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1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (−5, 7), (−1, 3)
(iii) (a, b), (− a, − b)
Solution:
(i) Distance between the two points is given by
√[(x₂ – x₁ )² + (y₂ – y₁ )² ]
Therefore, distance between (2, 3) an (4, 1) is given by,
d = √[(2–4)²+(3–1)²] = √[(2)²+(2)²] = √(4 + 4) =
√8 = 2√2 unit,
(ii) Distance between (-5, 7) an (-1, 3) is given by,
d = √[(-5 – (-1))² + (7 – 3)² ]
= √[(-4)² + (4)² ]
= √(16 + 16) = √32 = 4√2 unit,
(iii) Distance between (a, b) an (-a, -b) is given by,
d = √[(a – (-a))² + (b – (-b))² ]
= √[(2a)² + (2b)² ]
= √(4a + 4b ) = 2√(a +b )
2. Find the distance between the points (0, 0) and (36,15).
Solution:
Distance between points (0, 0) and (36, 15).
√[(36 – (0))² + (15 – 0)² ]
= √[(36)² + (15)² ]
= √(1296 + 225) = √1521 = 39unit.
3. Determine if the points (1, 5), (2, 3) and (−2,−11) are collinear.
Solution:
Let the points (1, 5), (2, 3), and (−2, −11) be
representing the vertices A, B, and C of the given
triangle respectively.
Let A = (1, 5), B = (2, 3), and C = (−2, −11)
AB = √[(1 – 2)² + (5 – 3)² ] = √[(-1)² + (2)² ]
= √(1 + 4) = √5 unit.
BC = √[(2 – (-2))² + (3 – (-11))² ] = √[(4)² + (14)² ]
= √(16 + 196) = √212 unit
CA = √[((-2) – 1)² + ((-11) – 5)² ] = √[(3)² + (16)² ] = √(9 + 256) = √265 unit.
Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.
4. Check whether (5, − 2), (6, 4) and (7, − 2) are the vertices of an isosceles triangle.
Solution:
Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given triangle respectively.
Let A = (5, -2), B = (6, 4), and C = (7, −2)
AB = √[(5 – 6)² + ((-2) – 4)²] = √[(-1)² + (-6)²]
= √(1 + 36) = √37 unit
BC = √[(6 – (7))² + (4 – (-2))² ]
= √[(-1) + (6) ] = √(1 + 36) = √37 unit
CA = √[(7 – 5) + ((-2) – (-2)) ] = √[(2) + (0) ]
= √(4 + 0) = 2unit.
Therefore, AB = BC
Thus ABC is an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends.
AB = √[(3 – 6)² + (4 – 7)² ] = √[(-3)² + (-3)² ]
= √(9 + 9)
= √18 unit,
BC = √[(6 – (9))² + (7 – 4)² ]
= √[(-3)² + (3)² ] = √(9 + 9)
= √18 unit
CD = √[(9 – 6)² + (4 – 1)² ] = √[(3)² + (3)² ]
= √(9 + 9)
= √18 unit
DA = √[(6 – (3))² + (1 – 4)² ] = √[(3)² + (-3)² ]
= √(9 + 9)
= √18 unit
Diagonal AC = √[(3 – 9) + (4 – 4) ] = √[(-6) + (0) ] = √(36 + 0) = 6 unit.
Diagonal BD = √[(6 – 6) + (7 – 1) ] = √[(0) + (6) ] = √(0 + 36) = 6 unit
It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length. Therefore, ABCD is a square and
hence, Champa was correct
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (− 1, − 2), (1, 0), (− 1, 2), (− 3, 0)
(ii) (− 3, 5), (3, 1), (0, 3), (− 1, − 4)
(iii)(4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let the points (−1, −2), (1, 0), (−1, 2), and
(−3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
AB = √[((-1) – (1))² + ((-2) – 0)² ]
= √[(-2)² + (-2)² ]
= √(4 + 4) = √8 = 2√2 unit
BC = √[((1) – (-1))² + ((0) – 2)² ] = √[(2)² + (-2)² ]
= √(4 + 4) = √8 = 2√2 unit
CD = √[((-1) – (-3))² + ((2) – (0))² ]
= √[(-2)² + (2)² ]
= √(4 + 4) = 2√2 unit
DA = √[((-3) – (-1))² + ((0) – (-2))² ]
= √[(-2)² + (-2)² ]
=√(4 + 4) = √8 = 2√2
Diagonal AC = √[((-1) – (-1))² + ((-2) – 2)² ]
= √[(0)² + (-4)² ] = √(0 + 16) = √16 = 4 unit
Diagonal BD = √[((1) – (-3))² + ((0) – 0)² ]
= √[(4)² + (0) ] = √16 = 4.
It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices are of a square.
