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NCERT Exemplar Class 11 Maths solutions chapter 10 are prepared by our experts for helping students to understand and grasp the very basics of coordinate geometry. NCERT Exemplar solutions for Class 11 Maths chapter 10 focuses on the representation of a straight line in coordinate geometry. This is done through the determination of the slope of the straight line with reference to the previously studied concept of algebra and geometry.
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NCERT Exemplar Class 11 Maths solutions chapter 10 provide detailed and simple explanations for NCERT problems that help students in their competitive exams as well. The solutions are drafted very carefully by a detailed study of the concepts and exam pattern of CBSE, keeping the important topics in mind for better understanding and preparation.
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Question:1
Answer:
Equation of line in intercept form=
Given that
x+y=a……(1)
If equation (1) passes through (1,-2) we get 1+(-2) =a
1-2=a
a=-1 Putting the value of'a'in equation 1we get x+y=-1
Equation of straight line is x+y+1=0
Question:2
Answer:
Given points are A(5,2) B(2,3) and C(3,-1).
Slopeof the line joining points B and C =
It is given that line passing through the point A is perpendicular to BC
m1*m2= -1
-4*m2=-1
m2=1/4
Equation of line passing through point A
Equation of line: y-y1 = m(x-x1)
y-2=1/4 (x-5)
4y-8=x-5
x-5-4y+8=0
Equation of straight line passing through point A is x-4y+3=0
Question:3
Find the angle between the lines and
Answer:
Given equations are
................(1)
and
............(2)
In equation 1 the slope is as it is in the form of y=mx+b and in equation 2 it is
Let θ be the angle between the given m1 and m2 two lines
Putting the values of m1 and m2 in above equation we get
θ=600
Question:4
Answer:
Equation of line in intercept form =
Given that a+b=14 b=14-a
So equation of line
........................(1)
If equation 1 passes through (3,4) then
42-3a+4a-14a+a2=0
a2-13a+42=0
a2-7a-6a+42=0
(a-6)(a-7)=0
a=6 or a=7
If a=6 then 6+b=14 b=8
If a=7 the 7+b=14 b=7
If a=6 and b=8 then equation of line is
4x+3y=24
If a=7 and b=7 then equation of line
x+y=7
Question:5
Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.
Answer:
Let x1, y1 be any point lying inthe equation x+y=4
x1+y1=4…......(1)
Distance of point x1,y1from the equation 4x+3y=10
4x1+3y1-10=±5
4x1+3y1-10=5 or 4x1+3y1-10=-5
4x1+3y1=(15)...….(2) or 4x1+3y1=5….....(3)
From equation 1 we have y1=4-x1….. (4)
Putting the values of y1in equation 2 we get 4x1+3(4-x1)=15
4x1+12-3x1=15
x1=15-12 =3
Putting the value of x1in equation 4 we get y1=1
Now, 4x1+34-x1=5
4x1+12-3x1=5
x1=5-12 x1=-7
Putting the value in equation 4 we get y1=4--7=4+7=11
Hence, the required points on the given line are (3,1) and (-7,11)
Question:6
Show that the tangent of an angle between the lines and is
Answer:
Equation of line in intercept form ............(i) and ..............(ii)
y=mx+b Slope of equation 1 is
Similarly for equation 2 ,
Since, the above equation is in y=mx+b form
Slope of the equation 2 is
Let θ be the angle between the given two lines
Putting the values of m1 and m2 in above equation we get
Question:7
Find the equation of lines passing through (1, 2) and making angle 30° with y-axis.
Answer:
Given that line passes through (1,2) making an angle of 300 with y axis.
Angle made by line with x axis is 600 Slope of the line, m=
Equation of line passing through x1,y1 and having slope 'm' is y-y1=m(x-x1)
Here, (x1, y1)=(1,2) and m=
Question:8
Answer:
. Given 2x+y=5…. (1) and x+3y=-8….(2)
Firstly, we find the point of intersection of both equations we get and
Hence, the point of intersection is
Now the slope of the equation 3x+4y=7 m=
Then the equation of the line passing through the point having slope is
y-y1=m(x-x1)
Question:9
Answer:
Given equation is ax+by+8=0 or ax+by= -8
Now dividing by-8 to both sides
So the intercepts are and
Now, the second equation which is given is 2x-3y+6=0 or 2x-3y=-6
Dividing by-6 on both sides
So, the intercepts are-3 and 2
Now, according to the question and
and
Question:10
Answer:
. Let a and b be the intercepts on the given line
Coordinates of A and B are (a,0)and (0,b) respectiively.
