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Class 11 Applied Maths Chapter 17 Solutions
Straight Line
EXERCISE- 17.1
Q.1 (i) A is a point on the y-axis whose ordinate is 5 and B is the point (-3, 1). Compute the length of AB.
(ii) The distance between A(1, 3) and B (x, 7) is 5. Find the values of x.
Ans. (i) ATQ, Coordinator of A : (0, 5), Coordinator of B : (-3, 1)
Using distance formula D = √(x2 – x1)^2 + (y2 – y1)^2, we get
Length of AB = √(-3 – 0)^2 + (1 – 5)^2
= √(-3)^2 + (-4)^2
= √9 + 16
= √25
= 5 Units
(ii) ATQ, Coordinate of A : (1, 3) ; Coordinate of B : (x, 7)
Distance between A & B = 5 units.
Using distance formula D = √(x2 – x1)^2 + (y2 – y1)^2, we get
5 = √(x – 1)^2 + (7 – 3)^2
Squaring both sides,
25 = x^2 + 1 – 2x + 4^2
25 = x^2- 2x + 1 + 16
0 = x^2 – 2x + 17 – 25
0 = x^2 – 2x – 8
0 = x^2 – 4x + 2x – 8
0 = x(x – 4) + 2(x – 4)
0 = (x – 4) (x + 2)
Therefore, x = 4 or -2
Q.2 (i) Find the abscissa of points whose ordinate is 4 and which are at a distance of 5 units from (5,0).
(ii) How many points are there on the x-axis whose distance from the point (2, 3) is less than 3 units?
Ans.(i) ATQ, Coordinate of A : (x, 4) ; Coordinate of B : (5, 0)
Distance between A & B = 5 units.
Using distance formula D = √(x2 – x1)^2 + (y2 – y1)^2, we get
5 = √(5 – x)^2 + (0 – 4)^2
Squaring both sides,
25 = 25 + x^2 – 10x + (-4)^2
25 = x^2- 2x + 1 + 16
0 = x^ – 10x + 16
0 = x^2 – 8x – 2x + 16
0 = x(x – 8) – 2(x – 8)
0 = (x – 8) (x – 2)
Therefore, x = 8 or 2
(ii) Let, a point on x-axis, whose distance from point (2, 3) is less than 3 units, be (x, 0)
Using, distance formula, we get,
D = √(x2 – x1)^2 + (y2 – y1)^2
D = √(x – 2)^2 + (0 – 3)^2
D = √(x^2 + 4 – 4x + 9
D = √(x^2 – 4x + 13
ATQ, D < 3
√(x^2 – 4x + 13 < 3
Squaring both sides
x^2 – 4x + 13 < 9
For no real value of ‘x’ the above condition is satisfied.
Hence, for no points on x-axis distance is less than 3 units from the point (2, 3).
Q.3 The points A(0, 3), B(-2, a), and C(-1, 4) are the vertices of a right-angled triangle at A, find the value of a.
Ans. Given, A triangle ABC right angles at A, has coordinates as A(0, 3), B(-2, a), C(-1, 4)
Using the Pythagoras theorem, we get
AB^2 + AC^2 = BC^2
Using distance formula D = √(x2 – x1)^2 + (y2 – y1)^2, we get,
[√(-2 – 0)^2 + (a – 3)^2]^2 + [√(-1 – 0)^2 + (4 – 3)^2]^2 = [√[-2 – (-1)]^2 + (a – 4)^2]^2
(-2)^2 + a^2 + 3^2 – 6a + (-1)^2 + (1)^2 = (-1)^2 +a^2 + 4^2 – 8a
4 + 9 + 1 – 6a = 16 – 8a
8a – 6a = 16 – 14
2a = 2
a = 1
Q.4 In what ratio does the point P (1/2, 6) divide the line segment joining the points A (3, 5) and B (-7,9)?
Ans. Given point P (1/2, 6) divides the line segment AB with coordinates of A (3, 5) & B (-7, 9).
