{"id":4613,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-10-1\/"},"modified":"2016-05-17T11:15:38","modified_gmt":"2016-05-17T05:45:38","slug":"ncert-solutions-class-11-maths-exercise-10-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-10-1\/","title":{"rendered":"NCERT Solutions class-11 Maths Exercise 10.1"},"content":{"rendered":"

Exercise 10.1<\/strong><\/p>\n

1. Draw a quadrilateral in the Cartesian plane, whose vertices are <\/strong> and <\/strong> Also, find its area.<\/strong><\/p>\n

Ans. <\/strong>Area of quadrilateral ABCD = Area of + Area of trapezium ABED<\/p>\n

= <\/p>\n

= <\/p>\n

= 26 + 34.8 = 60.8 sq. units<\/p>\n

<\/p>\n

\n

2. The base of an equilateral triangle with side <\/strong> lies along the <\/strong>axis such that the mid-point of the base is at origin. Find the vertices of the triangle.<\/strong><\/p>\n

Ans.<\/strong> Given: Length of side of equilateral triangle = . The base of triangle lies along axis and the mid-point of base is at origin so that the coordinates of vertices are and <\/p>\n

Let vertices of the third vertex be <\/p>\n

<\/p>\n

<\/p>\n

The vertices of triangle are , and <\/p>\n

\n

3. Find the distance between P<\/strong> and Q<\/strong> when (i) PQ is parallel to the <\/strong>axis (ii) PQ is parallel to the <\/strong>axis.<\/strong><\/p>\n

Ans. <\/strong>Given: P and Q are two points.<\/p>\n

PQ = <\/p>\n

(i) <\/strong>PQ is parallel to axis, then <\/p>\n

PQ = <\/p>\n

(ii) <\/strong>PQ is parallel to axis, then <\/p>\n

PQ = <\/p>\n

\n

4. Find the point on the <\/strong>axis, which is equidistant from the points (7, 6) and (3, 4).<\/strong><\/p>\n

Ans. <\/strong>Let P be any point on the axis which is equidistant from Q (7, 6) and R (3, 4).<\/p>\n

PQ = <\/p>\n

= <\/p>\n

= <\/p>\n

And PR = <\/p>\n

= <\/p>\n

= <\/p>\n

According to question, PQ = PR<\/p>\n

<\/p>\n

= <\/p>\n

Squaring both sides, = <\/p>\n

<\/p>\n

<\/p>\n

Therefore, the coordinates of the point are <\/p>\n

\n

5. Find the slope of a line, which passes through the origin and the mid-point of the line segment of joining the points P<\/strong> and B (8, 0).<\/strong><\/p>\n

Ans. <\/strong>Here, mid-point of the line segment joining P and Q (8, 0) is<\/p>\n

<\/p>\n

<\/p>\n

Since the line passes through points (0, 0) and .<\/p>\n

Therefore, Slope of the line = <\/p>\n

\n

6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and <\/strong> are the vertices of a right angled triangle.<\/strong><\/p>\n

Ans. <\/strong>Let A (4, 4), B (3, 5) and C be three vertices of a <\/p>\n

Slope of AB = <\/p>\n

Slope of BC = <\/p>\n

Slope of AC = <\/p>\n

Now, Slope of AB x Slope of AC = <\/p>\n

This shows that AB AC. Thus is right angled at point A.<\/p>\n

\n

7. Find the slope of the line, which makes an angle of <\/strong> with the positive direction of <\/strong>axis measured anticlockwise.<\/strong><\/p>\n

Ans. <\/strong>The line makes an angle of with the positive direction of axis.<\/p>\n

Now the line makes an angle of with the positive direction of axis.<\/p>\n

Slope of the line = <\/p>\n

\n

8. Find the value of <\/strong> for which the points <\/strong> and <\/strong> are collinear.<\/strong><\/p>\n

Ans. <\/strong>Let A B (2, 1) and C (4, 5) be three collinear points.<\/p>\n

Slope of AB = <\/p>\n

Slope of BC = <\/p>\n

According to question, Slope of AB = Slope of BC<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

9. Without using distance formula, show that the points <\/strong> and <\/strong> are the vertices of a parallelogram.<\/strong><\/p>\n

Ans. <\/strong>Let A B (4, 0), C (3, 3) and D be vertices of a quadrilateral ABCD.<\/p>\n

Slope of AB = Slope of BC = <\/p>\n

Slope of DC = Slope of AD = <\/p>\n

Here Slope of AB = Slope of DC<\/p>\n

AB DC<\/p>\n

And Slope of BC = Slope of AD<\/p>\n

BC AD<\/p>\n

Therefore, ABCD is a parallelogram.<\/p>\n

\n

10. Find the angle between the <\/strong>axis and the line joining the points <\/strong> and <\/strong><\/p>\n

Ans. <\/strong>Let A and B be two points. Let Q be the angle which AB makes with positive direction of axis.<\/p>\n

Slope of AB = <\/p>\n

Also Slope of AB = <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

11. The slope of a line is double of the slope of the another line. If tangent of the angle between them is <\/strong> find the slopes of the lines.<\/strong><\/p>\n

Ans. <\/strong>Given: . Let the slopes of two lines be and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Taking <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

and <\/p>\n

Taking <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

and <\/p>\n

Therefore, the slopes of lines are and or 1 and <\/p>\n

\n

12. A line passes through <\/strong> and <\/strong> If slope of the line is <\/strong> show that <\/strong><\/p>\n

Ans. <\/strong>Let A and B be two points. It is given that Slope of AB = <\/p>\n

Slope of AB = = (given)<\/p>\n

<\/p>\n

\n

13. Is three points <\/strong> and <\/strong> lies on a line, show that <\/strong><\/p>\n

Ans. <\/strong>Let A B and C be three points lie on the line.<\/p>\n

Slope of AB = Slope of BC = <\/p>\n

Slope of AB = Slope of BC (given)<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

14. 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?<\/strong><\/p>\n

<\/strong><\/p>\n

Ans.<\/strong> Given: The points on the line are A (1985, 92) and B (1995, 97).<\/p>\n

Slope of AB = <\/p>\n

Let the population in year 2010 be crores. Then C lies on the line.<\/p>\n

Slope of BC = <\/p>\n

Since points A, B and C lie on the line.<\/p>\n

Slope of AB = Slope of BC<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, population in 2010 will be 104.5 crores.<\/p>\n","protected":false},"excerpt":{"rendered":"

Exercise 10.1 1. Draw a quadrilateral in the Cartesian plane, whose vertices are and Also, find its area. Ans. Area of quadrilateral ABCD = Area of + Area of trapezium ABED = = = 26 + 34.8 = 60.8 sq. units 2. The base of an equilateral triangle with side lies along the axis such … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47,281],"tags":[],"class_list":["post-4613","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Exercise 10.1 | myCBSEguide<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-10-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions class-11 Maths Exercise 10.1 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Exercise 10.1 1. Draw a quadrilateral in the Cartesian plane, whose vertices are and Also, find its area. Ans. Area of quadrilateral ABCD = Area of + Area of trapezium ABED = = = 26 + 34.8 = 60.8 sq. units 2. The base of an equilateral triangle with side lies along the axis such ... 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