Download Question Bank Distance Formula, Section Formula and Equation of

Question Bank
Distance Formula, Section Formula and Equation of a
Straight Line
1. A (–3, 2), B (–5, –5), C (2, –3) and D (4, 4) are the four points in a
plane. Show that ABCD is a rhombus but not a square.
Solution. The given points are A (–3, 2), B (–5, –5), C (2, –3) and
D (4, 4).
∴ AB =
[ – 5 – (–3)]
2
+ (– 5 – 2) 2
=
(– 5 + 3) 2 + (– 7) 2
=
(– 2) 2 + (– 7) 2
=
=
BC =
=
CD =
=
DA =
4 + 49
53 units
[2 –
(– 5) ] + [ –3 – (– 5)] = (2 + 5)2 + (– 3 + 5)2
2
2
7 2 + 22 = 49 + 4 = 53 units
(4 – 2) 2 + [ 4 – (– 3]
2
= 22 + (4 + 3) 2
22 + 7 2 = 4 + 49 = 53 units
(– 3 – 4) 2 + (2 – 4) 2
= (7) 2 + (– 2) 2 = 49 + 4 = 53 units
∴ AB = BC = CD = DA.
∴ ABCD is either a rhombus or a square.
Diag. AC = [2– (– 3)]2 + (– 3 – 2) 2
=
(2 + 3) 2 + ( – 5) 2
=
52 + 52 = 25 + 25 = 50 = 5 2 units.
Diag. BD =
Math Class X
(2 + 3) 2 + ( – 5) 2
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Question Bank
2.
=
[4– (– 5)]2 + [4 – ( – 5)]2
=
(4 + 5) 2 + (4 + 5) 2
= 92 + 92 = 81 + 81
= 162 = 9 2 units
∴ Diag. AC ≠ Diag. BD
∴ ABCD is a rhombus but not a square. Proved
Find the co-ordinates of the circumcentre of ΔABC with vertices at
A(3, 0), B(–1, –6) and C(4, –1). Also, find its circum-radius.
Solution. The vertices of the given triangle are
A (3, 0), B (–1, –6) and C (4, –1).
Let O (x, y) be the circumcentre of ΔABC.
Then,
OA = OB
= OC
2
⇒ OA = OB2
= OC2
Now, OA2 = OB2
⇒ (x – 3)2 + (y – 0)2 = [x – (– 1)]2 + [y – (– 6)]2
⇒
(x – 3)2 + y2 = (x + 1)2 + (y + 6)2
⇒ x2 + y2 – 6x + 9 = x2 + y2 + 2x + 12y + 37
⇒ 8x + 12y + 28 = 0
⇒
2x + 3y = –7 ...(i)
2
And, OB = OC2
⇒ [ x – (– 1)]2 + [y – (– 6)]2 = (x – 4)2 + [y – (–1)]2
⇒ (x + 1)2 + (y + 6)2 = (x – 4)2 + (y + 1)2
⇒ x2 + y2 + 2x + 12y + 37 = x2 + y2 – 8x + 2y + 17
⇒ 10x + 10y + 20 = 0
⇒ x + y = –2
...(ii)
Solving (i) and (ii), we get x = 1 and y = –3.
∴ The circumcentre of ΔABC is O (1, –3).
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Question Bank
Circum-radius = OA
=
3.
=
(1 – 3) 2 + (– 3 – 0) 2
(– 2) 2 + (– 3) 2
= 4 + 9 13
= 13 units.
Find the co-ordinates of the centre of a circle which passes through
the points A (0, 0), B (–3, 3) and C (5, –1). Also, find the radius of
the circle.
Solution. Let P (x, y) be the centre of the circle passing through the
points A (0, 0), B (–3, 3) and C (5, –1).