Let, the ratio in which P divides AB be k: 1
Using section formula,
[x = (m1x2 + m2x1)/(m1 + m2) , y = (m1y2 + m2y1)/(m1 + m2)]
1/2 = (k x -7 + 1 x 3)/(k + 1) , 6 = (k x 9 + 1 x 5)/(k + 1)
K + 1 = 2 (-7k + 3) , 6(k + 1) = 9k + 5
K + 1 = -14k + 6 , 6k + 6 = 9k + 5
K + 14k = 6 – 1 , 6 – 5 = 9k – 6k
15k = 5 , 1 = 3k
k = 5/15 , k = 1/3
k = 1/3
Therefore,
The ratio would be 1 : 3 (internally)
Q.5 The mid-point of the line segment joining (2a, 4) and (-2, 3b) is (1, 2a + 1). Find the values of a and b.
Ans. Given the coordinate of mid-point P (1, 2a + 1) of the line segment joining Q (2a, 4) & R (-2, 3b).
Using the mid-point formula,
[x = (x1 + x2)/2 , y = (y1 + y2)/2]
x = (x1 + x2)/2
1 = (2a – 2)/2
2 = 2a – 2
2 + 2 = 2a
4 = 2a
4/2 = a
a = 2…………(i)
y = (y1 + y2)/2
2a + 1 = (4 + 3b)/2
from eq (i), we have a = 2
2 x 2 + 1 = (4 + 3b)/2
5 x 2 = 4 + 3b
10 – 4 = 3b
6/3 = b
Therefore, b = 2
Q.6 (i) The three vertices of a parallelogram taken in order are (-1, 1), (3, 1), and (2, 2) respectively. Find the coordinates of the fourth vertex.
(ii) If two vertices of a parallelogram are (3,2), (-1, 0) and its diagonal meet at (2,-5), find the other two vertices of the parallelogram.
Ans. (i) Given, three vertices of ||gm (ABCD) are A(-1, 1), B(3, 1), C(2, 2).
Let, the coordinates of vertex D be (p, q)
We know that, diagonals of ||gm bisect each other.
Therefore, Mid point of AC = Mid point of BD
Using, Mid point Formula, [x = (x1+x2)/2, y = (y1+y2)/2]
(-1 + 2)/2, (1 + 2)/2 = (3 + p)/2, (1 + q)/2
On equating, we get,
1/2 = (3 + p)/2, 3/2 = (1 + q)/2
1 = 3 + p, 3 = 1 + q
1 – 3 = p, 3 – 1 = q
So, p = -2, q = 2
Coordinates of vertex D would be (-2, 2).
(ii) Given, two vertices of ||gm PQRS are P(3, 2) Q(-1, 0). Also, diagonals of ||gm meet at (2, -5).
Let, the vertices of R & S be (a, b) & (c, d) respectively.
We know that, diagonals of ||gm bisect each other.
Mid point of PR = Mid point QS = (2, -5)
Using, mid point formula, [x = (x1+x2)/2, y = (y1+y2)/2]
(3+a)/2, (2+b)/2 = (-1+c)/2, (0+d)/2 = 2, -5
(3 + a)/2 = 2
3 + a = 4 , (2 + b)/2 = -5, (-1 + c)/2 = 2
a = 4 – 3, 2 + b = -10, -1 + c = 4
a = 1 , b = -10 – 2 = -12, c = 4 + 1 = 5
(0 + d)2 = -5
d = -5 x 2
d = -10
Therefore, Coordinate of R : (1, -12)
Coordinate of S : (5, -10)
Q.7 If (4,-3) and (-9,7) are two vertices of a triangle whose centroid is (1, 4), then find the third vertex.
Ans. Given, △ABC, coordinates of its two vertices are A(4, -3) & B(-9, 7). Coordinates of centroids are (1, 4)
Let, coordinate of third vertex , i.e. C be (p, q)
Using centroid of △ formula, [x = (x1+x2+x3)/3, y = (y1+y2+y3)/3],
1 = [4 + (-9) + p]/3 ; 4 = [-3 + 7 + q]/3
1 x 3 = -5 + p ; 4 x 3 = 4 + q
3 + 5 = p ; 12 – 4 = q
8 = p ; 8 = q
Therefore, the third vertex is (8, 8).