Using the section formula we find the value of a and b
and
and
and
Coordinates of A and B are and
Equation of line AB
Hence, trequired equation is 8x-5y+60=0
Question:11
Answer:
Given that length of the perpendicular from the origin is 4 units and line makes an angle with positive direction of x-axis
∠BAX=120
∠BA0=180-120=60
∠MAO=60
Now in triangle AMO ∠MAO+∠AOM+∠OMA=1800
600+θ+ 900=1800
θ=300 ∠AOM= 300
x cosθ+ysinθ=p
xcos 300+ysin300=4
Question:12
Answer:
Equation of hypotenuse is 3x+4y=4 and opposite vertex is (2,2)
Slope of the equation of hypotenuse is
Now let the slope of AC be m
Putting the values of m1and m2in the equation
4m+3=4-3m
4m+3m=4-3
7m=1 m=
OR
4m+3=-4+3m
4m-3m=-4-3
m= -7
If m=1/7
equation of AC is y-y1=m(x-x1)
y-2=1/7(x-2)
7y-14=x-2
x-7y-2+14=0
x-7y+12=0
If m= -7 then equation of AC is y-2=-(7)(x-2)
y-2=-7x+14
7x+y=16
The required equations are x-7y+12=0 and 7x+y=16
Question:13
Answer:
Let ABC be an equilateral triangle, BC is base, and altitude from A on BC meets at mid-point D.
Given:Equation of the base BC is x+y=2
Distance of point (x1,y1) from the equation Ax+By+C=0
Now, length of perpendicular from vertex A(2,-1) to the line x+y=2
Squaring both the sides, we get
Question:14
Answer:
Let the variable line be ax+by=1
Length of perpendicular from (2,0) to the line ax+by-1=0
Now perpendicular distance from B(0,2) =
Now, perpendicular distance from C(1,1)=
The algebraic sum of the perpendicular from the given points (2,0), (0,2)
and (1,1) to this line is zero.
d1+d2+d3=0
2a-1+2b-1+a+b-1=0
3a+3b-3=0
a+b-1=0
a+b=1
So, the equation ax+by=1 represents a family of straight lines passing
through a fixed point .
Comparing equation ax+by=1 and a+b=1, we get x=1 and y=1.
Coordinates of fixed point is (1,1)
Question:15
Answer:
Let the given line x+y=4 and the required line'l'intersect at B (a,b)
Slope of line'l'is
Given that
So, by distance formula for point A(1,2) and B(a,b) we get d
Point B(a,b)also satisfies x+y=4
a+b=4; b=4-a
Putting the value of b in equation (i )we get 3a2+3(4-a)2-6a-12(4-a)+13=0
3a2+48+3a2-24a-6a-48+12a+13=0
6a2-18a+13=0
Using the formula
Putting the value of a in the equation we get
Now putting the value of a and b in equation
θ= 600-450=150
Similarly, taking
θ= 600+450=1050
Question:16
Answer:
Equation of line in intercept form=xa+yb=1
where a and b are intercepts on the axes
Given that
Then,
This shows that the line is passing through the fixed point that is (k,k)
Question:17
Answer:
Let the line cut the x axis at (a,0) and Y axis at (0,b) Therefore
It is given that the point (-4,3) divides the line internally in 5: 3 ratio. Hence applying the formula for internal division of line segment, we get and
Hence Using slope intercept form we get
Question:18
Answer:
Given two lines are: x-y+1=0 and 2x-3y+5=0
Solving these two equations gives us points of intersection we get y=3 and x=2
(x,y)=(2,3)
Let m be the slope of the required line
Then, equation of the line is y-3=mx-2
y-3=mx-2m
mx-y-2m+3=0….(1)
Since, the perpendicular distance from the point 3,2to the line is75then
Squaring both the sides, we get
49m2+1=25(m+1)2
49m2+49=25m2+25+50m
25m2+25+50m-49m2-49=0
-24m2+50m-24=0
-12m2+25m-12=0
Factorising, we get (3m-4)(4m-3)=0
3m-4=0 or 4m-3=0
3m=4 or 4m=3
m=4/3 or 3/4
Putting the value of m=4/3 in equation 1 we get 4x/3-y-2(4/3)+3=0
4x/3-y-8/3+3=0
4x/3-y=-1/3
4x-3y+1=0
Putting the value of m=3/4 in the equation we get 3x/4-y-2(3/4)+3=0
3x/4-y-3/2+3=0
3x/4-y+3/2=0
3x-4y+6=0
Hence, the required equation are 4x-3y+1=0 and 3x-4y+6=0
Question:19
Answer:
Let the coordinates of a moving point P be (a,b)
It is given that the sum of the distance from the axes to the point is always 1 |x|+|y|=1
±x±y=1 -x-y=1, x+y=1, -x+y=1 and x-y=1
Hence, these equations gives us the locus of the point P which is a square.
Question:20
Answer:
Given lines are
If x≥0, then ..............(i)
If x<0 then .............(ii)
On adding both the equations we get
2y=4
y=2
Putting the value of y=2 in equation (ii), we get
Point of intersection of given lines is (0,2 )
Now we find the slopes of given lines Slope of equation (i) is
Comparing the above equation with y=mx+b, we get
and we know that
θ= 600
Slope of equation (ii) is , we get
θ=1800- 600= 1200
In ACB,
Hence, the coordinates of the foot of perpendicular
Question:21
Answer:
Equation of line in intercept form=
Since, p is the length of perpendicular drawn from the origin to the gievn line p=
Squaring both the sides, we have
Since, a2,b2 and p2 are in AP
2p2=a2+b2
From equation (i) and (ii) we get
Hence, proved.