Then,
PA = PB
= PC
2
⇒
PA = PB2
= PC2
Now, PA2 = PB2
⇒ (x – 0)2 + (y – 0)2 = (x + 3)2 + (y – 3)2
⇒ x2 + y2 = x2 + 9 + 6x + y2 + 9 – 6y
⇒ 6x – 6y = –18
⇒ x – y = –3
...(i)
2
2
And, PB = PC
⇒ (x + 3)2 + (y – 3)2 = (i – 5)2 + (y + 1)2
⇒ x2 + 9 + 6x + y2 + 9 – 6y = x2 + 25 – 10x + y2 + 1 + 2y
⇒
16x – 8y = 8
⇒
2x – y = 1
...(ii)
Subtracting (ii) from (i), we get
–x = – 4 ⇒ x = 4
Substituting x = 4 in (i), we get
4 – y = –3 ⇒ y = 7
Hence, centre of the circle is P (4, 7).
Radius of the circle = PA
= (4 – 0) 2 + (7 – 0) 2
=
Math Class X
16 + 49 = 65 units.
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Question Bank
4.
5.
KM is a straight line of 13 units. If K has the co-ordinates (2, 5) and
M has the co-ordinates (x, –7), find the possible values of x.
Solution.
We have, KM2 = (x – 2)2 + (–7 – 5)2
⇒
KM2 = x2 + 4 – 4x + 144
⇒
KM2 = x2 – 4x + 148
…(i)
But
KM = 13 (given)
⇒
KM2 = 169
…(ii)
∴ From (i) and (ii), we get
169 = x2 – 4x + 148
⇒
x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒
(x – 7) (x + 3) = 0
⇒
x = 7 or x = –3
Hence, the possible values of x are 7 and –3.
In what ratio does the point P (p, –1) divide the line segment joining
the points A (1, –3) and B (6, 2)? Hence, find the value of p.
Solution. Let P (p, –1) divide the line segment joining the points A
(1, –3) and B (6, 2) in the ratio k : 1, i.e., AP : PB = k : 1
⎡ k × 6 + 1 × 1 k × 2 × 1 × (–3) ⎤
,
∴⎢
⎥⎦
k
+
1
k +1
⎣
But P is (p, –1)
2k – 3
⇒
= –1
k +1
⇒
2k – 3 = – k – 1
⇒
3k = 2
2
⇒
k =
3
Math Class X
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Question Bank
∴ The required ratio is
2
: 1, i.e., 2 : 3 (internally).
3
6k + 1
=p
..(i)
k +1
2
Putting
k =
in (i), we get
3
2
6. + 1
5
5 3
3
=
= ×
ρ =
= 3.
2
5
1
5
+1
3
3
p = 3.
Hence,
The centre of a circle is C (–1, 6) and one end of a diameter is A
(5, 9). Find the co-ordinates of the other end.
Solution. Let the other end of
the diameter of the circle be
B (x, y) whose one end is the
point A (5, 9).
∴ The mid-point of AB is .
⎡5 + x 9 + y ⎤
⎢ 2 , 2 ⎥
⎣
⎦
The centre of the circle is C (–1, 6).
Since the centre of the circle is the mid-point of AB,
5+x
9+y
= – 1 and
=6
2
2
⇒
5 + x = –2 and
9 + y = 12
⇒
x = –7 and y = 3
∴ The co-ordinates of the other end of the diameter are
(–7, 3).
Two vertices of a ΔABC are A (6, –2) and B (4, 3). If the coordinates of its centroid be (3, –1), find the co-ordinates of the third
vertex of the triangle.
Solution. Let C (a, b) be the third vertex of ΔABC.
Also,
6.
7.
Math Class X
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Question Bank
8.
Then, the co-ordinates of its centroid are :
⎡10 + a 1 + b ⎤
⎡6 + 4 + a 2 + 3 + b⎤
,
i.e.,
G⎢
⎢ 3 , 3 ⎥
⎥⎦
3
3
⎣
⎣
⎦
But, the co-ordinates of its centroid are G (3, –1).
10 + a
∴
=3
3
1+b
= –1
and
3
⇒
10 + a = 9
and
1+ b=–3
⇒
a=–1
and
b=–4
Hence, third vertex of the triangle is C (–1, –4).
The mid points of the sides BC, CA and AB of ΔABC are D (2, 1),
E (–1, –3) and F (4, 5) respectively. Find the coordinates of A, B
and C.
Solution. Let the coordinates of A, B and C be
(x1, y1), (x2, y2) and (x3, y3) respectively.