Q.8 Show that the points:
(i) (2,-2), (8, 4), (5,7), (-1, 1) are the vertices of a rectangle.
(ii) (3,2), (0,5), (-3, 2), (0,-1) are the vertices of a square.
Ans. (i) Using, distance formula, we get,
AB = √(8-2)^2+[4-(-2)^2] = √6^2+6^2 = √2×6^2 = 6√2 units
BC = √(5-8)^2+(7-4)^2 = √(-3)^2+(-3)^2 = √9+9 = √2×9 = 3√2 units
CD = √(-1-5)^2+(1-7)^2 = √(-6)^2+(-6)^2 = √36+36 = √2×36 = 6√2 units
DA = √(-1-2)^2+[1-(-2)^2] = √(-3)^2+3^2 = √9+9 = √2×9 = 3√2 units
Here, AB = CD = 6√2 units, BC = DA = 3√2 units
Since, opposites sides are equal , ABCD is ||gm.
AC = √(2-5)^2+(-2-7)^2 = √(-3)^2+(-9)^2 = √9+81 = √90 = √9×10 = 3√10 units
BD = √(-1-8)^2+(1-4)^2 = √(-9)^2+(-3)^2 = √81+9 = √90 = √9×10 = 3√10 units
Since, ABCD is a||gm with equal diagonals
Therefore, ABCD is rectangle.
(ii) Using, distance formula, we get,
PQ = √(0-3)^2+(5-2)^2 = √(-3)^2+(3)^2 = √9+9 = √2×9 = 3√2 units
QR = √(-3-0)^2+(2-5)^2 = √(-3)^2+(-3)^2 = √9+9 = √2×9 = 3√2 units
RS = √(-3-0)^2+(2-(-1)^2 = √(-3)^2+(3)^2 = √9+9 = √2×9 = 3√2 units
SP = √(0-3)^2+(-1-2)^2 = √(-3)^2+(-3)^2 = √9+9 = √2×9 = 3√2 units
Here, All four sides of the quadrilateral PQRS are equal.
Therefore, PQRS is a Rhombus.
PR = √(-3-3)^2+(2-2)^2 = √(-6)^2+0^2 = √36 = 6 units
QS = √(0-0)^2+(-1-5)^2 = √(0)^2+(-6)^2 = √36 = 6 units
Since, PQRS is a Rhombus with equal diagonals.
Therefore, PQRS is a square.
Q.9 Find the coordinates of the point which is at a distance of 2 units from (5, 4) and 10 units from (11,-2).
Ans. ATQ, a point P, with coordinate (x, y) is at a distance of 2 units from A(5, 4) and 10 units from B(11, -2).
Using, distance formula D = √(x2 – x1)^2 + (y2 – y1)^2 between A & P, we get
2 = √(5 – x)^2 + (4 – y)^2
Squaring both sides,
4 = 25 + x^2 – 10x + 16+ y^2 – 8y
0 = x^2 + y^2 – 10x – 8y + 37…………(i)
Similarly, 10 = √(x – 11)^2 + (y – (-2))^2
Squaring both sides,
100 = x^2 + 121 – 22x + (y + 2)^2
0 = x^2 – 22x + y^2 + 4 + 4y + 121 – 100
0 = x^2 + y^2 – 22x + 4y + 25………(ii)
Equating (i) & (ii) we get,
x^2 + y^2 – 10x – 8y + 37 = x^2 + y^2 – 22x + 4y + 25
22x – 10x + 37 – 25 = 4y + 8y
12x + 12 = 12y
On dividing 12 on both sides, we get,
x + 1 = y
Substituting value of y in eq. (ii), we get
0 = x^2 + (x + 1)^2 – 22x + 4(x + 1) + 25
0 = x^2 + x^2 + 1 + 2x – 22x + 4 + 25
0 = 2x^2 – 16x + 30
0 = 2(x^2 – 8x + 15)
x^2 – 8x + 15 = 0
x^2 – 5x – 3x + 15 = 0
x(x – 5) – 3(x – 5) = 0
(x – 5) (x – 3) = 0
x = 5 or 3
If, x = 3, y = 4
If, x = 5, y = 6
Therefore, the coordinates are (3, 4) & (5, 6).