Question:22
A line cutting off intercept – 3 from the y-axis and the tangent at angle to the x-axis is 3/5, its equation is
A. 5y – 3x + 15 = 0
B. 3y – 5x + 15 = 0
C. 5y – 3x – 15 = 0
D. None of these
Answer:
. Given that tanθ=3/5
We know that slope of a line, m=tanθ Slope of line, m=3/5
Since, the lines cut off intercepts -3 on y axis so the line is passing
through the point (0,-3)
So, the equation of line is y-y1=m(x-x1)
y-(-3)=3/5(x-0)
y+3=3x/5
5y+15=3x
5y-3x+15=0
Hence, the correct option is (a)
Question:23
Slope of a line which cuts off intercepts of equal lengths on the axes is
A. – 1
B. – 0
C. 2
D.
Answer:
Equation of line in intercept form=
Given that a=b ;
x+y=a……(1)
y=-x+a
y=(-1)x+a
Since, the above equation is in y=mx+b form
So, the slope of the line is (-1) .
Question:24
The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is
A. x – y = 5
B. x + y = 5
C. x + y = 1
D. x – y = 1
Answer:
Given that straight line passing through the point (3,2) and is perpendicular to
line y=x
Let the equation of line L is (y-y1) =m(x-x1)
Since, L is passing through the point (3,2)
y-2=m(x-3)..…(i)
Now, given equation is y=x
Comparing it with y=mx+b
we get slope of the equation as 1.
It is also given that line L and y=x are perpendicular to each other.
We know that when two lies are perpendicular then m1*m2= -1 m*1=-1
m=-1
Putting this value of m in equation (i) we get y-2=-(1)(x-3)
y-2=-x+3
x+y=5
Hence, the correct option is b
Question:25
The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is
A. y – x + 1 = 0
B. y – x – 1 = 0
C. y – x + 2 = 0
D. y – x – 2 = 0
Answer:
Given that line passing through the point (1,2) and perpendicular to the line
x+y+1=0
Let the equation of line L is x-y+k=0…..(i)
Since, L is passing through the point (1,2)
1-2+k=0
k=1
Putting this value of k in equation )i) we get x-y+1=0 or y-x-1=0
Hence, the correct option is b
Question:26
The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is
A.
B.
C.
D. None of these
Answer:
First equation of line in intercept form=
bx-ay=ab……. (i)
Let the second equation of line having intercepts on the axes b, -a is
ax-by=ab…..(ii)
Now we find the slope in the first equation bx-ay=ab
ay=bx-ab
Slope of the equation
Now, we find the slope of equation (ii) ax-by=ab
by=ax-ab
Slope of the equation
Let θ be the angle between the given two lines
Putting the values of m1 and m2 in above equation, we get
is the required angle.
Hence, the correct option is (c)
Question:27
If the line passes through the points (2, –3) and (4, –5), then (a, b) is
A. (1, 1)
B. (– 1, 1)
C. (1, – 1)
D. (– 1, –1)
Answer:
Given points are (2, -3) and (4, -5)
Firstly the equation of line is found
We know that the equation of line when two points are given is y-y1=
Putting the values we get
y+3= -1( x-2)
y+3= -x+2
x+y=2-3
x+y=-1
in intercept form
Comparing the equaton with intercept form of equation that is
the value of a=-1 and b=-1
Hence, the correct option is (d)
Question:28
Answer:
(a) Given lines are:
and
Solving these lines, we get point of intersection as
therefore Distance of this point from the line
Question:29
The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line is
A. y + 2 = 0,
B. x – 2 = 0,
C.
D. None of these
Answer:
Given equation is and θ= 600
Slope of the equation
Slope of the equation
Let slope of the required line be m2
Then
Putting the values in the above equation we gettan
Taking+sign we get
Taking -ve sign we get
m2=0
Equation of line passing through (3,-2) with slope is y-y1=m(x-x1 )
and equation of line passing through (3, -2) with slope 0 is y-y1=m(x-x1)
y-(-2)=0(x-3)
y+2=0
Hence, the required equations are and y+2=0
Hence, the correct option is (a)
Question:30
The equations of the lines passing through the point (1, 0) and at a distance from the origin, are
A.
B.
C.
D. None of these.
Answer:
Let the equation of any line passing through the point (1,0) is y-y1=m(x-x1)
y-0=m(x-1)
y=mx-m
mx-m-y=0….........(i)
Distance from the origin of the line is Distance of point (x1,y1) from the equation Ax+By+C=0
Squaring both the sides we get
3(m2+1)=4m2
3m2+3=4m2
4m2-3m2=3
m2=3
Putting the value of m= in equation (i) we get x-y-=0
Now, putting the value of m= - in the same equation we get-x-y+=0
Hence, the correct option is (a)
Question:31
The distance between the lines y = mx + c1 and y = mx + c2 is
A.
B.
C.
