The diagram shows the points A (1, 2), B (4, 6) and D(-5, 2). • B(4,6) OX-5,2) 4(1.27 (a) Find the coordinates of the midpoint of AB, (b) Calculate the length of AB. (c) Calculate the gradient of the line AB. (d) Find the equation of the line AB. (c) The triangle ADC has line of symmetry x = 4. Find the coordinates of C.
Answers
Answer:
Given, the coordinates of the midpoints of △ABC are $$D(2,1), E(5,3), F(3,7)$$ respectively.
Let the coordinates of the vertices be A=(x
1
,y
1
),
B=(x
2
,y
2
),
C=(x
3
,y
3
)
∴ Coordinates of D is (
2
x
1
+x
2
,
2
y
1
+y
2
)=(2,1)
⇒
2
x
1
+x
2
=2
⇒x
1
+x
2
=4⟶(1)
and,
2
y
1
+y
2
=1
⇒y
1
+y
2
=2⟶(2)
Coordinates of E is (
2
x
2
+x
3
,
2
y
2
+y
3
)=(5,3)
⇒
2
x
2
+x
3
=5
⇒x
2
+x
3
=10⟶(3)
and,
2
y
2
+y
3
=3
⇒y
2
+y
3
=6⟶(4)
Coordinates of F is (
2
x
1
+x
3
,
2
y
1
+y
3
)=(3,7)
⇒
2
x
1
+x
3
=3
⇒x
1
+x
3
=6⟶(5)
and,
2
y
1
+y
3
=7
⇒y
1
+y
3
=14⟶(6)
Solving equations (1), (3), (5) we get,
x
1
=0, x
2
=4, x
3
=6
Solving equations (2), (4), (6) we get,
y
1
=5, y
2
=−3, y
3
=9
∴ coordinates of the vertices will be A=(0,5)
B=(4,−3)
C=(6,9)
∴ Lengths of AB=
(−4)
2
+8
2
=2
5
BC=
(−2)
2
+12
2
=2
37
CA=
6
2
+4
2
=2
13
The equations of the lines AB
⇒(y−5)=
4−0
−3−5
(x−0)
⇒2x+y=20
BC⇒(y+3)=
6−4
9+3
(x−4)
⇒6x−y=27
CA⇒(y−9)=
0−6
5−9
(x−6)
⇒2x−3y=−15