Q.10 Two vertices of an isosceles triangle are (2, 0) and (2,5). Find the third vertex if the lengths of equal sides are 3 units.
Ans. Given, △ABC with coordinate of vertex A(2, 0) & B(2, 5)
Length of equal sides is 3 units.
Let, the coordinate of vertex C be (p, q).
Using, distance formula D = √(x2 – x1)^2 + (y2 – y1)^2, we get,
AB = √(2 – 2)^2 + (5 – 0)^2 = √0^2 + 5^2 = √5^2 = 5 units.
Since, length of AB ≠ 3 units
Therefore, AC = BC = 3 units.
AC = √(p – 2)^2 + (q – 0)^2
AC = √p^2 + 4 – 4p + q^2
Similarly, BC = √(p – 2)^2 + (q – 5)^2
BC = √p^2 + 4 – 4p + q^2 + 25 – 10q
Now, AC = BC (Proved above)
√p^2 + q^2 – 4p + 4 = √p^2 + q^2 – 4p – 10q + 29
p^2 + q^2 – 4p + 4 = p^2 + q^2 – 4p – 10q + 4 + 25
0 = – 10q + 25
10q = 25
q = 25/10 = 5/2
We know that, AC = 3units
√p^2 + 4 – 4p + q^2 = 3
Squaring both sides,
p^2 + 4 – 4p + q^2 = 9
Put q = 5/2
p^2 – 4p + 4 + (5/2)^2 = 9
p^2 – 4p + 4 = 9 – 25/4
p^2 – 4p + 4 = (36 – 25)/4
4p^2 – 16p + 16 = 11
4p^2 – 16p + 16 – 11 = 0
4p^2 – 16p + 5 = 0
Q.11 (i) In what ratio is the line joining the point, s (3, 4) and (-2, 1) divided by the y-axis? Also, find the coordinates of the point of division.
(ii) Find the ratio in which the point P(k, 6) divides the line segment joining the points A(-4, 3) and B (2, 8). Also, find the value of k.
Ans. (i) ATQ, line joining points P(3, 4) and Q(-2, 1) is divided by y- axis.
Hence, coordinate of point of division would be (0, y).
Let, the ratio of division PQ be k : 1.
Using Section formula, we get,
0 = [k x (-2) + 1 x 3]/(k + 1)
0 = -2k + 3
2k = 3
Therefore, k = 3/2
Ratio of division = 3 : 2 (internally)
Now, y = [3 x 1 + 2 x 4]/(3 + 2)
= (3 + 8)/5
= 11/5
Therefore, coordinate of point of division would be (0, 11/5).
(ii) ATQ, Point P(k, 6) divided the line segment joining A(-4, 3) & B(2, 8).
Let, the ratio of division be k : 1.
Using, Section formula, we get,
y = [m1y2 + m2y1]/[m1 + m2]
6 = [k x 8 +1 x 3]/[k + 1]
6 (k + 1) = 8k + 3
6k + 6 = 8k + 3
6 – 3 = 8k – 6k
3 = 2k
Therefore, k = 3/2
Ratio of division = 3 : 2 (Internally)
Now, k = [3 x 2 + 2 x (-4)]/[3 + 2]
= [6 + (-8)]/5
= -2/5
So, k = -2/5
Q.12 If A(-1, 3), B(1,-1) and C(5, 1) are three vertices of a triangle ABC, find the length of the median through A.
Ans. Given, ΔABC with vertices A(-1, 3), B(1, -1), & C(5, 1).
To find length of median through A, we need mid point of BC.