D is the mid-point of BC
x + x3
y + y3
∴ 2
= 2 and 2
=1
2
2
⇒ x2 + x3 = 4 ...(i)
and y2 + y3 = 2 ... (ii)
E is the mid point of AC
x1 + x3
y + y3
= –1 and 1
= –3
∴
2
2
⇒ x1 + x3 = –2 ...(iii) and y1 + y3 = – 6 ... (iv)
F is the mid point of AB
x + x2
y + y2
= 4 and 1
=5
∴ 1
2
2
⇒ x1 + x2 = 8 ...(v) and y1 + y2 = 10 ... (vi)
From (i) and (iii), we get x1 – x2 = – 6 ... (vii)
From (ii) and (iv), we get y1 – y2 = – 8 ... (viii)
Math Class X
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Question Bank
9.
From (vii) and (v), we get 2x1 = 2 ⇒ x1 = 1 ... (vii)
From (viii) and (vi), we get 2y1 = 2 ⇒ y1 = 1
From (vii) and (viii), we get x2 = 7 and y2 = 9
From (i) and (ii), we get x3 = –3 and y3 = –7
Hence, coordinates of A, B and C are (1, 1), (7, 9) and (–3, –7)
respectively.
Show that the line segment joining the points
A (–5, 8) and B (10, – 4) is trisected by the coordinate axes. Also,
find the points of trisection of AB.
Solution. Let the x-axis cut the line segment AB at P(x, 0) which
divide it in the ratio p : 1. Also, let the y-axis cut the line segment
AB at Q(0, y) which divide it in the ratio q : 1.
P (x, 0) divides AB in the ratio p : 1.
p × (–4) + 1 × 8
∴ 0=
p+1
⇒ 4p = 8 ⇒ p = 2
∴ P (x, 0) divides AB in the ratio 2 : 1. ... (i)
2 × 10 + 1 × (–5)
15
∴ x=
=
=5
2+1
3
∴ Coordinates of P are (5, 0).
Q (0, y) divides AB in the ratio q : 1.
y × 10 + 1 × (–5)
∴ 0=
q+1
1
⇒ 10q = 5 ⇒ q =
2
1
∴ Q (0, y) divides AB in the ratio : 1 i.e., 1 : 2 ... (ii)
2
1 × (–4) + 2 × 8
=4
∴y=
1+2
∴ Coordinates of Q are (0, 4).
AP 2
AQ
1
From (i),
=
and from (ii)
,=
QB
2
PB 1
Math Class X
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Question Bank
∴ AP = 2PB and 2AQ = QB
... (iii)
Now, AP + PB = AB and AQ + QB = AB
⇒
3PB = AB and 3AQ = AB
[From (iii)]
1
1
⇒
PB = AB and AQ = AB
3
3
⇒
PB = AQ
... (iv)
Now,
QB = QP + PB
⇒
2AQ = QP + AQ
[From (iii) and (iv)
⇒
AQ = QP
... (v)
From (iv) and (v), we get
AQ = QP = PB. Proved.
The points of trisection of AB are (5, 0) and (0, 4).