(ii) Let the points (− 3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
AB = √[((-3) – (3))² + ((5) – 1)² ]
= √[(-6)² + (4)² ]
= √(36 + 16) = √52 = 2√13 unit
BC = √[((3) – (0))² + ((1) – 3)² ] = √[(3)² + (-2)² ]
= √(9 + 4) = √13 unit
CD = √[(0 – (-1)) + (3 – (-4)) ]
= √[(1) + (7) ]
= √(1 + 49)
= √50 = √(25×2) = 5√2 unit
DA = √[((-1) – (-3))² + ((-4) – 5)² ]
= √[(2)² + (-9)² ]
= √(4 + 81)
= √85 unit.
It can be observed that all sides of this quadrilateral arebof different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as
square, rectangle, etc.
(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be
representing the vertices A, B, C, and D of the given quadrilateral respectively.
AB = √[((4) – (7))² + ((5) – 6)² ] = √[(-3)² + (-1)² ]
= √(9 + 1) = √10
BC = √[(7 – 4)² + (6 – 3)² ]
= √[(3)² + (3)² ] = √(9 + 9) = √18 unit.
CD = √[(4 – 1)² + (3 – 2)² ] = √[(3)² + (1)² ]
= √(9 + 1) =√10 unit.
DA = √[(1 – 4)² + (2 – 5)² ] = √[(-3)² + (-3)² ]
= √(9 + 9)
= √18 unit.
Diagonal AC = √[((4) – (4))² + ((5) – 3)² ]
= √[(0) + (2) ] = √(0 + 4) = √4 = 2 unit.
Diagonal BD = √[(7 – 1)² + (6 – 2)² ]
= √[(6)² + (4)² ] = √(36 + 16) = √52 = 13√2 unit.
It can be observed that opposite sides of this
quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.
7. Find the point on the x-axis which is equidistant from (2, − 5) and (− 2, 9).
Solution:
We have to find a point on x-axis. Therefore, its y-coordinate will be 0. Let the point on x-axis be (x, 0)
By the given condition, these distances are equal in measure.
√[(x-2)²+(5)²] = √[(x+2)²+9²]
(x-2)² + 25 = (x+2)² + 81
x² + 4 – 4x + 25 = x² + 4 + 4x +81
8x = 25 – 81
8x = -56
x = -7
Therefore, the point (-7, 0)
8. Find the values of y for which the distance between the points P (2, − 3) and Q (10, y) is 10 units.
Solution:
It is given that the distance between (2, − 3) and Q (10, y) is 10.
⇒√[(2 – 10)² + (-3 – y)²] = 10²
⇒(-8)² + (y + 3)² = 10²
⇒64 + (y + 3)² = 100
⇒(y + 3)² = 100 – 64 = 36
⇒y + 3 = ±6
⇒y = +6 – 3 = +3 OR y = -6 – 3 = -9
Therefore, y = -9 or +3.
9. If Q (0, 1) is equidistant from P (5, − 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
Solution:
PQ = QR
√[(5 – 0)²+(-3 -1)²] = √[(0 – x)²+(1 – 6)²]
√[5²+(-4)²] = √[ x²+(-5)²]
√[25 + 16] = √[ x² + 25]
41 = x² + 25
16 = x²
x = 4
Therefore, point R is (4, 6) or (-4, 6)
When point R is (4,6)
PR = √[((5) – (4))² + ((-3) – 6)² ]
= √[(9)² + (-9)² ]
= √(81 + 81) = 9√2 unit
QR = √[((0) – (4))² + ((1) – 6)² ]
= √[(-4)² + (-5)² ]
= √(16 + 25) = √41 unit.
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (− 3, 4).
Solution:
Point (x, y) is equidistant from (3, 6) and (−3, 4).