D. 0
Answer:
Given equations are y=mx+c1….(i) and y=mx+c2…….(ii)
Firstly, we find the slope of both the equations
Since, both of them have the same slope they are parallel lines.
We know that distance between two parallel lines Ax+By+C1=0 and Ax+By+C2=0
Hence, the correct option is b.
Question:32
The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by
A.
B.
C.
D.
Answer:
Given equations are y=3x+4….(i)
Comparing this equation with y=mx+b form , the slope of the equation is 3.
Let the equation of any line passing through the point (2,3) is y-y1=m(x-x1)
y-3=m(x-2)……(ii)
Given that equation (i) is perpendicular to equation (ii)
And we know that, if two lines are perpendicular then m1m2= -1
3*m2=-1
m2=-1/3 which is the slope of the required line
Putting the value of slope in equation ii we get y-3=-1/3(x-2)
3y-9=-x+2
x+3y-9-2=0
x+3y-11=0……(iii)
Now we have to find the coordinates of foot of the perpendicular
Solving equation (i) and (iii) we get x+3(3x+4)-11=0
x+9x+12-11=0
10x+1=0
x=-1/10
Putting the value of x in equation i , we get y=3(-1/10)+4
y= -3/10+4
y=37/10
So the required coordinates are (-1/10,37/10)
Hence, the correct option is (b)
Question:33
If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be
A. 2x + 3y = 12
B. 3x + 2y = 12
C. 4x – 3y = 6
D. 5x – 2y = 10
Answer:
Let the given line meets the axes at A(a,0) and B(0,b)
Given that (3,2) is the midpoint
a=6 and
b=4
Intercept form of the line AB is
Putting the value of a and b in above equation, we get
2x+3y=12
Hence, the correct option is (a)
Question:34
Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is
A. y + 2 = x + 1
B. y + 2 = 3 (x + 1)
C. y – 2 = 3 (x – 1)
D. y – 2 = x – 1
Answer:
Given equation of the line is y=3x-1
Now we find the slope of the above equation by comparing it with y=mx+b form
So m=3
Now, we find the equation of line passing through the point (1,2) and parallel to the given line
with slope=3
y-y1=m(x-x1)
y-2=3(x-1)
Hence, the correct option is (c)
Question:35
Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are
A. y = x, y + x = 1
B. y = x, x + y = 2
C. 2y = x, y + x = 1/3
D. y = 2x, y + 2x = 1
Answer:
It is given that the lines x=0, y=0 , x=1 and y=1 form a square of side 1 unit
Let us form a square OABC having corners O(0,0) from the given lines
with A(1,0), B(1,1) and C(0,1)
Now we have to find the equation of the diagonal AC
Equation of a line is found out by
y=-1(x-1)
y=-x+1
x+y=1
Equation of diagonal OB is
y=x
y=x
Hence, the correct option is (a)
Question: 36
For specifying a straight line, how many geometrical parameters should be known?
A. 1
B. 2
C. 4
D. 3
Answer:
Equation of straight line in intercept form=
where a and b are the intercepts on the axis.
In intercept form we need 2 parameters a and b to specify a straight line
Slope-Intercept Form y=mx+c where m=tanθ
and θ is the angle made with positive x axis and c is the intercept on y axis
So, we need two parameters 'm' and 'c' to specify a straight line
Hence, the correct option is b
Question:37
The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis. Then the final coordinates of the point are
A. (4, 3)
B. (3, 4)
C. (1, 4)
D.
Answer:
Let Q(x,y) be the reflection of P(4,1) about the line y=x, then
midpoint of PQ
which lies on y=x
4+x=1+y
x-y+3=0……(i)
Now, we find the slope of given equation y=x
Since this equation is in y=mx+b form
So the slope=m=1
Slope of
Since, PQ is perpendicular to y=x
And when two lines rae perpendicular then m1m2=-1
y-1=-(x-4)
x+y-5=0…..(ii)
On adding equation i and equation ii we get x-y+3+x+y-5=0
2x-2=0
x-1=0
x=1
Putting the value of x=1 in equation i we get 1-y+3=0 -y+4=0 y=4
It is given that translation through a distance of 2 units along the positive x axis takes place
The point after translaton is (1+2,4)=(3 ,4)
Hence, the correct option is (b)
Question:38
A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is
A. (1, –1)
B. (1, 1)
C. (0, 0)
D. (0, 1)
Answer:
Given equations are 4x+3+10=0 …..(i)
5x-12y+26=0…......(ii)
and 7x+24y-50=0
Let p,qbe the point which is equidistant from the givenlines
Now, we find the distance of (p,q) from the given lines
Distance of point (x1,y1) from the equation Ax+By+C=0
Distance of (p,q) from equation i is
