Straight Lines MCQ Questions & Answers in Geometry | Maths

The sides given are $$y = \sqrt 3 \left( {x - 1} \right) + 2$$    and $$y = - \sqrt 3 x.$$   Lines are at an angle $${60^ \circ }$$  to each other.
Now any line parallel to obtuse angle bisector will make equilateral triangle with these lines as its two sides.

Therefore, infinitely many lines.

Let the line cuts the axes at points $$A\left( {a,\,0} \right)$$   and $$B\left( {0,\,b} \right).$$   Now, given that $$\left( { - 4,\,3} \right)$$  divides $$AB$$  in the ratio $$5 : 3.$$  Then, $$ - 4 = \frac{{3a}}{8}$$   and $$3 = \frac{{5b}}{8}.$$
Therefore, $$a = \frac{{ - 32}}{3}$$   and $$b = \frac{{24}}{5}.$$   Then using the intercept form $$\frac{x}{a} + \frac{y}{b} = 1,$$   the equation of line is
$$\eqalign{ & - \frac{{3x}}{{32}} + \frac{{5y}}{{24}} = 1 \cr & {\text{or }}9x - 20y + 96 = 0 \cr} $$

$$\eqalign{ & x = \frac{{a\,\cos \,t + b\,\sin \,t + 1}}{3} \Rightarrow a\,\cos \,t + b\,\sin \,t = 3x - 1 \cr & y = \frac{{a\,\sin \,t - b\,\cos \,t}}{3} \Rightarrow a\,\sin \,t - b\,\cos \,t = 3y \cr} $$
Squaring & adding, $${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$

The intersection point of two lines is $$\left( {\frac{{ - c}}{{a + b}},\,\frac{{ - c}}{{a + b}}} \right)$$
Distance between $$\left( {1,\,1} \right)$$  and
$$\eqalign{ & \left( {\frac{{ - c}}{{a + b}},\,\frac{{ - c}}{{a + b}}} \right) < 2\sqrt 2 \cr & \Rightarrow 2{\left( {1 + \frac{c}{{a + b}}} \right)^2} < 8 \cr & \Rightarrow 1 + \frac{c}{{a + b}} < 2 \cr & \Rightarrow a + b - c > 0 \cr} $$

$$\eqalign{ & 2x = 3y = - z \cr & {\text{or }}\frac{x}{3} = \frac{y}{2} = \frac{z}{{ - 6}} \cr & 6x = - y = - 4z \cr & {\text{or }}\frac{x}{2} = \frac{y}{{ - 12}} = \frac{z}{{ - 3}} \cr & \cos \,\theta = \frac{{{x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2}}}{{\sqrt {x_1^2 + x_2^2 + x_3^2} .\sqrt {y_1^2 + y_2^2 + y_3^2} }} \cr & \cos \,\theta = \frac{{\left( {6 - 24 + 18} \right)}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 6} \right)}^2}} .\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }} \cr & \cos \,\theta = 0 \cr & \theta = {90^ \circ } \cr} $$
So lines are perpendicular.

Let $$R\left( {h,\,k} \right)$$   be the variable point. Then,

$$\eqalign{ & \angle RPQ = \theta {\text{ and }}\angle RQP = \phi ,\,{\text{so that }}\theta - \phi = 2\alpha \cr & {\text{Let, }}RM \bot PQ,{\text{ so that}} \cr & RM = k,\,MP = a - h{\text{ and }}MQ = a + h \cr & {\text{Then, }}\tan \,\theta = \frac{{RM}}{{MP}} = \frac{k}{{a - h}}, \cr & \tan \,\phi = \frac{{RM}}{{MQ}} = \frac{k}{{a + h}} \cr & {\text{Therefore, from }}2\alpha = \theta - \phi ,\,{\text{we have}} \cr & \tan \,2\alpha = \tan \left( {\theta - \phi } \right) = \frac{{\tan \,\theta - \tan \,\phi }}{{1 + \tan \,\theta \,\tan \,\phi }} = \frac{{k\left( {a + h} \right) - k\left( {a - h} \right)}}{{{a^2} - {h^2} + {k^2}}} \cr & \Rightarrow {a^2} - {h^2} + {k^2} - 2hk\,\cot \,2\alpha = 0 \cr} $$
Therefore, the locus of $$R\left( {h,\,k} \right)$$   is $${x^2} - {y^2} + 2xy\,\cot \,2\alpha - {a^2} = 0$$
Hence, (A) is the correct answer.

$$\eqalign{ & {\left( {x - {a_1}} \right)^2} + {\left( {y - {b_1}} \right)^2} = {\left( {x - {a_2}} \right)^2} + {\left( {y - {b_2}} \right)^2} \cr & \left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + \frac{1}{2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) = 0 \cr & c = \frac{1}{2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) \cr} $$

The distance between the parallel straight lines given by
$$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$        is $$2\sqrt {\frac{{{g^2} - ac}}{{a\left( {a + b} \right)}}} $$
Here, $$a = 8,\,b = 2,\,c = 15,\,g = 13.$$
So, required distance
$$ = 2\sqrt {\frac{{169 - 120}}{{80}}} = 2 \times \frac{7}{{4\sqrt 5 }} = \frac{7}{{2\sqrt 5 }}$$

Let the lines be $$y = {m_1}x$$   and $$y = {m_2}x$$   then
$$\eqalign{ & {m_1} + {m_2} = - \frac{{2c}}{7}{\text{ and }}{m_1}{m_2} = - \frac{1}{7} \cr & {\text{Given }}{m_1} + {m_2} = 4\,{m_1}{m_2} \cr & \Rightarrow \frac{{2c}}{7} = - \frac{4}{7}\,\,\,\, \Rightarrow c = 2 \cr} $$

Total number of integral points inside the square $$OABC\,\, = 40 \times 40 = 1600$$
Number of integral points on $$AC$$

$$=$$ Number of integral points on $$OB$$
$$=40$$   [namely (1, 1), (2, 2) ... (40, 40)]
Number of integral points inside the $$\Delta OAC$$
$$ = \frac{{1600 - 40}}{2} = 780$$