√[(x – 3)²+(y – 6)²] = √[(x – (-3))²+(y – 4)²]
⇒[(x – 3)² + (y – 6)²] = [(x + 3))² + (y – 4)²]
⇒x²–6x+9+y² -12y+36 = x²+6x+9 + y² – 8x+16
⇒-6x -12y + 45 = 6x – 8x + 25
⇒20 = 12x + 4y
⇒3x + y = 5
⇒3x + y – 5 = 0
1. Find the slope of the curve whose inclination is
(i) 90˚
Solution:
θ = 90˚
Slope = m = tan θ
m = tan 90˚
m = undefined
(ii) 45˚
Solution:
θ = 45˚
Slope = m = tan θ
m = tan 45˚
m = 1
(iii) 30˚
Solution:
θ = 30˚
Slope = m = tan θ
m = tan 30˚
m = 1/√3
(iv) 0˚
Solution:
θ = 0˚
Slope = m = tan θ
m = tan 0˚
m = 0
2. Find the angles of inclination of straight lines whose slopes are
(i)√3
Solution:
Slope = tan θ
√3 = tanθ
We know that, tan 60˚ = √3
θ = 60˚
(ii) 1
Solution:
Slope = tan θ
1 = tan θ
We know that, tan 45˚ = 1
θ = 45˚
(iii) 1/√3
Solution:
Slope = tan θ
1/√3 = tanθ
We know that, tan 30˚ = 1/√3
θ = 30˚
3. Find the slope of the line joining the points
(i) (-4, 1) and (-5, 2)
Solution:
Let (x₁ ,y₁ ) = (-4, 1) and (x₂ ,y₂ ) (-5, 2)
Slope = (x₂-x₁)/(y₂-y₁) = (-5+4)/(2-1) = -1/1 = -1
(ii) 4, -8) and (5, -2)
Solution:
Let (x ,y ) = (4, -8) and (x ,y ) = (5, -2)
Slope = (x₂-x₁)/(y₂-y₁) = 6
(iii) (0, 0) and (√3, 3)
Solution:
Let (x ,y ) = (0, 0) and (x ,y ) = (√3, 3)
Slope = (x₂-x₁)/(y₂-y₁) = √3
(iv) (-5, 0) and (0, -7)
Solution:
Let (x ,y ) = (-5, 0) and (x ,y ) = (0, -7)
Slope = (x₂-x₁)/(y₂-y₁) = 5/-7
(v) (2a, 3b) and (a, -b)
Solution:
Let (x ,y ) = (2a, 3b) and (x ,y ) = (a, -b)
Slope = (x₂-x₁)/(y₂-y₁)= (a-2a)/(-b-3b)= -a/-4b= a/4b
4. Find whether the lines drawn through the two pairs of points are parallel or perpendicular
(i)(5, 2), (0, 5) and (0, 0), (-5, 3)
Solution:
For line 1:
Let (x ,y ) = (5, 2) and (x ,y )= (0, 5)
Slope = m₁ = (x₂-x₁)/(y₂-y₁) =
For line 2:
Let (x ,y ) = (0, 0) and (x ,y )= (-5, 3)
Slope = m₂ = / = / = /
We have m₁ = m₂
Therefore, (5, 2), (0, 5) and (0, 0), (-5, 3) lines are
parallel.
(ii) (3, 3), (4, 6) and (4, 1), (6, 7)
Solution:
For line 1:
Let (x ,y ) = (3, 3) and (x ,y )= (4, 6)
Slope = m₁ = / = / = / = 3
For line 2:
Let (x ,y ) = (4, 1) and (x ,y )= (6, 7)
Slope = m₂ = / = / = / = 3
We have m₁ = m₂
Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) parallel.
(iii) (4, 7), (3, 5) and (-1, 7), (1, 6)
Solution:
For line 1:
Let (x ,y ) = (4, 7) and (x ,y )= (3, 5)
Slope = m₁ = / = / = / = 2
For line 2:
Let (x ,y ) = (-1, 7) and (x ,y )= (1, 6)
Slope = m₂ = / = / = /
We have m₁ ≠ m₂
m₁ .m₂ = 2 x ( / ) = – 1
Therefore, (4, 7), (3, 5) and (-1, 7), (1, 6) are
perpendicular.
(iv) (-1, -2), (1, 6) and (-1, 1), (-2, -3)
Solution:
For line 1:
Let (x ,y ) = (-1, -2) and (x ,y )= (1, 6)
Slope = m₁ = / = / = / = 4
For line 2:
Let (x ,y ) = (-1, 1) and (x ,y )= (-2, -3)
Slope = m₂ = / = / = / = 4
We have m₁ = m₂
Therefore, (3, 3), (4, 6) and (4, 1), (6, 7) are parallel.
5. Find the slope of the line perpendicular to the line
joining the points
(i)(1, 7) and (-4, 3)
Solution:
Let (x ,y ) = (1, 7) and (x ,y )= (-4, 3)
Slope = m = / = / = / = /
The slope of the line perpendicular to the line joining
points (1, 7) and (-4, 3) is ¹/ₘ .[since m₁ .m₂ = -1]
(ii) (2, -3) and (1, 4)
Solution:
Let (x ,y ) = (2, 3) and (x ,y )= (1, 4)
Slope = m = / = / = / = -1
The slope of the line perpendicular to the line joining
points (2, -3) and (1, 4) is 1. [Since m .m = -1]
6. Find the slope of the line parallel to the line joining
the points
(i) (-4, 3) and (2, 5)
Solution:
Let (x ,y ) = (-4, 3) and (x ,y )= (2, 5)
Slope = m = / = / = / = /
The slope of the line parallel to the line joining points
(-4, 3) and (2, 5) is / .
(ii) (1, -5) and (7, 1)
Solution:
Let (x ,y ) = (1, -5) and (x ,y )= (7, 1)
Slope = m = / = / = / = 1
The slope of the line parallel to the line joining points (1,
-5) and (7, 1) is 1.