Distance of (p,q) from equation ii is
Distance of (p,q) from equation iii is
Given that (p,q) is equidistant from the given lines =
On putting the value of (p,q) as (0,0) we get
Hence, the correct option is c
Question:39
A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
A. 1/3
B. 2/3
C. 1
D. 4/3
Answer:
Given line is 3x+y=3 It can be re written as y=-3x+3
Comparing it with the y=mx+b form of equation we get the slope as m= -3
So slope of the perpendicular line will be1/3
The line passes through (2,2) and has a slope1/3 is
3y-6=x-2
3y=x+4
So, the y intercept is
Hence, the correct option is (d)
Question:40
The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
A. 1 : 2
B. 3 : 7
C. 2 : 3
D. 2 : 5
Answer:
Given lines are 3x+4y+5=0…....(i)
3x+4y-5=0…....(ii)
and 3x+4y+2=0….......(iii)
Since the coefficient of x and y are same equation i, ii and iii are parallel to each other
We know that in case of i and iii
distance between two parallel lines is
Similarly in case of ii and iii distance between two parallel lines is
Ratio between the distance is
Hence, the correct option is b
Question:41
One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is
A. (–1, –1)
B. (2, 2)
C. (–2, –2)
D. (2, –2)
Answer:
Let ABC be an equilateral triangle with vertex A (a,b)
Let AD be perpendicular to BC and let (p,q) be the coordinates of D
Given that the centroid P lies at the origin (0,0)
We know that, the centroid of a triangle divides the median in the ratio 1:2
Now, using the section formula, we get
and
a+2p=0 and b+2q=0…A…
a+2p=b+2q
2p-2q=b-a …i
It is given that BC=x+y-2=0
Since, the above equation passes through p,q
p+q-2=0
Now we find the slope of line AP
Equation of line BC is x+y-2=0
y= -x+2
y=-(1)x+2
Since the above equation is in y=mx+b form
So, slope of line BC is mBC=-1
Since both lines rae perpendicular
b=a
Putting this value in equation (i )we get 2p-q=b-b=0
p=q
Now putting this value in equation (i) we get p+q-2=0
2p=2
p=1
q=1
Putting the value of p and q in equation A , we get a+2*1=0 and b+2*1=0
a=-2 and b=-2
So, the coordinates of vertex A (a,b) is (-2,-2)
Hence, the correct option is c
Question:42
Given that a,b,c are in AP
2b=a+c
a-2b+c=0….i
Now comparing equation iwith the given equation ax+by+c=0 we get x=1, y=-2
So, the line will pass through (1,-2 )
Question:43
Answer:
Equation of straight line in intercept form= where a and b are intercepts
Given that a=b
x+y=a….(i)
If equation ipasses through the point (1, -2) we get 1+(-2)=a
1-2=a
a=-1
Putting the value of a in equation i we get x+y=-1
x+y+1=0
Question:44
Answer:
Given equation is x-2y=3
x-3=2y
The slope of the equation can be found by comparing with y=mx+b form
So,
We have to find an equation which is passing through the point (3,2)
A line passing through the point x1,y1 has an equation y-y1=m(x-x1)
So, here x1=3 and y1=2
y-2=m(x-3)....…(ii)
Now, it is given that the angle between the given two lines is 450
Putting the values of m1and m2 in above equation we get tan450 =
2m-1=2+m or-(2m-1)=2+m
2m-m=2+1 or -2m+1-m=2
m=3 or-3m=1 or m=-1/3
Putting the value of m=3 in equation (ii) we get y-2=3(x-3)
y-2=3x-9
3x-y-9+2=0
3x-y-7=0
Putting the value of m=-1/3 in equation (ii) we get y-2= -1/3(x-3)
3(y-2)=3-x
3y-6=3-x
x+3y-6-3=0
x+3y-9=0
Question:45
Answer:
Given line 3x-4y-8=0 and given points are (3,4) and (2,-6)
For point (3,4)
3(3)-4(4)-8=9-16-8
=9-24= -15<0
Forpoint (2, -6) 3(2)-4(-6)-8 =6+24-8
=30-8=22>0
So the points (3,4)and (2,-6) are situated on the opposite sides of 3x-4y-8=0
Question:46
Answer:
Given point is (3, -2) and equation of line is 5x-12y=3
Let (p,q) be any moving point
Distance between them (p,q) and (3,-2)
(d1)2=(p-3)2+(q+2)2
Now, distance of the point (p,q) from the given line 5x-12y-3=0
=
Taking numerical values only, we have (p-3)2+(q-2)2=
13[(p-3)2+(q+2)2]=5p-12q-3
13[p2+9-6p+q2+4+4q]=5p-12q-3
On solving we get 13p2+13q2-83p+64q+172=0
A point moves so that square of its distance from the point (3,-2) is numerically equal to its distance from the line 5x-12y=3 .
The equation of its locus is 13p2+13q2-83p+64q+172=0
Question:47
Answer:
Given equation of the line is x sinθ+y cosθ=p…..(i)
Let Ph,kbe the midpoint of the given line where it meets the two axis at a,0and 0,b.
