Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 23 - The Straight Lines

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Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 11 Science Maths Chapter 23 The Straight Lines are provided here with simple step-by-step explanations. These solutions for The Straight Lines are extremely popular among Class 11 Science students for Maths The Straight Lines Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 11 Science Maths Chapter 23 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 23.102:

Question 1:

Find the values of α so that the point P (α², α) lies inside or on the triangle formed by the lines x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0.

Answer:

Let ABC be the triangle of sides AB, BC and CA whose equations are x − 5y + 6 = 0, x − 3y + 2 = 0 and x − 2y − 3 = 0, respectively.
On solving the equations, we get A (9,3), B (4, 2) and C (13, 5) as the coordinates of the vertices.

It is given that point P (α², α) lies either inside or on the triangle. The three conditions are given below.

(i) A and P must lie on the same side of BC.

(ii) B and P must lie on the same side of AC.

(iii) C and P must lie on the same side of AB.

If A and P lie on the same side of BC, then

$(9 - 9 + 2) (α^{2} - 3 α + 2) \geq 0 \Rightarrow (α - 2) (α - 1) \geq 0$

$\Rightarrow α \in (- \infty, 1] \cup [2, \infty)$             ... (1)

If B and P lie on the same side of AC, then

$(4 - 4 - 3) (α^{2} - 2 α - 3) \geq 0 \Rightarrow (α - 3) (α + 1) \leq 0$

$\Rightarrow α \in [- 1, 3]$                           ... (2)

If C and P lie on the same side of AB, then

$(13 - 25 + 6) (α^{2} - 5 α + 6) \geq 0 \Rightarrow (α - 3) (α - 2) \leq 0$

$\Rightarrow α \in [2, 3]$                         ... (3)

From (1), (2) and (3), we get:

$α \in [2, 3]$

slope of the line = −5

slope of line joining by points (h, k) and (4,−13)

\frac{k + 13}{h - 4}

Answer:

Line with equal intercept from the axes is of the form $\frac{x}{a} + \frac{y}{a} = 1$
Since this line passes through (1, −2)

$i . e . \frac{1}{a} - \frac{2}{a} = 1 i . e . \frac{- 1}{a} = 1 i . e . a = - 1$
∴ equation of line is –x –y = 1
i.e. x + y + 1 = 0

Page No 23.136:

Question 4:

The locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is __________.

Answer:

Given line x sinθ + y cosθ = P meet axes at $(0, \frac{P}{\cos θ}) and (\frac{P}{\sin θ}, 0)$

$i . e . Let A (\frac{9}{\sin θ}, 0) and B (0, \frac{P}{\cos θ})$

Let P(h, k) be the midpoint of AB

$i . e . h = \frac{P}{2 \sin θ}, k = \frac{P}{2 \cos θ} i . e . \sin θ = \frac{p}{2 h}, \cos θ = \frac{p}{2 k} i . e . \sin θ = \frac{P}{2 h} and \cos θ = \frac{P}{2 k} ∴ \sin^{2} θ + \cos^{2} θ = \frac{P^{2}}{4 h^{2}} + \frac{P^{2}}{4 k^{2}} i . e . 1 = \frac{P^{2}}{4 h^{2}} + \frac{P^{2}}{4 k^{2}} i . e . locus is \frac{P^{2}}{4 x^{2}} + \frac{P^{2}}{4 y^{2}} = 1 i . e . \frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{4}{P^{2}}$

Page No 23.136:

Question 5:

The points (3, 4) and (2, –6) are situated on the ________ of the line 3x – 4y – 8 = 0.

Answer:

For given line 3x − 4y − 8 = 0
For (3, 4) and (2, −6) are such that 3(3) − 4(4) − 8 = 9 − 16 − 8 < 0 and 3(2) − 4(−6) − 8 = 6 + 24 − 8 > 0
i.e. (3, 4) and (2, −6) lie on opposite sides of 3x − 4y − 8 = 0

Page No 23.136:

Question 6:

If the points (–2, 1) and (–1, – 2) are on the opposite sides of the line 2x + y + a = 0, then the set of values of a is _________.