7. A line passing through the points (2, 7) and (3, 6) is
parallel to a line joining (9, a) and (11, 3). Find a.
Solution:
Line 1:
Let (x ,y ) = (2, 7) and (x ,y )= (3, 6)
Slope = m = / = / = / = -1
Line 2:
Let (x ,y ) = (9, a) and (x ,y )= (11, 3)
Slope = m = / = / = /
Since m = m as line (2, 7) and (3, 6) is parallel to (9,
a) and (11, 3).
We have, -1 = /
-2 = 3 – a
-2 – 3 = – a
– 5 = -a
a = 5
8. A line joining through the points (1, 0) and (4, 3) is
perpendicular to the line joining (-2, -1) and (m, 0).
Find the value of m.
Solution:
Line 1:
Let (x ,y ) = (1, 0) and (x ,y )= (4, 3)
Slope = m = / = / = / = 1
Line 2:
Let (x ,y ) = (-2, -1) and (x ,y )= (m, 0)
Slope = m = / = / = /
Since m m = -1 as A line joining through the points (1,
0) and (4, 3) is perpendicular to the line joining (-2, -1)
and (m, 0).
We have, m .m = -1
1.( / ) = -1
/ = -1
1 = -(m+2)
1 = -m – 2
-m = 1 + 2
m = -3
2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Solution:
Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Let us assume base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a),
while the coordinates of point B are (0, –a).
It is known that the line joining a vertex of an equilateral
triangle with the mid-point of its opposite side is
perpendicular.
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ∆AOC, we obtain
(AC)² = (OA)² + (OC)²
⇒ (2a)² = (OA)² + a²
⇒ 4a² – a² = (OA)²
⇒ (OA)² = 3a²
⇒ OA =√3a
∴Coordinates of point A = (±√3a, 0)
Therefore, the vertices of the given equilateral triangle
are
(0, a), (0, –a), and (√3a, 0)
or
(0, a), (0, –a), and (-√3a, 0).
3. Find the distance between P(x , y ) and Q(x , y )
when:
(i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.
Solution:
The given points are P(x₁ , y₁ ) and Q(x₂ , y₂ )
(i) When PQ is parallel to the y-axis, x₁ = x₂ .
In this case, distance between P and Q
= √[(x₂ – x₁ )² +(y₂ – y₁ )² ]
= √(y₂ – y₁ )²
=|y₂ – y₁ |
(ii)When PQ is parallel to the x-axis, y = y .
In this case, distance between P and Q
= √[(x₂ – x₁ )² +(y₂ – y₁ )² ]
= √(x₂ – x₁ )²
=|x₂ – x₁ |.
4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let (a, 0) be the point on the x-axis that is equidistant from the points (7, 6) and (3, 4).
√[(7-a)² +(6-0)² ] = √[(3-a)² +(4 – 0)² ]
⇒ √[49+a² -14a +36] = √[9+a² – 6a + 16]
⇒ √[a – 14a + 85] = √[a – 6a + 25]
On squaring both sides, we obtain
a – 14a + 85 = a – 6a + 25
⇒ –14a + 6a = 25 – 85
⇒ –8a = –60
a = ⁶⁰/₈ = ¹⁵/₂
Thus, the required point on the x-axis is ( ¹⁵/⁴ , 0).
5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).
Solution:
The coordinates of the mid-point of the line segment joining the points P (0, –4) and B (8, 0) are
( ⁰⁺⁸/₂ , ⁻⁴⁺⁰/₂ ) = (4, -2).
It is known that the slope (m) of a non-vertical line passing through the points (x₁ , y₁ ) and (x₂ , y₂ ) is given by
m = (y₂ – y₁ )/(x₂ – x₁ ) ,
Therefore, the slope of the line passing through (0, 0) and (4, –2) is
(-2-0)/(4-0) = ⁻²/₄ = ⁻¹/₂
Hence, the required slope of the line is ⁻¹/₂ .
6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Solution:
Given: The vertices of the triangle are A (4, 4), B (3, 5), and C (–1, –1).
We know that the slope (m) of a non-vertical line
passing through the points (x₁ , y₁) and (x₂ , y₂) is m = (y₂ – y₁ )/(x₂ – x₁ ) ,
Slope of AB (m₁ ) = ¹/-1 = -1
Slope of BC (m₂ ) = ⁻⁶/-4 = ³/₂
Slope of CA (m₃ ) = ⁵/₅ = 1
It is observed that m₁.m₃ = –1
This shows that line segments AB and CA are
perpendicular to each other i.e., the given triangle is
right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and (–1, –1) are the
vertices of a right-angled triangle.
7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Solution:
If a line makes an angle of 30° with the positive
direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° + 30° = 120°.
Thus, the slope of the given line is tan 120°
= tan(180°– 60°) = –tan 60° = – √3
8. Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.