Since (a,0) lies on equation (i) then asinθ+0=p a= …..(ii)
(0,b) also lies on the equation i then 0+bcosθ=p
……(iii)
Since, P (h,k) is the midpoint of the given line
2h=a and
2k=b
Putting the value of a=2h in equation (ii) we get
…..(iv)
Putting the value of b=2k in equation (ii) we get
….(v)
Squaring and adding equation (iv) and (v), we get
or
or 4x2y2=p2y2+p2x2
or 4x2y2=p2(x2+y2) is the locus of the mid-points of the portion of the line intercepted between the axes
Question:48
Answer:
Let ABC be a triangle with vertices A(x1,y1), B (x2,y2) and C (x3, y3) where xi, yi, i=1,2,3 are integers
Then area of
Since, xi and yi all are integers but is a rational number.
So, the result comes out to be a rational number. i.e . Area of ABC=a rational number
Suppose, ABC be an equilateral triangle, then area of ABC is=
It is given that vertices are integral coordinates, it means the value of coordinates is in whole
number. Therefore, the value of (AB)2 is also an integer.
(positive integer)
But, is an irrational number
Area of triangle ABC=an ir-rational number
This contradicts the fact that the area is a rational number
Hence, the given statement is true.
Question:49
Answer:
Given points are A(-2,1), B (0,5) and C(-1,2)
There are two ways to find that given points are collinear or not .
The first is if 3 points are collinear, then slope of any two pairs of points will be equal.
Second way is that if the value of area of triangle formed by the 3 points is zero, then the points
are collinear.
Slope of AB
Slope of BC
Slope of CA is
Since, the slopes are different.
So, the given points are not collinear
Hence, the given statement is False
Question:50
Answer:
Let the equation of line y=mx+c….(i)
So, slope of the above equation is m
Given equation of line is x secθ+y cosecθ=a sin 2θ.
Since the slope of the equation is m'=
Given that equation (i) is perpendicular to x secθ+y cosec θ=a sin 2θ
m*m'= -1
Putting the value of m in equation (i)we get
ysecθ=x cosec θ+c secθ
x cosec θ-y secθ=k….(ii)
If equation (ii) passes through the point (a cos3θ, a sin3θ)
(a cos3θ )cosecθ-(a sin3θ) secθ=k
a[(cos2θ-sin2θ)(cos2θ+sin2θ)]=x cosθ-ysinθ
a[cos2θ-sin2θ]=xcosθ-ysinθ
a[cos2θ]=xcos θ-y sinθ
The given statement is FALSE
Question:51
Answer:
Given equations are x+2y-10=0….(i) and 2x+y+5=0…..(ii)
The point of intersection is obtained by solving them together i.e.
If the given line 5x+4y=0 passes through the point then
0=0
So, the given line passes through the point of intersection of the given lines.
Hence, the given statement is True.
Question:52
Answer:
Let ABC be an equilateral triangle with vertex (2,3) and the equation of the opposite side is x+y=2
In the case of equilateral triangle θ=60
Let the slope of line AB is m and the slope of the given equation x+y=2 is m2=-1
We know that
Putting the values of m1 and m2 in the above equation, we get
Rationalizing both the equations we get
So, the slope of line AB is
Thus, the equations of the other two lines joining the point (2, 3) are
Hence, the given statement is True.
Question:53
Answer:
Given two lines are 4x+y-1=0…(i ) and 7x-3y-35=0…..(ii)
Now, point of intersection of these lines can be find out as x=2 and y=-7
To find the equation of the line joining the point (3,5) and (2,-7)
y-5=12x-3
y-5=12x-36
12x-y-31=0……iv
Now, the distance of equation (iv) from the point (0,0) is d=
Now the distance of
equation iv from the point (8,34) is
Hence, the equation of line 12x-y-31=0 is equidistant from (0,0)and (8,34)
Hence, the given statement is True.
Question:54
Answer:
Equation of line ……i
Equation of line passing through the origin and perpendicular to the given line ….ii
Now the foot of perpendicular from origin on the line (i) is the point of intersection of lines (i) and ii
So, to find its locus we have to eliminate the variable a and b
Squaring and adding both the equations we get
x2+y2=c2
Hence, the given statement is true.
Question:55
Answer:
Given that ax+2y+1=0 bx+3y+1=0 and cx+4y+1=0 are concurrent .
For the lines to be concurrent
Now expanding along first column we get
a[3-4]-b[2-4]+c[2-3]=0
-a+2b-c=0
2b=a+c and we know that if a, b ,c are in AP then
This means given lines are in AP not in GP
Hence, the given statement is False.
Question:56
Answer:
Given points are (3,-4), (-2,6), (-3,6) and (9,-18)
Now we find the slope since the lines are perpendicular, the product of the slopes is -1 i.e. m1m2= -1
Slope of the line joining the points (3,-4) and (-2,6)
Here, x1=3, x2=-2 , y1=-4 and y2=6
Now, slope of the line joining the points -(3,6) and (9,-18)
Here, x1=-3, x2=9, y1=6 and y2=-18
m1=m2=-2
and m1m2= -2*(-2)=-4 ≠-1
So, the lines are parallel and not perpendicular
Hence, the given statement is False
Question: 57
Match the questions given under Column C1 with their appropriate answers given under the Column C2.