Using, x = (x1 + x2)/2, y = (y1 + y2)/2 (mid point formula), we get,
x = (1 + 5)/2, y = (-1 + 1)/2
x = 6/2, y = 0/2
x = 3, y = 0
Therefore, coordinate of D (Mid point of BC) = (3, 0)
Length of AD (Median through A) = √(x2 – x1)^2 + (y2 – y1)^2
AD = √[3-(-1)]^2+[0+3]^2
= √[3+1]^2+[-3]^2
= √4^2 + 9
= √16+9
= √25
= 5 units.
Q.13 If a vertex of a triangle is (1, 1) and the mid-points of the two sides through this vertex are (-1, 2) and (3,2), then find the coordinates of the centroid of the triangle.
Ans. Given, coordinate of vertex B & C be (p, q) & (r, s) respectively.
Using, mid point formula, we get,
For AB : Mid point : (-1, 2), Coordinate of A : (1, 1)
Coordinate of B : (p, q)
-1 = (1 + p)/2, 2 = (1 +q)/2
-2 = 1 + p ; 4 = 1 + q
-2 – 1 = p ; 4 – 1 = q
-3 = p ; 3 = q
Coordinate of vertex B : (-3, 3)
For AC : Mid point : (3, 2)
Coordinate of A : (1, 1)
Coordinate of C : (r, s)
3 = (1 + r)/2 ; 2 = (1 + s)/2
6 = 1 + r ; 4 = 1 + s
6 – 1 = r ; 4 – 1 = s
5 = r ; 3 = s
Coordinate of AC : (5, 3)
Coordinate of centroid :
(x1 + x2 + x3)/3, (y1 + y2 + y3)/3
x = [1 + (-3) + 5]/3, y = [1 + 3 + 3]/3
x = (6 – 3)/3 = 3/3 = 1
y = 7/3
Q.14 Find the coordinates of the center of the circle inscribed in a triangle whose angular points are (-36, 7), (20, 7) and (0,-8).
Ans. Using distance formula D = √(x2 – x1)^2 + (y2 – y1)^2, we get
c = AB = √(20 + 36)^2 + (7 – 7)^2 = √56^2 + 0^2 = √56^2 = 56 units
b = AC = √(0 + 36)^2 + (-8 – 7)^2 = √36^2 + (-15)^2 = √1296 + 225 = √1521 39 units
a = BC = √(20 – 0)^2 + (7 + 8)^2 = √20^2 + 15^2 = √400 + 225 = √625 = 25 units
Coordinate of incentre (p, q)
p = (ax1 + bx2 + cx3)/(a + b + c)
= (25 x -36 + 39 x 20 + 56 x 0)/(25 + 39 + 56)
= (-900 + 780)/120
= -120/120 = 1
q = (ay1 + by2 + cy3)/(a + b + c)
= (25 x 7 + 39 x 7 + 56 x 8)/(25 + 39 + 56)
= (175 + 273 – 448)/120
= 0/120 = 0
Q.15 Find the area of the triangle formed by the points (p + 1, 1), (2p+1,3), (2p+2, 2p) and show that these points are collinear if p = 2 or -1/2.
Ans. Using, Area = 1/2|[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]| (Area of triangle formula), we get,
Area = 1/2|[(p + 1)(3 – 2p) + (2p + 1)(2p – 1) + (2p + 2)(1 – 3)]|
Area = 1/2|[3p + 3 – 2p^2 – 2p + 4p^2 – 1 – 4p – 4]|
Area = 1/2|[2p^2 – 3p – 2]| sq. units
Now, put p = 2
Area = 1/2|[2 x 2^2 – 3 x 2 – 2]|
Area = 1/2 |[8 – 6 – 2]|
Area = 1/2 x 0 = 0 sq. units
Therefore, points are collinear p = 2
Now, put p = -1/2
Area = 1/2|[2 x (-1/2)^2 – 3 x -1/2 – 2]|
Area = 1/2|[2 x 1/4 + 3/2 – 2]|
Area = 1/2|[4/2 – 2]|
Area = 1/2|[2 – 2]|
Area = 1/2 x 0 = 0 sq. units
Q.16 Find the area of the quadrilateral, the coordinates of whose vertices are (2, 1), (6, 0), (5,-2) and (-3,-1).
