Straight Lines and Pair of Straight Lines

For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the triangle inequality theorem.

Let's apply this to our lengths: $5$, $5r$, and $5r^2$. We have three inequalities to check:

1. $5 + 5r > 5r^2$
2. $5 + 5r^2 > 5r$
3. $5r + 5r^2 > 5$

Solving these inequalities will give us a range of values for $r$. Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.

1. 1. $1 + r > r^2$

We can rewrite this as: $r^2 - r - 1 < 0$

$\Rightarrow\left[r-\left(\frac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{1+\sqrt{5}}{2}\right)\right]<0$

The roots of this quadratic equation are $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$, so the inequality holds for $r$ in the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$.

1. 2. $1 + r^2 > r$

We can rewrite this as: $r^2 - r + 1 > 0$

This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$.

1. 3. $r + r^2 > 1$

We can rewrite this as: $r^2 + r - 1 > 0$

Taking the intersection of the intervals from the three inequalities, we get $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ as the valid range for $r$.

Given the solution of the inequalities $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$, we need to determine which of the options are outside this interval.

1. Option A: $\frac{7}{4}$

The decimal representation of $\frac{7}{4}$ is 1.75, which is outside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ or approximately (-0.618, 1.618). Hence, r cannot be equal to $\frac{7}{4}$.

1. Option B: $\frac{5}{4}$

The decimal representation of $\frac{5}{4}$ is 1.25, which is inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{5}{4}$.

1. Option C: $\frac{3}{4}$

The decimal representation of $\frac{3}{4}$ is 0.75, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{4}$.

1. Option D: $\frac{3}{2}$

The decimal representation of $\frac{3}{2}$ is 1.5, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{2}$.

Therefore, the only value that r cannot be equal to, among the given options, is $\frac{7}{4}$. So, the correct answer is:

Option A: $\frac{7}{4}$

The number of integral points lie inside the triangle are

1. If x = 1, then y may be 1, 2, 3, ....., 39

2. If x = 2, then y may be 1, 2, 3, ....., 38

3. If x = 3, then y may be 1, 2, 3, ....., 37

$ \vdots $

39. If x = 39, then the value of y is 1.

Hence, the number of interior points are

Since the point of intersection lies on fourth quadrant and equidistant from the two axes,

i.e., let the point be (k, $-$k) and this point satisfies the two equations of the given lines.

$\therefore$ 4ak $-$ 2ak + c = 0 ......... (1)

and 5bk $-$ 2bk + d = 0 ..... (2)

From (1) we get, $k = {{ - c} \over {2a}}$

Putting the value of k in (2) we get,

$5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0$

or, $ - {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0$ or, $ - {{3bc} \over {2a}} + d = 0$

or, $ - 3bc + 2ad = 0$ or, $3bc - 2ad = 0$

$x + \sqrt 3 y = \sqrt 3 $ or $y = - {1 \over {\sqrt 3 }}x + 1$

Let $\theta$ be the angle which the line makes with the positive x-axis.

$\therefore$ $\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)$ or $\theta = \pi - {\pi \over 6}$

$\therefore$ $\angle ABC = {\pi \over 6}$; $\therefore$ $\angle DBE = {\pi \over 6}$

$\therefore$ the equation of the line BD is,

$y = \tan {\pi \over 6}x + c$ or $y = {x \over {\sqrt 3 }} + c$ ..... (1)

The line $x + \sqrt 3 y = \sqrt 3 $ intersects the x-axis at $B(\sqrt 3 ,0)$ and, the line (1) passes through $B(\sqrt 3 ,0)$.

$\therefore$ $0 = {{\sqrt 3 } \over {\sqrt 3 }} + c$ or, c = $-$1

Hence, the equation of the reflected ray is,

$y = {x \over {\sqrt 3 }} - 1$ or $y\sqrt 3 = x - \sqrt 3 $

The point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 is

$ = \left( {{{3 \times 2 + 2 \times 1} \over {3 + 2}},{{3 \times 4 + 2 \times 1} \over {3 + 2}}} \right)$

$ = \left( {{{6 + 2} \over 5},{{12 + 2} \over 5}} \right) = \left( {{8 \over 5},{{14} \over 5}} \right)$

Since the line 2x + y = k passes through this point,

$\therefore$ $2 \times {8 \over 5} + {{14} \over 5} = k$ or ${{30} \over 5} = k$ or, k = 6