10. G is the centroid of a triangle with vertices A(–a, 0), B(0, a) and
C(a, b). Prove that
GA2 + GB2 + GC2 = (AB2 + BC2 + CA2)
Solution. Coordinates of the Solution. Coordinates of the centroid G
of ΔABC are
⎛ –a + 0 + α 0 + a + β ⎞
,
⎜
⎟
3
3
⎝
⎠
⎛α –a β +a⎞
i.e., ⎜
,
⎟
3
3 ⎠
⎝
2
2
⎛α –a
⎞
⎛β + a
⎞
2
2
+ a⎟ + ⎜
– 0⎟
Now, GA = GA = ⎜
⎝ 3
⎠
⎝ 3
⎠
2
2
⎛ α + 2a ⎞
⎛β + a⎞
=⎜
⎟ +⎜
⎟ ... (i)
⎝ 3 ⎠
⎝ 3 ⎠
2
2
⎛α –a
⎞
⎛β + a
⎞
2
2
GB = GB = ⎜
– 0⎟ + ⎜
– a⎟
3
3
⎝
⎠
⎝
⎠
2
2
⎛α –a⎞
⎛ β – 2a ⎞
=⎜
+
⎟
⎜
⎟ ... (ii)
3
3
⎝
⎠
⎝
⎠
Math Class X
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Question Bank
2
2
⎛α –a
⎞
⎛β + a
⎞
GC = ⎜
– α⎟ + ⎜
– β⎟
⎝ 3
⎠
⎝ 3
⎠
2
2
⎛ –2α – a 2 ⎞
⎛ a – 2β ⎞
=⎜
⎟ +⎜
⎟ ... (iii)
3
3
⎝
⎠
⎝
⎠
From (i), (ii) and (iii), we get
1
GA2 + GB2 + GC2 = [(α + 2a)2 + (β + a)2 + (α – a)2 + (β – 2a)2
9
+ (–2α – a)2 + (a – 2β)2 ]
1 2
=
[α + 4a2 + 4aα + β2 + a2 + 2aβ + α2
9
+ a2 – 2aα + β2 + 4a2 – 4aβ + 4α2 + a2
+ 4aα + a2 + 4β2 – 4aβ]
1
= [6α2 + 6β2 + 6aα – 6aβ + 12a2]
9
2 2
[α + β2 + 2a2 + aα – aβ)
...(vii)
=
3
…(iv)
AB2 = (0 + a)2 + (a – 0)2 = 2a2
2
2
2
2
2
BC = (0 – a) + (a – β) = α + (α – β)
.... (v)
2
2
2
2
2
CA = (a + α) + (β – 0) = (α + a) + β
.... (vi)
From (iv), (v) and (vi), we get
AB2 + BC2 + CA2
= 2a2 + α2 + a2 + β2 – 2aβ + α2 + a2 + 2aα + β2
= 2α2 + 2β2 + 4α2 + 2aα – 2aβ
... (viii)
= 2 (α2 + β2 + 2a2 + aα – aβ)
From (vii) and (viii), we get
1
GA2 + GB2 + GC2 = (AB2 + BC2 + CA2).
3
2
Math Class X
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Question Bank
11. A straight line passes through the points P (–1, 4) and Q (5, –2). It
intersects the co-ordinate axes at points A and B. M is the mid-point
of the segment AB. Find :
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M.
Solution. Given points are
P (–1, 4) and Q (5, –2).
(i) The slope of the line
–2–4
= –1.
PQ =
5 – (– 1)
The line passes through
P (–1, 4) and has slope –1.
Its equation is y – 4 = –1 [x –(–1)]
[∴ y – y1 = m (x – x1)]
⇒ x + y – 3 = 0.
(ii) The line PQ meets x-axis, i.e., y = 0 where x + 0 – 3 = 0
⇒x=3
∴ The co-ordinates of A are (3, 0).
The line PQ meets y-axis, i.e., x = 0 where 0 + y – 3 = 0 ⇒ y = 3
∴ The co-ordinates of B are (0, 3).
(iii) Since M is mid-point of the segment AB, its co-ordinates
are
⎛3 + 0 0 + 3⎞
⎛3 3⎞
,
⎜
⎟ , i.e, ⎜ , ⎟ .
2 ⎠
⎝ 2
⎝2 2⎠
12. In the adjoining figure, AB and CD are the lines 2x – y + 6 = 0 and
x – 2y = 4 respectively.
(i) Write down the co-ordinates
of A, B, C and D;
(ii) Prove that ΔOAB and
ΔODC are similar;
(iii) Is figure ABCD cyclic?
Math Class X
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Question Bank
Give reasons for your answer.
Solution.
(i) Equation of AB is 2x – y + 6 = 0
AB cuts y-axis at the point, where x = 0.
Putting x = 0, we get y = 6.
∴ Co-ordinates of A are A (0, 6).
Again, AB cuts x-axis at the point, where y = 0.
Putting y = 0, we get x = –3.
∴ Co-ordinates of B are B (–3, 0).
Equation of CD is x – 2y = 4.
CD cuts y-axis at the point, where x = 0.
Putting x = 0 in its equation, we get y = –2
∴ Co-ordinates of C are C (0, –2).
