{"id":4615,"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-10-3\/"},"modified":"2016-05-17T11:19:35","modified_gmt":"2016-05-17T05:49:35","slug":"ncert-solutions-class-11-maths-exercise-10-3","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-10-3\/","title":{"rendered":"NCERT Solutions class-11 Maths Exercise 10.3"},"content":{"rendered":"

Exercise 10.3<\/strong><\/p>\n

1. Reduce the following equations into slope-intercept form and find their slopes and the <\/strong>intercepts.<\/strong><\/p>\n

(i) <\/strong><\/p>\n

(ii) <\/strong><\/p>\n

(iii) <\/strong><\/p>\n

Ans. (i)<\/strong> Given: <\/p>\n

<\/p>\n

\u2026\u2026\u2026.(i)<\/p>\n

Comparing with we have and <\/p>\n

(ii)<\/strong> Given: <\/p>\n

<\/p>\n

\u2026\u2026\u2026.(i)<\/p>\n

Comparing with we have and <\/p>\n

(iii)<\/strong> Given: <\/p>\n

\u2026\u2026\u2026.(i)<\/p>\n

Comparing with we have and <\/p>\n

\n

2. Reduce the following equations into intercept form and find their intercepts on the axis:<\/strong><\/p>\n

(i) <\/strong><\/p>\n

(ii) <\/strong><\/p>\n

(iii) <\/strong><\/p>\n

Ans. (i)<\/strong> Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Comparing with , we have and <\/p>\n

(ii)<\/strong> Given: <\/p>\n

<\/p>\n

<\/p>\n

Comparing with , we have and <\/p>\n

(iii)<\/strong> Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Comparing with , we have and <\/p>\n

\n

3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular <\/strong>axis:<\/strong><\/p>\n

(i) <\/strong><\/p>\n

(ii) <\/strong><\/p>\n

(iii) <\/strong><\/p>\n

Ans. (i)<\/strong> Given: <\/p>\n

<\/p>\n

Dividing both sides by we have<\/p>\n

<\/p>\n

<\/p>\n

Putting and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

= <\/p>\n

Equation of line in normal form is <\/p>\n

Comparing with we have and <\/p>\n

(ii)<\/strong> Given: <\/p>\n

<\/p>\n

Dividing both sides by we have <\/p>\n

Putting and <\/p>\n

<\/p>\n

Equation of line in normal form is <\/p>\n

Comparing with we have and <\/p>\n

(iii)<\/strong> Given: <\/p>\n

Dividing both sides by we have<\/p>\n

<\/p>\n

<\/p>\n

Putting and <\/p>\n

<\/p>\n

<\/p>\n

Equation of line in normal form is <\/p>\n

Comparing with we have and <\/p>\n

\n

4. Find the distance of the point <\/strong> from the line <\/strong><\/p>\n

Ans. <\/strong>Given: A line <\/p>\n

<\/p>\n

<\/p>\n

Now, perpendicular distance of the point from the line is<\/p>\n

= = = 5 units<\/p>\n

\n

5. Find the points on the <\/strong>axis, where distances from the line <\/strong> are 4 units.<\/strong><\/p>\n

Ans. <\/strong>Let the coordinates of the point on axis be <\/p>\n

Now, perpendicular distance of the point from the line is<\/p>\n

= = <\/p>\n

According to question, <\/p>\n

<\/p>\n

or <\/p>\n

or <\/p>\n

or <\/p>\n

or <\/p>\n

Therefore, the points on axis are (8, 0) and <\/p>\n

\n

6. Find the distance between parallel lines:<\/strong><\/p>\n

(i) <\/strong> and <\/strong><\/p>\n

(ii) <\/strong> and <\/strong><\/p>\n

Ans. (i)<\/strong> Given: Two equations and <\/p>\n

Here, and <\/p>\n

Distance between two parallel lines = <\/p>\n

= = = units<\/p>\n

(ii)<\/strong> Given: Two equations and <\/p>\n

Here, and <\/p>\n

Distance between two parallel lines = <\/p>\n

= = units<\/p>\n

\n

7. Find equation of the line parallel to the line <\/strong> and passing through the point <\/strong><\/p>\n

Ans. <\/strong>Given: Equation of a line which is parallel to the line is .<\/p>\n

Since the line passes through point .<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the equation of required line is .<\/p>\n

\n

8. Find the equation of the line perpendicular to the line <\/strong> and having <\/strong> intercept 3.<\/strong><\/p>\n

Ans. <\/strong>Given: Equation of a line which is perpendicular to the line is .<\/p>\n

Since the line passes through point .<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the equation of required line is .<\/p>\n