Solution:
If points A (x, –1), B (2, 1), and C (4, 5) are collinear,then
Slope of AB = Slope of BC
⇒(1+1)/(2-x) = (5-1)/(4-2)
⇒ 2/(2-x) = 4/2
⇒ 2/(2-x) = 2
⇒ 2 = 4 – 2x
⇒ 2x = 2
⇒ x = 1
Thus, the required value of x is 1.
9. Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.
Solution:
Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be
respectively denoted by A, B, C, and D.
Slope of AB = ¹/₆
Slope of CD = ¹/₆
Slope of AB = Slope of CD
Thus AB and CD are parallel to each other
Now slope of BC = -3
Slope of AD = –3
⇒ Slope of BC = Slope of AD
⇒ BC and AD are parallel to each other.
Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.
Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.
10. Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).
Solution:
The slope of the line joining the points (3, –1) and (4, –2) is m = -1
Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by tan θ = –1
⇒ θ = tan⁻¹(-1) = (90° + 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.
11. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.
Solution:
Let m₁ and m₂ be the slopes of the two given lines such
that m₁ = 2m₂.
We know that if θ is the angle between the lines l₁ and l₂ with slopes m₁ and m₂ , then
1+ 2m₂² = ±3m₂
taking ⁺sign,
2m₂² – 3m₂ + 1 = 0
2m₂² – 2m₂ – m₂ + 1 = 0
2m₂(m₂ – 1) – 1(m₂ – 1) = 0
(m₂ – 1)(2m₂ – 1) = 0
m₂ = 1 or m₂ = ¹/₂
If m = 1, then slopes of the lines are 1 and 2.
If m=¹/₂, then slopes of the lines are ¹/₂ and 1.
Hence the slopes of the lines are -1 and -2 or ⁻¹/₂ and -1(by taking negative sign)
or 1 and 2 or ¹/₂ and 1
12. A line passes through (x , y ). IF slope of the line is m , show that k – y = m(h – x )
Solution:
The slope of the line passing through (x , y ) and (h, k) is (k – y )/ (h – x ).
It is given that the slope of the line is m
[(k – y )/ (h – x )] = m
k – y = m(h – x )
13. If three point (h, 0) , (a, b) and (0, k) lie on a line, show that a/h + b/h = 1
Solution:
IF the points A(h, 0) , B(a, b) and C(0, k) lie on a line, then
Slope of AB = slope of BC
⇒(b-0)/(a-h) = (k-b)/(0-a)
⇒ b/(a-h) = (b-k)/a
⇒ ab = (b - k)(a – h)
⇒ ka + bh = kh
On dividing both sides by kh, we obtain
⇒ ka/kh + bh/kh = kh/kh
⇒a/h + b/h = 1.
EXERCISE
Calculate the Following
a. Distance between the point (1,3) and ( 2,4)
b. Mid-point of line segment AB where A(2,5) and B( -5,5)
c. Area of the triangle formed by joining the line
segments (0,0) ,( 2,0) and (3,0)
d. Distance of point (5,0) from Origin
e. Distance of point (5,-5) from Origin
f. Coordinate of the point M which divided the line
segment A(2,3) and B( 5,6) in the ratio 2:3
g. Quadrant of the Mid-point of the line segment A(2,3) and B( 5,6)
h. the coordinates of a point A, where AB is the diameter of circle whose center is (2,−3) and B is (1, 4)
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1. Find a point on the y – axis which is equidistant from the point A (6, 5) and B (-4, 3).
2. Show that the points (p,p), (-p,-p) and (-p√3, p√3) are the vertices of an equilateral triangle. Also find its area.
3. Check whether the points (4,5), (7,6) and (6,3) are collinear.
4. Find the value of q for which the points (7, -2), (5, 1), and (3, q) are collinear.
5. Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also find its area.
6. If two vertices of an equilateral triangle be (0,0), (3,√3), find the third vertex.
7. If P (2, -1), Q (3, 4), R (-2, 3) and S (-3, -2) be four points is a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.
8. Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle.
9. Show that the points A(1, -2), B(3, 6), C(5, 10) and D(3, 2) are the vertices of a parallelogram.
10. Prove that the points A (1,7), B (4,2), C (-1,-1) and D(-4, 4) are the vertices of a square.
11. Prove that the points (3, 0), (6, 4) and (-1, 3) are vertices of a right angled isosceles triangle.
12. Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (12, -6).
13. If two opposite vertices of a square are (5, 4) and (1, -6), find the co- ordinates of its remaining two vertices.
14. Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.
15. Find the value of x such that PQ = QR where the co-ordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.
Answers
7) 54
12) (3,-3)
13) (8, -3) and (-2, 1)
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If the slope = m and the line passes through the point (x₁,y₁), the equation of the line is
y - y₁ = m(x - x₁ ), which is known
as point-slope form of a straight line.