Again, CD cuts x-axis at the point, where y = 0.
Putting y = 0 in its equation, we get x = 4.
∴ Co-ordinates of D are D (4, 0).
(ii) Thus, OA = 6 units, OB = 3 units, OC = 2 units and
OD = 4 units.
∴ AB = OA 2 + OB2 = 62 + 32 = 45 = 3 5 units
CD = OC2 + OD 2 = 22 + 42 = 20 = 2 5 units
OA
6
3 OB
3
AB
3 5 3
∴
.
= = ,
= and
=
OD
4
2 OC
2
CD
2 5 2
OA
OB
AB
Thus, in ΔOAB and ΔODC, we have =
=
=
.
OD
OC
CD
∴ ΔOAB ~ ΔODC. Proved
(iii) Let P (h, k) be a point equidistant from A, B, C and D, then,
PA = PB = PC = PD ⇒ PA2 = PB2 = PC2 = PD2.
∴ (h – 0)2 + (k – 6)2 = (h + 3)2 + (k – 0)2 = (h – 0)2 + (k + 2)2
= (h – 4)2 + (k – 0)2
⇒ h2 + k2 – 12k + 36 = h2 + 6h + 9 + k2 = h2 + k2 + 4k + 4
= h2 + k2 – 8h + 16
⇒
2h + 4k = 9 and k + 2h = 3
Math Class X
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Question Bank
⇒
k = 2 and h =
1
.
2
⎛1 ⎞
Thus, a circle can be drawn with centre P ⎜ , 2 ⎟ and radius PA to
⎝2 ⎠
pass through A, B, C and D.
Hence, the points A, B, C, D are cyclic.
13. P (3, 4), Q (7, –2) and R (–2, –1) are the vertices of Δ PQR. Write
down the equation of the median of the triangle through ∠R.
Solution. Let M be the mid-point of PQ.
⎛3 + 7 4 – 2⎞
Then, co-ordinates of M are ⎜
,
⎟ = (5, 1)
2 ⎠
⎝ 2
∴ Equation of the median RM is given by :
1 – (– 1)
y – (–1) =
[x – (2)]
5 – (– 2)
2
⇒
y + 1 = (x + 2)
7
7y + 7 = 2x + 4
2x – 7y = 3
14. A line passes through the point (–1, 2) and cuts off negative
intercept –a on the x-axis and positive intercept b on the y-axis such
that a : b = 2 : 1. Find the slope and the equation of the line.
Solution. Let the line cut the x-axis at A and the y-axis at B.
a:b=2:1
Let OA = 2a and OB = a
So, the coordinates of A are (–2a, 0) and that of B are (0, a).
Now equation of the line through the points (–2a, 0) and (0, a) is
given by
a–0
(x + 2a)
y–0=
0 + 2a
⇒
2y = x + 2a
⇒
–x + 2y = 2a ... (i)
Also, (i) passes through (–1, 2)
Math Class X
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Question Bank
∴ 1 + 4 = 2a ⇒ a =
5
2
∴ Equation of the line is given by –x + 2y = 2 ×
∴ –x + 2y = 5 ⇒ y =
x
5
+
2
2
5
2
1
and equation of the line is 2y – x = 5
2
15. The equations of the lines PQ and PR are 3x – 4y = –1 and x + y = 2
respectively. Find the equation of the line PM, if the coordinates of
M are (–2, 5).
Solution. To get the coordinates of the point P, we solve the given
equations simultaneously.
Multiplying x + y = 2 by 4 and adding to 3x – 4y = –1,
We get 4x + 4y = 8
3x – 4 y = – 1
7x = 7 ⇒ x = 1
∴ Slope of the line
Substituting x = 1 in x + y = 2, we get
1+2=2⇒x=1
∴ Coordinates of P are (1, 1).
Now equation of PM is given by y – y1 =
y2 – y1
(x – x1)
x2 – x1
5–1
(x – 1)
–2 – 1
(4 x – 4)
⇒ y–1=
–3
⇒ –3y + 3 = 4x – 4
⇒ 4x + 3y = 7
⇒ y–1=
Math Class X
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Question Bank
16. Find the equation of the perpendicular from the point P (–1, –2) on
the line 3x + 4y – 12 = 0. Also, find the co-ordinates of the foot of
perpendicular.
