NCERT full solution for class 11 Maths Chapter 9 Exercise 9.2 | cbse24

Q1:-Write the equations for the x-and y-axes.

Equation of the x-axis: Along the x-axis, the value of y is always 0, regardless of x. Therefore, the equation is:

$$y=0\mathrm{<br>}\mathrm{<br>}$$

Equation of the y-axis: Along the y-axis, the value of $x$ is always 0, regardless of $y$. Therefore, the equation is:

$$x=0$$

Q2:- Passing through the point (– 4, 3) with slope 1/2.

Q3:- Passing through (0, 0) with slope m. 

Q4:-To find the equation of the line passing through the point $(2, 2\sqrt{3})$ and inclined at an angle of $75^\circ$

Q5:-Intersecting the x-axis at a distance of 3 units to the left of the origin with slope –2.

Point of Intersection: Since the line intersects the x-axis 3 units to the left of the origin, the end of intersection with the x-axis is $(-3,0)$.

Slope: The slope of the line is given as $m=-2$.

Equation of the Line: Using the point-slope form of the equation of a line:

$$y-y_1=m(x-x_1)$$

where $(x_1,y_1)=(-3,0)$ and $m=-2$ we get:

$$y-0=-2(x+3)$$

Simplifying this:

$$y=-2(x+3)$$
$$y=-2x-6$$

Thus, the equation of the line is:

$$y = -2x - 6$$

Q6:- Intersecting the y-axis 2 units above the origin and making an angle of $30^\circ$ with the positive direction of the x-axis.

Point of Intersection: Since the line intersects the y-axis 2 units above the origin, the point of intersection is $(0,2)$.

The slope of the Line: The slope $m$ of the line is related to the angle $\theta$ that the line makes with the x-axis. The slope is given by:

$$m=\tan (\theta )$$

where $\theta =30^{\circ }$

We know that:

$$\tan (30^{\circ })=\frac{1}{\sqrt{3}}\approx 0.577\mathrm{<br>}$$

Equation of the Line: Using the point-slope form of the equation of a line:

$$y-y_1=m(x-x_1)$$

where $(x_1,y_1)=(0,2)$  and $m\approx 0.577$ , we get:

$$y-2=0.577(x-0)$$

Simplifying this:

$$y=0.577x+2$$

Thus, the equation of the line is:

$$y = 0.577x + 2$$

Q7:-Passing through the points (–1, 1) and (2, – 4).

Q8:-The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find the equation of the median through the vertex R

Q9:-Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Q10:- A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line

Q11:-Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the points (2, 3)

Q12:-Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Q13:-Find the equation of the line through the point (0, 2) making an angle 2π/ 3 with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin. 

Q14:-The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line. 

Q15:-The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Q16:-The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Q17:-P (a, b) is the mid-point of a line segment between axes. Show that the equation of the line is (x/a)+(y/b)=2.

Q18:-Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find an equation of the line.

Q19:-By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear

Read Also:-

• Straight line Exercise 9.1