1. Determine the equation of the straight line whose inclination is 60° and which passes through the point (3, 4).
2. Find the equation of the straight line whose slope is 3 and passes through the point (-1, 2)
3. Find the equation of the straight line whose slope is -2 and passes through the point (3, 0)
4. Find the equation of the straight line whose inclination is 45° and passes through the point (4, -1)
5. A straight line passes through the point (- 1, 4) and makes an angle 60° with the positive direction of the x-axis. Find the equation of the straight line.
6. Find the equation of the straight line whose inclination is 60° and passes through the origin.
7. Find the equation of the line which is inclined at 135° with the negative direction of the x-axis and passes through the point (1, 2)
8. Find the equation of the line which is inclined at 45° with the positive direction of the y-axis and passes through the point (3, 2)
9. Find the equation of the straight line parallel to the y-axis and passes through the point (1, 3).
10. Find the equation of the straight line parallel to the y-axis and passes through the point (-2, 3).
11. Find the equation, of the straight line whose inclination is 600 and which passes through the mid-point of the line-segment joining the points (3, - 4) and (7, 8).
Answers:
1. y - 4 = √3(x- 3)
2. 3x - y + 5 = 0
3. 2x + y - 6 = 0
4. x - y - 5 = 0
5. y - 4 = √3(x + 1)
6. √3x - y = 0
7. x - y + 1 = 0
8. x - y - 1 = 0
9. x - 1 = 0
10. x + 2 = 0
11. y - 2 = √3(x - 5)
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If a straight line passes through the points (x₁,y₁) & (x₂,y₂) then its equation is
y - y₁ = [(y₂-y₁)/(x₂-x₁)] (x-x₁),
and the slope of the straight line is :
m= [(y₂-y₁)/(x₂-x₁)]
1. Find the equations of the straight lines joining each of the following pair of points:
(i) (- 3, - 4) and (2, 5)
(ii) (0, b) and (- a, 0)
(iii) (at₁² , 2at₂ ) and (at₂² , 2at₁ )
(iv) (a cos α, a sin α) and (a cos β, a sin β).
2. Find the equation and the slope of the line joining the points
(i) (1, 6), (6, 1)
(ii) (-2, 1), (3, -2)
(iii) Origin and (-3, 1)
(iv) (3, 4), (-2, 4)
(v) (7, 0), (0, 3)
3. Find the equation and the slope of the line joining the points A on the x-axis and B on the y-axis if
(i) OA = 4, OB = 5
(ii) OA = -2, OB= 3
(iii) OA = -1, OB = -2, where O is the origin.
Answers:
1. (i) 9x - 5y + 7 = 0
(ii) bx - ay + ab = 0
(iii) y(t₁ + t₂) - 2x = 2at₁t₂
(iv) xcos[¹/₂(α+β)]+ysin[¹/₂(α+β)]=acos[¹/₂(α+β)]
2. (i) x + y - 7 = 0
(ii) 3x + 5y + 1 = 0
(iii) x + 3y = 0
(iv) y = 4
(v) 3x + 7y - 21 = 0
3. (i) 5x + 4y - 20 = 0
(ii) 3x - 2y + 6 = 0
(iii) 2x + y + 2 = 0
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1. Find the equation of the line
(i) whose slope is 2 and which cuts off an intercept 2 on the y-axis
(ii) whose inclination is 45° and which cuts off an intercept -1 on the y-axis.
2. Find the equation of the line
(i) passing through (1, 3) and making an intercept 5 on the y- axis
(ii) passing through (4, -2) and making an intercept -3 on the y-axis
3. Find the equation of the line
(i) passing through (-2, 5) and cutting the y-axis at A on the positive side of the y-axis such that OA = 4, O being the origin.
(ii) passing through (1, -2) and cutting the y-axis at B on the negative side of the y-axis such that OB = 4, O being the origin.
4. Find the equation of the line parallel to the x-axis at a distance
(i) 8 on the positive side of the y-axis
(ii) 5 on the negative side of the y-axis
5. A and B are two points on the x-axis. A is on the positive side of the x-axis at a distance 5 and B is on the negative side at a distance 3 from the origin O. P is the midpoint of AB. Find the equation of the line PC which cuts an intercept 2 on the y- axis. Also, find the slope of PC.
6. Find the slope and the y-intercept of the line whose equation is:
(i) y = x + 3
(ii) 3y = √3x – 1
(iii) 11x – 5y + 2 = 0
(iv) 2y = 3(x + 1)
7. If the inclination of the line y – 1 = ax + a² is 45°, find its y-intercept.
Answer:
1. (i) y = 2x + 2
(ii) y = x - 1
2. (i) y + 2x = 5
(ii) x – 4y = 12
3. (i) x + 2y = 8
(ii) y = x - 3
4. (i) y = 8
(ii) y + 5 = 0
5. 2x + y = 2; slope = -2
6. (i) slope = 1, y-intercept = 3
(ii) slope = 1/√3, y-intercept = -1/3
(iii) slope = 11/5, y-intercept = 2/5
(iv) slope = 3/2, y-intercept = 3/2
7. 2 unit
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1. Determine the equation of the straight line whose gradient is ⁻³/₂ and which intersects the y-axis at a distance of 4 units from the origin.