Column C1 | Column C2 |
a) The coordinates of the points P and Q on | i) (3,1), (-7,11) |
b) The coordinates of the points on | ii) -1/3, 11/3, 4/3, 7/3 |
c) The coordinates of the points on the line joining A | iii) 1, 12/5,-3,16/5 |
Answer:
Let P (x1,y1) be any point on the given line x+5y=13
x1+5y1=13
5y1=13-x1….(i)
Distance of the point P(x1,y1)from the equation 12x-5y+26=0
2=|x1+1|
2= ±(x1+1)
So x1=1….(ii) or x1=-3…. (iii)
Putting the value in equation (i) we get 5y1=13-1=12
Putting the value of x1=-3 in the same equation we get 5y1=13-(-3)=16
Hence, the required points on the given line areand -3,165
Hence, (a)-(iii)
bLet P x1, y1 be any point lying in the equation x+y=4
x1+y1=4…. i
Now, the distance of the point from the equation is
4x1+3y1-10= ±5
either 4x1+3y1-10=5 or 4x1+3y1-10=-5
4x1+3y1=15 ….(ii) or 4x1+3y1=5…..(iii)
From equation i we have y1=4-x1….(iv)
Putting the value of y1 in equation ii we get 4x1+34-x1=15
4x1+12-3x1=15
x1=3 Putting the value of x1 in equation (iv) we get y1=4-3=1
Putting the value of y1 in equation iii we get 4x1+34-x1=5
4x1+12-3x1=5
x1=5-12=-7
Putting the value of x1 in equation (iv) , we get y1=4-(-7)
y1=4+7=11
Hence, the required points on the given line are (3,1) and (-7,11)
Hence, b-i
c Given that AP=PQ=QB and given points are A(-2,5) and B(3,1)
Firstly, we find the slope of the line joining the points (-2,5) and (3,1)
Slope of line joining two points=
Now equation of line passing through the point (-2, 5) y-5=-4/5[x-(-2)]
5y-25=-4(x+2)
4x+5y-17=0 Let P (x1,y1) and Q (x2,y2) be any two points on the AB
P(x1,y1) divides the line AB in the ratio 1:2
Now, Q (x2,y2) is the midpoint of PB
Hence, the coordinates of Q (x2,y2) is (4/3,7/3)
Hence, c-ii
Question:58
The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are
Column C1 | Column C2 |
a) parallel to y-axis is | i) |
b) perpendicular to 7x+y-4=0 | ii) |
c) Passes through (1,2) is | iii) |
d) parallel to x-axis is | iv) |
Answer:
a ) Given equation is (2x+3y+4)+ λ(6x-y+12)=0
2x+3y+4+6λx-λy+12λ=0
(2+6λ)x+(3-λ)y+4+12λ=0…. (i)
If equation i is parallel to y-axis, then 3-λ=0
λ=3
Hence, iv.
b) Given equation is 2x+3y+4+λ6x-y+12=0
2x+3y+4+6λx-λy+12λ=0
(3-λ)y= -4-12λ-(2+6λ)x
Since, the above equation is in y=mx+b form
So the slope of equation (i) is
Now the second equation is 7x+y-4=0…..(ii)
y=-7x+4
So, the slope of equation (ii) is m2=-7
Now equation i is perpendicular to equation (ii)
m1m2=-1
(2+6λ)*7=-(3-λ)
On solving we get λ=-17/41
Hence, (b)-(iii)
c) Given equation is (2x+3y+4)+λ(6x-y+12)=0
If the above equation passes through the point (1,2) then [2*1+3*2+4]+λ[6*1-2+12]=0
2+6+4+λ(6+10)=0
12+16λ=0
12=-16λ
λ=-12/16=-3/4
Hence, (c)-(i)
d) Given equation is (2x+3y+4)+λ(6x-y+12)=0
2x+3y+4+6λx-λy+12λ=0
(2+6λ)x+(3-λ)y+4+12λ=0…..(i)
If equation (i) is parallel to x axis, then 2+6λ=0 6λ=-2
λ=-1/3
(d)-(ii)
Question:59
The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and
Column C1 | Column C2 |
a) through the point (2,1) is | i) 2x-y =4 |
b)perpendicular to the line x+2y+1=0 is | ii)x+y-5=0 |
c)parallel to the line 3x+4y+5=0 is | iii)x-y-1=0 |
d) equally inclined to the axes is | iv) 3x-4y-1=0 |
Answer:
.a) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii )
Equation of line passing through equation iand (ii), we get 2x-3y+λ4x-5y-2=0….(iii)
If the above equation passes through the point (2,1) we get (2*2-3*1)+λ(4*2-5*1-2)=0
(4-3)+λ(8-5-2)=0
1+λ=0
λ=-1
Putting the value of λ in equation iii we get (2x-3y)+(-1)(4x-5y-2)=0
2x-3y-4x+5y+2=0
-2x+2y+2=0
x-y-1=0
Hence, (a)-(iii)
b) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii)
Equation of line passing through equation i and ii, we get 2x-3y+λ4x-5y-2=0…(iii)
2x-3y+4λx-5λy-2λ=0
x(2+4λ)-y(3+5λ)-2λ=0
-y(3+5λ)= -(2+4λ)+2λ
Slope of equation iii is m1=
Now, we find the slope of thegiven line x+2y+1=0…..(iv)
2y=-x-1 y=-1/2x+(-1/2)
So slope of equation (iv) is m2=(-1/2)
We know that, if two lines are perpendicular to each other then the product of their slopes is equal to-1
So, m1*m2=-1
2+4λ=2*(3+5λ)
2+4λ=6+10λ
-6λ=4
λ=-4/6= -2/3
Putting the value of λ in equation (iii) we get (2x-3y)+(-2/3)(4x-5y-2)=0
6x-9y-8x+10y+4=0
-2x+y+4=0