Ans. Using, Area of triangle formula, we get,
Area of ABC = 1/2|2(0+2) + 6(-2-1) + 5(1 – 0)|
= 1/2|4 – 18 + 5|
= 1/2 |-9|
= 9/2
Area of ADC = 1/2|2(-1+2) – 3(-2-1) + 5(1+1)|
= 1/2|2 + 9 + 10|
= 21/2
Area of quadrilateral = 9/2 + 21/2
= 30/2 = 15 sq. units.
Q.17 The coordinates of A, B, and C are (6,3), (-3, 5) and (4,-2) respectively, and P is any point (x, y). Show that the ratio of the area of the triangles PBC and ABC is |(x+y-2)/7|.
Ans. Using, Area of triangle formula, we get,
Area of ABC = 1/2|[6(5+2)+(-3)(-2-3)+4(3-5)|
Area of ABC = 1/2|6×7+(-3)(-5)+4x-2|
Area of ABC = 1/2|[42+15-8]|
= 1/2|49|
Area of PBC = 1/2|[x(5+2)+(-3)(-2-y)+4(y-5)|
Area of PBC = 1/2|[7x+6+3y+4y-20]|
Area of PBC = 1/2|[7x+7y-14]|
Now, Area of PBC/Area of ABC = {1/2|[7x+7y-14]|}/{1/2|[49]|}
Area of PBC/Area of ABC = |7(x+y-2)/49|
= |(x + y – 2)/7|
Q.18 A, B are two points (3, 4) and (5,-2), find a point P such that | PA | = | PB | and area of APAB = 10.
Ans. Let, the coordinate of point P be (x, y)
ATQ, |PA| = |PB|
Using, distance formula, we get,
√(x-3)^2+(y-4)^2 = √(x-5)^2+(y+2)^2
x^2 + 9 – 6x + y^2 + 16 – 8y = x^2 + 25 – 10x + y^2 + 4 + 4y
10x – 6x + 25 = 25 + 8y + 4y + 4
4x = 12y + 4
So, x = 3y + 1……………(eq. i)
Area of Triangle = 1/2|[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
10 = 1/2|[x(4+2) + 3(-2-y) + 5(y-4)|
20 = |[6x – 6 – 3y + 5y – 20]|
20 = |[6(3y+1) – 6 + 2y – 20]| [From eq i]
20 = |[18y + 6 – 6 + 2y – 20]|
20 = |20y – 20]|
±20 = 20 (y – 1)
±1 = y – 1
Taking + sign,
1 = y – 1,
y = 2, x = 7 (From eq i)
Taking – sign,
-1 = y – 1
-1 + 1 = y
y = 0, x = 1 (from eq i)
Therefore, two points are (7, 2) & (1, 0).
FAQ’s related to Class 11 Applied Maths Chapter 17 on Straight Line:
Q.1 What is the general equation of a straight line?
Ans. The general equation of a straight line is given by: Ax + By + C = 0 where A, B, and C are constants, and x and y are variables representing the coordinates of any point on the line.
Q.2 What is the slope-intercept form of a straight line?
Ans. The slope-intercept form of a straight line is: y = mx + c where m is the slope of the line, and c is the y-intercept (the point where the line crosses the y-axis).
Q.3 How do you find the slope of a line given two points?
Ans. The slope mmm of a line passing through two points (x1, y1) and (x2, y2) is calculated using the formula: m = (y2 – y1)/(x2 – x1)
Q.4 What is the equation of a horizontal line?
Ans. The equation of a horizontal line is : y = k where k is a constant representing the y-coordinate of all points on the line.
Q.5 What is the equation of a vertical line?
Ans. The equation of a vertical line is: x = k where k is a constant representing the x-coordinate of all points on the line.
These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 17
In Class 11 Applied Maths chapter 17, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 17 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 17, we delve deep into advanced mathematical concepts that are crucial for understanding.
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