\n

9. Find the angles between the lines <\/strong> and <\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

<\/p>\n

<\/p>\n

Also <\/p>\n

<\/p>\n

<\/p>\n

Let be the angle between the lines.<\/p>\n

<\/p>\n

= = = <\/p>\n

and <\/p>\n

and <\/p>\n

\n

10. The line through the points <\/strong> and (4, 1) intersects the line <\/strong> at right angle. Find the value of <\/strong><\/p>\n

Ans. <\/strong>Slope of the line passing through the points and (4, 1) = <\/p>\n

Also slope of the line is <\/p>\n

Since both lines are perpendicular to each other.<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

11. Prove that the line through the point <\/strong> and parallel to the line <\/strong> is <\/strong>.<\/strong><\/p>\n

Ans. <\/strong>Equation of the line parallel to the line is \u2026..(i)<\/p>\n

Since line (i) passes through , therefore \u2026..(ii)<\/p>\n

Subtracting eq. (ii) from eq. (i), we have <\/p>\n

\n

12. Two lines passing through the point (2, 3) intersects each other at an angle of <\/strong> If slope of the line is 2, find equation of the other line.<\/strong><\/p>\n

Ans. <\/strong>Given: and <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Taking <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Equation of required line is <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Taking <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Equation of required line is <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

\n

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and <\/strong><\/p>\n

Ans. <\/strong>Mid-point of the line segment joining the points 93, 4) and = = (1, 3)<\/p>\n

Slope of the line joining points (3, 4) and = <\/p>\n

Slope of the required line is <\/p>\n

Therefore, the required line passes through point (1, 3) having slope <\/p>\n

Equation of the required line <\/p>\n

<\/p>\n

<\/p>\n

\n

14. Find the coordinates of the foot of perpendicular from the point <\/strong> on the line <\/strong><\/p>\n

Ans. <\/strong>Let Q be the foot of perpendicular drawn from P on the line <\/p>\n

<\/p>\n

Equation of a line perpendicular to is <\/p>\n

Since the line passes through <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, Q is a point of intersection A<\/p>\n

of the lines and <\/p>\n

Solving both the equations, we have and <\/p>\n

Therefore, coordinates of foot of perpendicular are <\/p>\n

\n

15. The perpendicular from the origin to the line <\/strong> meets it at the point <\/strong> Find the value of <\/strong> and <\/strong><\/p>\n

Ans. <\/strong>Equation of the line PQ<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Slope of the required line which is perpendicular to this line is <\/p>\n

Equation of the line AB is<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Comparing with we have and <\/p>\n

\n

16. If <\/strong> and <\/strong> are the lengths of perpendiculars from the origin to the line <\/strong> and <\/strong> respectively, prove that <\/strong><\/p>\n

Ans. <\/strong>Length of perpendicular from origin to line is<\/p>\n

<\/p>\n

= = <\/p>\n

And Length of perpendicular from origin to line is<\/p>\n

<\/p>\n

= <\/p>\n

= = <\/p>\n

Now, <\/p>\n

= <\/p>\n

= <\/p>\n

\n

17. In the triangle ABC with vertices A (2, 3), B<\/strong> and C (1, 2), find the equation and length of altitude from the vertex A.<\/strong><\/p>\n

Ans. <\/strong>Slope of BC <\/p>\n

<\/p>\n

Since AD BC, therefore slope of AD = 1<\/p>\n

Equation of AD is<\/p>\n

<\/p>\n

<\/p>\n

And Equation of BC is<\/p>\n

<\/p>\n

<\/p>\n

Length of AD = <\/p>\n

= units<\/p>\n

\n

18. If <\/strong> is the length of perpendicular from the origin to the line whose intercepts on the axes are <\/strong> and <\/strong> then show that <\/strong><\/p>\n

Ans. <\/strong>Given: Line <\/p>\n

<\/p>\n

<\/p>\n

Now, is the length of perpendicular from origin to .<\/p>\n

= <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Exercise 10.3 1. Reduce the following equations into slope-intercept form and find their slopes and the intercepts. (i) (ii) (iii) Ans. (i) Given: \u2026\u2026\u2026.(i) Comparing with we have and (ii) Given: \u2026\u2026\u2026.(i) Comparing with we have and (iii) Given: \u2026\u2026\u2026.(i) Comparing with we have and 2. Reduce the following equations into intercept form and find … 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-4615","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Exercise 10.3 | 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-10-3\/\" \/>\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 10.3 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Exercise 10.3 1. Reduce the following equations into slope-intercept form and find their slopes and the intercepts. (i) (ii) (iii) Ans. (i) Given: \u2026\u2026\u2026.(i) Comparing with we have and (ii) Given: \u2026\u2026\u2026.(i) Comparing with we have and (iii) Given: \u2026\u2026\u2026.(i) Comparing with we have and 2. Reduce the following equations into intercept form and find ... 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