2. Find the value of k, given that the line
y=2(x–k) passes through the point (-4, 4).
3. Find the equation of the straight line which makes an angle 450° with the positive direction of x-axis and intersects the x- axis at a distance of (-3) units from the origin.
4. If the line y/2= 3x - 6 passes through the point (p, 2p), find the value of p.
5. Determine the equation of the straight line whose inclination is 60° and which passes through the point (3, 4).
6. Find the point midway between the point
(-1, 3) and the point intersection of the lines
4x + y - 10= 0 and 2x + 3y - 8 = 0.
7. Find the equation of the straight line which cuts off an intercept (- 4) from the x-axis and passes through the point (2, - 1).
8. If the slope of the line joining the points
(2k, - 2) and (1, - k) be (- 2), find k.
9. Find the co-ordinates of the point on the line 7x - 6y = 20 for which the ordinate is double the abscissa.
10 . A straight line passes through the point
(2, 3) and is such that the sum of its intercepts on the coordinate axes is 10.
Find the equation of the straight line.
11. Prove that the line lines x - 2 = 0, x + 1 = 0,
y = 0 and y - 3 = 0 form a square. Find the equations of its diagonals.
Answers:
1. 3x + 2y = 8
2. k = -6
3. x - y + 3 = 0
4. p = 3
5. y - 4 = √3(x - 3)
6. (³/₅ ,∞ )
7. x + 6y + 4 = 0
8. k = ⁴/₅
9. (- 4, - 8)
10 . 3x + 2y = 12 or x + y = 5
11. x + y - 2 = 0 and x - y + 1 = 0
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1. Find the equation of the straight line passing through the points (3, - 4) and (1, 2) and hence show that the three points (3 ,-4), (1, 2 ) and
(2, - 1) are collinear.
2. Show that the points (1, - 1), (5, 5) and (- 3, - 7) are collinear. Also find the equation of the line on which they lie.
3. Prove that the points (2, -3), (1, 2) and (0, 7) are collinear. Also, find the equation of the line on which the points lie.
4. Prove that the points (3, 1), (5, -5) and (- 1, 13) are collinear. Find the equation of the straight line on which the points lie.
5. If the point (h, -2) is collinear with the points (2, 2) and (-3, 1) then find the value of h.
Also, find the slope of the line containing the three points.
6. Show that the points (-1, 3), (0, 2) and (1, 1) are collinear. Also find the equation of the line on which they lie.
Answers:
1. 3x + y = 5
2. 3x - 2y = 5
3. 5x + y - 7 =0
4. 3x + y = 10.
5. h = - 18 and slope = ¹/₅
6. x + y - 2 = 0
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1) Determine the ratio in which the point P (m, 6) divides the join of A (-4, 3) and B (2, 8). Also find the value of m.
2) The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the co-ordinates of P and Q are (p, -2) and (5/3, q) respectively. Find the values of p and q.
3) The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + K = 0. Find the value of K.
4) A (4, 2), B (6, 5) and C (1, 4) are the vertices of Δ ABC.
i)The median from A meets BC in D. Find the co- ordinates
of the point D.
ii)Find the co- ordinates of point P on AD such that AD : PD
= 2 : 1
iii)Find the co- ordinates of the points Q and R on medians
BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF
= 2 : 1
iv) What do you observe?
5) If the points A (6, 1), B (8, 2), C (9, 4) and d (k, p) are the vertices of a parallelogram taken in order, then find the value of k and p.
6) Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the co-ordinates of the third vertex.
7) If C be the centroid of a triangle PQR and X be any other point in the plane, prove that
XP²+ XQ²+ XR² = CP²+CQ²+CR²+3CX²
8) If P(x, y) is any point on the line joining the points A (a, 0), B (0, b), then show that x/a + y/b = -1 9) If (-2, 3) (4, -3) and (4, 5) are the mid points of the sides of a triangle, find
the co-ordinates of its centroid.
10) Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
11) A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the co- ordinates [ 5/3,-1/3 ]. Find the co- ordinates of the third vertex C of the triangle.
12) If D, E and F are the mid points of sides BC, CA and AB respectively of Δ ABC, then testing co- ordinate geometry prove that
Area of Δ DEF = (1/4) (Area of Δ ABC)
13) If A (4, 6), B (3, -2) and C (5, 2) are the vertices of Δ ABC, then verify the fact that a median of a triangle ABC divides it into two triangles of equal areas.
14) Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear.
15) For what value of x will the points (x, -1), (2,1) and (4,5) lie on a line?