Question 10:

If the lines ax + 2y + 1 = 0, bx + y + 1 = 0 and cx + 4y + 1 = 0 are concurrent, then a, b, c are in __________.

Answer:

check ans
Given lines are,
$a x + 2 y + 1 = 0 . . . (1) b x + y + 1 = 0 . . . (2) c x + 4 y + 1 = 0 . . . (3) are concurrent$

Solving (1) and (2),

\frac{x}{2 - 1} = \frac{- y}{b - a} = \frac{1}{a - 2 b} i . e . x = \frac{1}{a - 2 b}, y = \frac{a - b}{a - 2 b}

Answer:

Since 2x + y − 5 = 0 and 4x + 2y − 15 = 0 are parallel
i.e. 2x + y − 5 = 0
2x + y − $\frac{15}{2}$ = 0 are parallel
Distance between these parallel lines is

$\begin{array}{rcl} \frac{|5 - \frac{15}{2}|}{\sqrt{4 + 1}} & = & \frac{|10 - 15|}{2 \sqrt{5}} = \frac{5}{2 \sqrt{5}} \\ = & \frac{\sqrt{5}}{2} \end{array}$
i.e. Distance between parallel sides which is length of side of square
$∴ Area of square is {(\frac{\sqrt{5}}{2})}^{2} = \frac{5}{4} sq . units$

Page No 23.137:

Question 17:

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are two parallel tangents to a circle, then the length of its diameter is _____________.

Answer:

Given two parallel tangents 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0
$i . e . 3 x - 4 y - \frac{7}{2} = 0$
Distance between these two parallel tangents is $\frac{|4 + \frac{7}{2}|}{\sqrt{9 + 16}}$

= \frac{|8 + 7|}{2 \times 5}

i.e distance between parallel tangents

Hence, the equation of the required line is

The required line is shown in the following figure.

The line which is inclined at an angle of 135° with the positive direction of y-axis makes an angle of 45° with x-axis.
Here, $m = \tan 45^{\circ} = 1$

Thus, the equation of the required line passing through (2, 0) with slope 1 is

$y - 0 = 1 (x - 2) \Rightarrow x - y - 2 = 0$

Page No 23.29:

Question 10:

Find the equation of the straight line which divides the join of the points (2, 3) and (−5, 8) in the ratio 3 : 4 and is also perpendicular to it.

Answer:

Let the required line divide the line joining the points $A (2, 3) and B (- 5, 8)$ at P (x₁, y₁).
Here, AP : PB = 3 : 4

$∴ P (x_{1}, y_{1}) = (\frac{4 \times 2 - 5 \times 3}{3 + 4}, \frac{4 \times 3 + 3 \times 8}{3 + 4}) = (- 1, \frac{36}{7})$

Now, slope of AB = $\frac{8 - 3}{- 5 - 2} = - \frac{5}{7}$
Let m be the slope of the required line.
Since, the required line is perpendicular to the line joining the points $A (2, 3) and B (- 5, 8)$

$∴ m \times Slope of the line joining the points A (2, 3) and B (- 5, 8) = - 1 \Rightarrow m \times (\frac{- 5}{7}) = - 1 \Rightarrow m = \frac{7}{5}$

Substituting

m = \frac{7}{5}, x_{1} = - 1 and y_{1} = \frac{36}{7}

y - y_{1} = m (x - x_{1})

we get,

y - \frac{36}{7} = \frac{7}{5} (x + 1) \Rightarrow 35 y - 180 = 49 x + 49 \Rightarrow 49 x - 35 y + 229 = 0

Hence, the equation of the required line is

49 x - 35 y + 229 = 0

∴ 2 (3 + \frac{2 r}{\sqrt{5}}) + 3 (5 + \frac{r}{\sqrt{5}}) = 14 \Rightarrow 6 + \frac{4 r}{\sqrt{5}} + 15 + \frac{3 r}{\sqrt{5}} = 14 \Rightarrow \frac{7 r}{\sqrt{5}} = - 7 \Rightarrow r = - \sqrt{5}

Hence, the distance of the point (3, 5) from the line 2x + 3y = 14 is

\sqrt{5}

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