-ordinates (0,3), (4,4) and (8,0) respectively. Find the equation of the line through A and perpendicular to BC.
Answers
Answer:
Answer
Correct option is
A
x−y+3=0
B
x−4y+14=0
D
4x−y+1=0
A median of a triangle is a line segment that joins the vertex of a triangle to the midpoint of the opposite side.
Mid
point of two points (x
1
,y
1
) and (x
2
,y
2
) is calculated by the formula (
2
x
1
+x
2
,
2
y
1
+y
2
)
Using this formula,
mid point of AB =(
2
4−2
,
2
7+3
)=(1,5)
mid point of BC =(
2
−2+0
,
2
3+1
)=(−1,2)
mid point of CA =(
2
0+4
,
2
1+7
)=(2,4)
Equation
of a line joining two points (x
1
,y
1
) and (x
2
,y
2
) is given by the formula y−y
1
=(
x
2
−x
1
y
2
−y
1
)(x−x
1
)
Equation of Median passing through
A is the equation passing through A (4,7) and Midpoint of BC (−1.2) is y−7=(
−1−4
2−7
)(x−4)
=>y−7=
−5
−5
(x−4)
=>y−7=x−4
=>x−y+3=0
Equation of Median passing through B is the equation passing through B
(−2,3) and Midpoint of AC (2,4) is y−3=(
2−(−2)
4−3
)(x−(−2))
=>y−3=
4
1
(x+2)
=>4y−12=x+2
=>x−4y+14=0
Equation of Median passing through C is the equation passing through C
(0,1) and Midpoint of AB (1,5) is y−1=(
1−0
5−1
)(x−0)
=>y−1=
1
4
(x)
=>y−1=4x
=>4x−y+1=0
Step-by-step explanation:
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Step-by-step explanation:
Given :-
A,B and C have co-ordinates (0,3), (4,4) and (8,0) respectively.
To find :-
Find the equation of the line through A and perpendicular to BC?
Solution :-
Given that :-
The coordinates of A = (0,3)
The coordinates of B = (4,4)
The coordinates of C = (8,0)
We know that
Slope of a line segment joining the points (x1, y1) and (x2, y2) is (y2-y1)/(x2-x1)
Let (x1, y1) = (0,3) => x1 = 0 and y1 = 3
Let (x2, y2) = (4,4) => x2 = 4 and y2 = 4
The slope of the linesegment BC
=> (0-4) / (8-4)
=> -4/4
=> -1
The slope of the line (m1) = -1
Let the slope of the required line be m2
We know that
If the two lines having slopes m1 and m2 are perpendicular to each other then m1×m2 = -1
=> -1×m2 = -1
=> m2 = -1/-1
=> m2 = 1
Now,
We know that
The equation of a slope and point is
y-y1 = m(x-x1)
Let (x1, y1) = (0,3) => x1 = 0 and y1 = 3
m = m2 = 1
The equation of the line through A and perpendicular to BC
=> y-3 = 1(x-0)
=> y-3 = 1(x)
=> y-3 = x
=> x-y+3 = 0
Answer:-
The equation of the line through A and perpendicular to BC is x - y + 3 = 0
Used formulae:-
→Slope of a line segment joining the points (x1, y1) and (x2, y2) is (y2-y1)/(x2-x1)
→If the two lines having slopes m1 and m2 are perpendicular to each other then m1×m2 = -1
→The equation of a slope and point is
y-y1 = m(x-x1)