**NCERT Solutions Class 11 Maths Chapter 10 Straight Lines** – Here are all the NCERT solutions for Class 11 Maths Chapter 10. This solution contains questions, answers, images, explanations of the complete chapter 10 titled Of Straight Lines taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 10 Straight Lines After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines in one place.

## NCERT Solutions Class 11 Maths Chapter 10 Straight Lines

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 10 Straight Lines , Maths, Class 11.

Class | 11 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 10 |

Chapter Name |
Straight Lines |

### NCERT Solutions Class 11 Maths chapter 10 Straight Lines

Class 11, Maths chapter 10, Straight Lines solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Question & Answer

Q.1:Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Ans :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 \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \text { and }\left(x_{3}, y_{3}\right)\) is \(\frac{1}{2} | x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)|\) Therefore, area of ΔABC \(\begin{aligned} &=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)| \text { unit }^{2} \\ &=\frac{1}{2}|-4(12)+5(-2)| \text { unit }^{2} \\ &=\frac{1}{2}|-48-10| \text { unit }^{2} \\ &=\frac{1}{2}|-58| \text { unit }^{2} \\ &=\frac{1}{2} \times 58 \text { unit }^{2} \\ &=29 \text { unit }^{2} \end{aligned}\) Area of ΔACD \(\begin{aligned} &=\frac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5+5)| \text { unit }^{2} \\ &=\frac{1}{2}|-4(-3)+5(-7)-4(10)| \text { unit }^{2} \\ &=\frac{1}{2}|12-35-40| \text { unit }^{2} \end{aligned}\) \(\begin{array}{l}{=\frac{1}{2}|-63| \text { unit }^{2}} \\ {=\frac{63}{2} \text { unit }^{2}}\end{array}\) Thus, area (ABCD) \(=\left(29+\frac{63}{2}\right) \text { unit }^{2}=\frac{58+63}{2} \text { unit }^{2}=\frac{121}{2} \text { unit }^{2}\)

Q.2: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.

Ans :Let ABC be the given equilateral triangle with side 2a.Accordingly, AB = BC = CA = 2a Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. i.e., BO = OC = a, where O is the origin. Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, -a). It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the y-axis. On applying Pythagoras theorem to \(\Delta\)AOC, we obtain \(\begin{array}{l}{(A C)^{2}=(O A)^{2}+(O C)^{2}} \\ {\Rightarrow(2 a)^{2}=(O A)^{2}+a^{2}} \\ {\Rightarrow 4 a^{2}-a^{2}=(O A)^{2}} \\ {\Rightarrow(O A)^{2}=3 a^{2}} \\ {\Rightarrow O A=\sqrt{3} a}\end{array}\) \(\therefore \text { Coordinates of point } \mathrm{A}=\left( \begin{array}{c}{ \pm \sqrt{3} a, 0 )}\end{array}\right.\) \(\begin{array}{l}{\text { Thus, the vertices of the given equilateral triangle are }(0, a),(0,-a), \text { and }(\sqrt{3} a, 0) \text { or }} \\ {(0, \text { a), }(0,-a), \text { and }}\end{array}\) \((-\sqrt{3} a, 0)\).

Q.3:Find the distance between \(\mathrm{P}\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Ans :The given points are \(\mathrm{P}\left(x_{1}, y_{1}\right)_{\text { and }} \mathrm{Q}\left(x_{2}, y_{2}\right)\). (i) When PQ is parallel to the y-axis, \(x_{1}=x_{2}\). In this case, distance between P and Q \(\begin{array}{l}{=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}} \\ {=\sqrt{\left(y_{2}-y_{1}\right)^{2}}} \\ {=\left|y_{2}-y_{1}\right|}\end{array}\) (ii) When PQ is parallel to the x-axis, \(\mathrm{y}_{1}=\mathrm{y}_{2}\) . In this case, distance between P and Q \(\begin{array}{l}{=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}} \\ {=\sqrt{\left(x_{2}-x_{1}\right)^{2}}} \\ {=\left|x_{2}-x_{1}\right|}\end{array}\)

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

Ans :Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4). \(\begin{array}{l}{\text { Accordingly, } \sqrt{(7-a)^{2}+(6-0)^{2}}=\sqrt{(3-a)^{2}+(4-0)^{2}}} \\ {\Rightarrow \sqrt{49+a^{2}-14 a+36}=\sqrt{9+a^{2}-6 a+16}} \\ {\Rightarrow \sqrt{a^{2}-14 a+85}=\sqrt{a^{2}-6 a+25}}\end{array}\) On squaring both sides, we obtain \(a^{2}-14 a+85=a^{2}-6 a+25\) ⇒ –14a + 6a = 25 – 85 ⇒ –8a = –60 \(\Rightarrow a=\frac{60}{8}=\frac{15}{2}\) Thus, the required point on the x-axis is \(\left(\frac{15}{2}, 0\right)\) .

Q.5: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)

Ans :The coordinates of the mid-point of the line segment joining the points P (0, –4) and B (8, 0) are It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, x_{2} \neq x_{1}\). Therefore, the slope of the line passing through (0, 0) and (4, –2) is \(\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\). Hence, the required slope of the line is \(-\frac{1}{2}\).

## NCERT / CBSE Book for Class 11 Maths

You can download the NCERT Book for Class 11 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 11 Maths
- Click here to buy NCERT Book for Class 11 Maths

### All NCERT Solutions Class 11

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### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 | Class 2 | Class 3 |

Class 4 | Class 5 | Class 6 |

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Class 10 | Class 11 | Class 12 |

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