16) If the points (p, q) (M, n) and (p – m, q –n) are collinear, show that pn = qm.
17) Find k so tht the point P (-4, 6) lies on the line segment joining A (k, 10) and B (3, -8). Also, find the ratio in which P divides AB.
18) Find the area of the quadrilaterals, the co- ordinates of whose vertices are
i)(-3, 2), (5, 4), (7, -6) and (-5, -4)
ii)(1, 2), (6, 2), (5, 3) and (3, 4)
iii) (-4, -2), (-3, -5), (3, -2), (2, 3)
19) Show that the following sets of points are collinear
i) (2, 5), (4, 6) and (8, 8)
ii) (1, -1), (2, 1) and (4, 5)
20) The area of a triangle is 5. Two of its vertices are (2, 1) and (3, 2). The third vertex lies on y = x + 3. Find the third vertex.
21) Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that find x.
22) If three points (x₁ , y₁), (x₂ , y₂ ), (x₃ , y₃ ) lie on the same line, prove that
(x₂-x₁)(y₃-y₁)=(x₃-x₁)(y₂-y₁)
23) If the point P (m, 3) lies on the line segment joining the points A [-2/5, 6] and B (2, 8), find the value of m.
24) If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a) then prove that x + y = a + b
25) Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units.
26) If G be the centroid of a triangle ABC, prove that AB² + BC² + CA² = 3(GA² + GB² + GC² )
27. Find the co-ordinates of circumcenter of a ? ABC where
A( 1, 2) , B( 3, –4) and C( 5, –6 )
28. The coordinates of the mid-point of the line joining the points (2p + 2, 3) and (4, 2q + 1) are (2p, 2q). Find the value of p and q.
Answer
1. 3 : 2, m = (-2)/5
2. p = 7/3, q = 0
3. K = -8
4. (i) [ 7/2,9/2 ]
(ii) [11/3,11/3 ]
(iii) [ 11/3,11/3 ]
5. k = 7, p = 3
6. (-4, -7)
9. [ 2, 5/3 ]
10. (3, 1)
11. (4, -4)
15. x = 1
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1. Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)
2. Two opposite vertices of a square are (-1, 2) and (3, 2). Find thee co-ordinates of other two vertices.
3. Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus. Also, find its area.
4. Find a point on y – axis which is equidistant from the points (5, -2) and (-3, 2).
5. Find a relation between a and b such that the point (a,b) is equidistant from the points (3, 6) and (-3, 4).
6. If the coordinates of the mid points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of its vertices.
7. The coordinates of one end point of a diameter of a circle are (4, -1) and the co-ordinates of the centre of the circle are (1, -3). Find the co- ordinates of the other end of the diameter.
8. If A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of triangle ABC, find the length of median through A and the co- ordinates of the centroid.
9. If the mid- point of the line joining (3, 4) and (z,7) is (x,y) & 2x + 2y + 1 = 0 find the value of z.
10. Find the ratio in which the line segment joining (-2, -3) and (5, 6) is divided by (i) x – axis (ii) y –axis. Also, find the co- ordinates of the point of division in each case.
11. Prove that (4, 3), (6, 4), (5, 6) and (3, 5) are the angular points of a square.
12. Three vertices of a parallelogram are (a +b, a – b), (2a + b, 2a – b), (a – b, a + b).Find the fourth vertex.
13. If two vertices of a parallelogram are (3, 2), (-1, 0) and the diagonals cut at (2, -5), find the other vertices of the parallelogram.
14. Find the lengths of the medians of a Δ ABC having vertices at A (0, -1), B (2, 1) and C (0, 3).
15. Find the lengths of the medians of a Δ A (5, 1), B (1, 5) and C (-3, -1).
16. Find the co- ordinates of the points which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.
17. Show that following points are vertices of a rectangle:
(a) (2 , –2) , ( 8, 4) , ( 5, 7 ) , (– 1, 1)
(b)(–4 , –1) , (–2 , 4) , ( 4, 0 ) , ( 2, 3 )
18. If centre of circle passing through (a,–8), (b,–9) and (2,1) is (2,–4), find the value of a and b.
Answer
1. (3, -2)
2. (1, 0) and (1, 4)
3. Area = 24 sq. units
4. (0, -2)
5. 3a + b = 5
6. (1, -4), (3, 2), (-1, 2)
7. Co- ordinates are (-2, -5)
8. Median = √65 units, co- ordinate = [ 1/3,5/3 ]
9. z =-15
10 (i) 1 : 2; [1/3,0]
(ii) 2 : 5; [ 0, -3/7 ]
12. (-b, b)
13. (1, -12), (5, -10)
14. AD = √ 10 units, BE = 2units, CF = √ 10 units
15. AD = √ 37 units, BE = 5 units, CF = 2 √13 units
16. (-3, 1.5), (-2, 3), (-1, 4.5)