Solution.
The given line is 3x + 4y – 12 = 0 ... (i)
⇒ 4y = –3x + 12
–3
⇒ y=
x+3
4
3
∴ The slope of the line (i) = –
4
From P (–1, –2), draw PN perpendicular to the given line.
4 ⎡
1⎤
∵
m
∴ The slope of the line PN =
=
–
⎢
2
⎥
m1 ⎦
3 ⎣
4
is
The equation of the line through P (–1, –2) and having slope
3
4
y – (–2) = [x – (–1)]
3
⇒
3y + 6 = 4x + 4
⇒ 4x – 3y – 2 = 0
... (ii)
which is the required equation of the perpendicular from P to the
given line.
To find the co-ordinates of N (the foot of perpendicular), solve (i)
and (ii) simultaneously.
Multiplying (i) by 3 and (ii) by 4, and on adding, we get
44
25x – 44 = 0 ⇒ x =
25
Multiplying (i) by 4 and (ii) by 3, and on subtracting, we get
42
25y – 42 = 0 ⇒ y =
25
⎛ 44 42 ⎞
Hence, the co-ordinates of the foot of perpendicular are ⎜ ,
⎟.
25
25
⎝
⎠
Math Class X
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Question Bank
17. ABCD is a rhombus. The co-ordinates of the points A and C are
(3, 6) and (–1, 2) respectively. Write down the equation of the
diagonal BD.
Solution. We know that the diagonals of a rhombus bisect each
other at right angles.
Let the diagonals AC and BD of the
given rhombus ABCD intersect at M.
⎛ 3 + (– 1) 6 + 2 ⎞
,
⎜
⎟ = (1, 4)
2
2 ⎠
⎝
Now, the diagonal BD is a line passing through (1, 4) and
perpendicular to AC.
y – y2
6–2
4
=1
The slope of AC = 1
=
=
x1 – x2
3 – (– 1)
4
1
∴ The slope of BD = –
[∵ m1 × m2 = – 1]
1
So, by the point-slope form, the equation of BD is
y – y1 = m (x – x1) ⇒ y – 4 = (–1) (x – 1) ⇒ y – 4 = –x + 1
or x + y = 5.
18. Find the equation of a straight line parallel to the line 2x – 5y + 1 = 0
and passing through the point which divides the line joining (–5, 4)
and (3, 1) in the ratio 2 : 1.
Solution. Let P(x, y) be the point which divide the join of
(–5, 4) and (3, 1) in the ratio 2 : 1.
2 × 3 + 1 × (– 5)
2×1+1×4
and y =
Then, x =
2 +1
2 +1
1
⇒ x = = and y = 2
3
Also, equation of the given line is 2x – 5y + 1 = 0
2
1
⇒ y = x+ =0
5
5
So, M = mid-point of AC =
Math Class X
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Question Bank
2
.
5
The required line p is parallel to the given line.
2
∴ Gradient of the required line = .
5
2
⎛1 ⎞
Now, the required line passes through P ⎜ , 2 ⎟ and have gradient
5
⎝3 ⎠
2 ⎛
1⎞
∴ Its equation is given by y – 2 =
–
x
⎜
⎟
5 ⎝
3⎠
⇒ 15y – 30 = 6x – 2 ⇒ 6x – 15y + 28 = 0
19. The points A (2, 3), B (3, 5) and C (–1, –1) are the vertices of the
triangle ABC. Find the equation of the altitude of the triangle
through A.
Solution. Let the slope of the altitude AD be m.
y – y2
–1–5
–6
3
The slope of BC = 1
=
=
=
x1 – x2
–1–3
–4
2
Since AD ⊥ BC,
3
2
∴ m. = –1 ⇒ m = –
2
3
Thus, AD is a line of slope and
passing through the point A (2, 3).
By the point-slope form, the equation
of AD is
2
y – 3 = – (x – 2) or 3 (y – 3) = –2 (x – 2)
3
⇒
3y – 9 = –2x + 4 ⇒ 2x + 3y = 13.
