NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.2

Content Curator

Class 11 Maths NCERT Solutions Chapter 10 Straight Lines Exercise 10.2 is based on forms of Equation of a Line. Following concepts are covered under this topic:

Horizontal and vertical lines
Point-slope form
Two-point form
Slope-intercept form

Download PDF NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

Check out the solutions of Class 11 Maths NCERT solutions Chapter 10 Straight Lines Exercise 10.2

Also check other Exercise Solutions of Class 11 Maths Chapter 10 Straight Lines

Also check:

Chapter Related Topics
Horizontal and vertical lines	Analytical Geometry	Intercept
Slope Formula	Slope Intercept Form Formula	Point Gradient Formula
Perpendicular Line Formula	Point Slope Form Formula	Different Forms of Equation of Line

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Arithmetic

CBSE CLASS XII Related Questions

1.
Sports car racing is a form of motorsport which uses sports car prototypes.The competition is held on special tracks designed in various shapes.
The equation of a sports car racing track is given as: \[ f(x)= \begin{cases} x^4-4x^2+4, & 0\leq x<3,\\ x^2+40, & x\geq 3 \end{cases} \] Based on this information:
2.
Evaluate: \[ \cos\left[\sin^{-1}(-1)-\tan^{-1}(-\sqrt{3})\right]. \]
View Solution
3.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]
- Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
4.
If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is:
- \(\frac{1}{3}\)
- \(\frac{1}{9}\)
- \(3\)
- \(9\)
View Solution
5.
Find the domain of \[ q(x)=\cos^{-1}(4x^2-3). \] Hence, find the value of \(x\) for which \[ q(x)=0. \] Also, write the range of \[ 3q(x)-\pi. \]
View Solution
6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]
- local maximum value is 2
- local minimum value is \( -2 \)
- local maximum value is \( -2 \)
- local minimum value \( < \) local maximum value
View Solution