NCERT Solutions For Class 11 Maths chapter-10 Straight Lines Exercise 10.1

NCERT Solutions for Class-11 Maths Chapter-10 Straight LInes

NCERT Solutions For Class 11 Maths Chapter-10 Straight LInes Exercise 10.1 prepared by  the expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.

 

NCERT Solutions for Class-11 Maths Exercise 10.1

Question1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7),(5, – 5) and (– 4, –2). Also, find its area.

Solution :
Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).

Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as

= To find the area of quadrilateral ABCD, we draw one diagonal, say AC.

Accordingly, area (ABCD) = area (ΔABC) + area (ΔACD)

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

Question2. 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 :
 .

Question3. Find the distance between P(x₁ ,y₁) and Q(x₂ ,y₂) when : 

(i) PQ is parallel to the y-axis, 

(ii) PQ is parallel to the x-axis.

Solution :Here, the points are P(x₁,y₁) and Q(x₂,y₂).

When PQ is parallel to the y -axis, we have x₁ = x₂

The distance between P and Q is,

√((x₂−x₁)²+(y₂−y₁)²)

 = √(y₂−y₁)² (since x₁ = x2.)

=|y²−y¹|

Therefore, the distance between P(x₁,y₁)
 and Q(x₂,y₂) when PQ
is parallel to the y

-

axis is |y₂−y₁|
.

(ii) PQ is parallel to x - axis, then

The points are P(x₁,y₁) and Q(x₂,y₂) When PQ is parallel to the x-axis, we know y₁=y₂
Distance between P and Qis

Therefore, the distance between P(x1,y1) and Q(x2,y2) when PQ is parallel to the x axis is |x₂−x₁|

Question4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Solution :Let A(7,6) and B(3,4) be the given points.
Assume C(a,0) as the point on the X - axis that is equidistance from the points (7,6)
and (3,4).In general, Distance between two points is

 

=

And PR = .Now, Distance between A and C = Distance between B and C

 

Question5. 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 midpoint of the line segment joining two points are,

(x₁+x₂ / 2  , y₁+y_{2  /} 2)

Then, 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)

We know, the slope (m)
of a non-vertical line passing through the points (x1,y1) and (x2,y2)

 is,

m= y₂−y_{1 /}  x₂−x₁,

where

x₂ ≠ x₁

Thus, the slope of the line passing through (0,0,
, and (4,−2)

 is,

 ⇒−2−04−0

 =−24

 =−12

 

Therefore, the slope of a line, which passes through the origin, and the mid-point of the

segment joining the points P(0,−4)
 and B(8,0) is −12.
.

Question6. 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 :
The vertices of the given triangle are P(4,4),Q(3,5), and R(−1,−1).

We know, the slope (m) of a non-vertical line passing through the points (x1,y1) and (x2,y2) is,

Question7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Solution :Given that, the line makes an angle of 30^∘ with the positive direction of y -axis

measured anticlockwise.

Plot the figure,

(Image will be uploaded soon)

From the figure, the angle made by the line with the positive direction of the x-axis

measured anticlockwise is,

 90^∘+30^∘=120^∘

Now, the slope of the given line is tan120^∘

Rewritetan120^∘as tan(180^∘−60^∘)

=−tan60^∘

=−√3

Therefore, the slope of the line, which makes an angle of 30∘
 with the positive direction of y-axis measured anticlockwise is −√3

Question8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear.

Solution :.Given that, the points are (x,−1),(2,1) and (4,5)

If points P(x,−1),Q(2,1) and R(4,5) are collinear, then Slope of PQ= Slope of QR

⇒(1−(−1)) / (2−x) = (5−1) / (4−2) Open bracket and simplify,

⇒(1+1)  / (2−x) = 42

⇒2 / (2−x) = 2

⇒2=4−2x

⇒2x=2

⇒x=1

Therefore, the value of x for which the points (x,−1),(2,1) and (4,5) are collinear is 1.

Question9. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Solution :
Let points (−2,−1),(4,0),(3,3) and (−3,2) be respectively denoted by P,Q,R and S.

Question10. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).

Solution :
Let A (3,–1) and B (4,–2) be two points. Let Q be the angle which AB makes with positive direction of axis.

The points are (3, −1) and (4, −2).

The slope (m) of a non-vertical line passing through the points (x₁,y₁) and (x₂,y₂) is,

m= (y₂−y_{1)  / (} x2−x1),where x₂ ≠ x₁ The slope of the line joining the points (3,−1) and (4,−2)is,

 m=−2−(−1)  / 4−3

 =−2+1

 =−1
The inclination (θ) of the line joining the points (3,−1) and (4,−2) is,
tanθ = m

Substitute the value of m
tanθ = −1
⇒ θ = (90^∘+45^∘)

= 135^∘

Therefore, the angle between the x-axis and the line joining the points (3,−1)  and (4,−2)
 is 135^∘

Question11. The slope of a line is double of the slope of the another line. If tangent of the angle between them is 1/3 find the slopes of the lines.

Solution :Take m,m1 as the slopes of the two lines.
Given that, slope of a line is double of the slope of another line.
That is, m1=2m If θ is the angle between the lines l1 and l2 with slopes m1 and m2
We have,

 

 

Question12. A line passes through (x₁,y₁) and (h,k) .If slope of the line is m, show that k−y₁ = m(h−x₁)

Solution :.Given that, the slope of the line is m

The slope of the line passing through (x_1, y₁) and (h,k) is k−y₁ / h−x₁

.Now,
k−y₁ / h−x₁=m

⇒k−y1 = m(h−x1)

Therefore, if the line passes through (x1,y1)
 and (h,k) with a slope of the m

,

k−y₁ =  m(h−x₁)

Question13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that 

Solution :The points P(h,0),Q(a,b), and R(0,k)

 lie on a line. Slope of PQ= Slope of QR

Cross multiply,

⇒−ab=(k−b)(a−h) Open brackets,

⇒−ab=ka−kh−ab+bh

⇒ka+bh=kh

Divide both sides by kh
,

Question14. Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?

Solution :
Given: The points on the line are A (1985, 92) and B (1995, 97).

⇒ Slope of AB = (97 - 92) / (1995 -1985)

⇒ 5/10

⇒ 1/2

LAssume y as the population in the year 2010. Then, according to the given graph, line  AB must pass through the point C(2010,y)
∴ Slope of AB= Slope of BC

Since points A, B and C lie on the line.