20. The points A (1, 2), B (3, – 4) and C (5, – 6) are the vertices of a
ΔABC. Find the equation of the right bisectors of the sides BC and
CA. Hence, find the circumcentre of ΔABC.
Solution. Let the right bisector of BC intersect it at D and the right
bisector of AC intersect it at E.
⇒ Gradient of the given line, m1 =
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Question Bank
Then, coordinates of D are
⎛3 + 5 – 4 – 6⎞
,
⎜
⎟ i.e., (4, –5)
2 ⎠
⎝ 2
And coordinates of E are
⎛5 + 1 – 6 + 2⎞
,
⎜
⎟ i.e., (3, – 2)
2 ⎠
⎝ 2
–6+4
=–1
Gradient m1 of BC =
5–3
∴ Gradient of right bisector of BC, m2 = 1
[∴ m1 × m2 = –1]
∴ Equation of the right bisector of BC, which passes through
(4, –5) and has gradient 1 is given by y + 5 = 1 (x – 4)
or x – y = 9
2+6
Gradient m3 of BC =
=–1
1–5
1
∴ Gradient m4 of the right bisector of AC =
2
[∴ m3 × m4 = –1]
∴ Equation of the right bisector of AC, which passes through (3, –
1
1
2) and has gradient
is given by y + 2 = (x – 3)
2
2
or 2y + 4 = x – 3
or x – 2y = 7
To find the coordinates of the circumcentre of ΔABC, we solve x – y
= 9 and x – 2y = 7 simultaneously.
x – y
=
9
x – 2y =
7
–
+
–
⇒ y
=
2
Substituting y = 2 in x – y = 9, we get x = 9 + 2 = 11
Hence, the equations of the right bisectors of the sides BC and AC
are x – y = 9 and x – 2y = 7 respectively. The coordinates of the
circumcentre of ΔABC are (11, 2)
Math Class X
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Question Bank
21. Using coordinate geometry, prove that the diagonals of a square
bisect each other perpendicularly.
Solution. Let OABC be the square,
where coordinates of O, A, B and
C are (0, 0), (a, 0), (a, a) and (0, a) respectively.
Coordinates of the mid-point of AC are
⎛a +0 0+a ⎞
⎛a a⎞
,
i,e.,
⎜
⎟
⎜ , ⎟
2 ⎠
⎝ 2
⎝2 2⎠
Also the coordinates of the mid-point OB are
⎛a +0 0+a ⎞
⎛a a⎞
,
i,e.,
⎜
⎟
⎜ , ⎟ i.e., .
2 ⎠
⎝ 2
⎝2 2⎠
⎛a a⎞
⎜ , ⎟ are coordinates of the mid-point of AC as well as OB.
⎝2 2⎠
⎛a a⎞
Hence, AC and OB bisect each other at ⎜ , ⎟ ... (i)
⎝2 2⎠
a–0
=–1
Slope of AC (m1) =
0–a
a–0
=1
Slope of OB (m2) =
a–0
m1 × m2 = –1 × 1 = –1
Hence, AC and OB are perpendicular to each other. ... (ii)
From (i) and (ii), we conclude that the diagonals of a square bisect
each other perpendicularly Proved.
22. Three sides of a parallelogram have the equations
2x – y = 2, x + y = 3 and y = 2x + 3. Find the equation of the fourth
side if it passes through the origin.
Solution. We have
2x – y = 2 ⇒ y = 2x – 2
y = 2x + 3
and x + y = 3 ⇒ y = 3 – x
We see that the gradient of y = 2x – 2 is 2 and that of y = 2x + 3 is
also 2.
⇒ y = 2x – 2 and y = 2x + 3 are parallel sides of the parallelogram.
Math Class X
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Question Bank
Hence, the required line will be parallel to y = 3 – x.
Gradient of y = 3 – x is – 1
⇒ Gradient of the required line is also –1.
The required line also passes through the origin i.e., through
(0, 0).
∴ Equation of the fourth side of the parallelogram is given by
y – 0 = –1 (x – 0)
⇒y+x=0
Math Class X
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Question Bank