2x-y=4
b-i
c) Given equations are 2x-3y=0….(i) and 4x-5y=2…..(ii)
Equation of line passing through equation iand (ii), we get (2x-3y)+ λ(4x-5y-2)=0….iii
2x-3y+4λx-5λy-2λ=0
x(2+4λ)-y(3+5λ)-2λ=0
So slope of equation iii is m1=
Now, we find the slope of the given line 3x-4y+5=0
3x+5=4y y=3/4x+5/4
So slope of equation is 3/4
If 2 lines are parallel then their slopes are also equal So
4(2+4λ)=3(3+5λ)
8+16λ=9+15λ
λ=1
Putting the values of λ in equaaion iii we get (2x-3y)+(4x-5y-2)=0
2x-3y+4x-5y-2=0
6x-8y-2=0
3x-4y-1=0
Hence, c-iv
d) Given equations are 2x-3y=0…(i) and 4x-5y=2….(ii)
Equation of line passing through equation (i) and (ii), we get (2x-3y)+λ(4x-5y-2)=0
x(2+4λ)-y(3+5λ)-2λ=0
-y(3+5λ)= -(2+4λ)+2λ
Slope of equation iii is
Since the equation is equally inclined with axes, it means that the line makes equal angles with both the
coordinate axes.
It will make an angle of 450 or 1350
m2=tan450 andtan1350
= 1 and-1
2+4λ=-3+5λ or 2+4λ=3+5λ
So λ=-5/9 or-1
Putting the value in equation iii we get (2x-3y)+(-5/9)(4x-5y-)2=0
18x-27y-20x+25y+10=0
-2x-2y+10=0
x+y-5=0
If λ=-1, then the required equation is (2x-3y)+(-1)( 4x-5y-2)=0
2x-3y-4x+5y+2=0
-2x+2y+2=0
x-y-1=0
d-ii, iii
The students trying to ace their examinations can access NCERT Exemplar Class 11 Maths solutions chapter 10 PDF download from here. This will help them in getting a helpful approach towards the preparation of your exams that are fabricated by our experts through thorough study.
The NCERT Exemplar solutions for Class 11 Maths chapter 10 tries to aid students with properly examined solutions from the perspective of exams and its usage in various other fields.
Also read - NCERT Solutions for Class 11 Maths Chapter 10
The students will get a brief introduction of coordinate geometry in two dimensions along with the main focus on one of the simplest figures, which is a straight line. NCERT Exemplar solutions for Class 11 Maths chapter 10 covers different concepts relating to a straight line that will teach students about the calculation of angle between two lines, the distance between two parallel lines, determining the slope of a line and other forms of the equation of a line. A straight line is the simplest shape of geometry, but it is also the most important part as it acts as the base for the introduction of other shapes such as curves that will help students in the further lessons. The learners will also benefit from NCERT Exemplar Class 11 Maths solutions chapter 10 through its use and application in different fields, including physics, architecture, geometry, and much more.
Also, check -
Some of the important topics for students to review from this chapter are:
The learners should revise all the basics and concepts of coordinate geometry learned previously for better understanding.
NCERT Exemplar Class 11 Maths chapter 10 solutions covers the calculation of inclination of the line along with gradient or slope of a line in two-dimensional planes.
The students should study all the solutions of NCERT books along with examples to get a better understanding of concepts.
The students should cover various forms of the equation of lines from an examination perspective.
Check Chapter-Wise NCERT Solutions of Book
Chapter-1 | |
Chapter-2 | |
Chapter-3 | |
Chapter-4 | |
Chapter-5 | |
Chapter-6 | |
Chapter-7 | |
Chapter-8 | |
Chapter-9 | |
Chapter-10 | Straight Lines |
Chapter-11 | |
Chapter-12 | |
Chapter-13 | |
Chapter-14 | |
Chapter-15 | |
Chapter-16 |
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Yes, NCERT Exemplar Class 11 Maths chapter 10 solutions are very important from the perspective of exams including Board exams as well as other competitive exams.
Yes, NCERT Exemplar Class 11 Maths solutions chapter 10 are very important for the Board examination as it has the relevant weightage of marks along with its importance in other competitive exams.
The important topics to be covered in this chapter are the slope of a line, general equation of a line, the distance of a point from a line and various forms of the equation of a line in coordinate geometry.
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