Straight Lines MCQ Questions & Answers in Geometry | Maths

Equation of bisectors of lines, $$xy=0$$   are $$y = \pm x$$

Put $$y = \pm x$$   in the given equation
$$\eqalign{ & m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0 \cr & \therefore m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0 \cr & \Rightarrow 1 - {m^2} = 0\,\,\, \Rightarrow m = \pm 1 \cr} $$

The coordinates of points $$P,\,Q,\,R$$   are $$\left( { - 1,\,0} \right),\,\left( {0,\,0} \right),\,\left( {3,\,3\sqrt 3 } \right)$$     respectively.

Slope of $$QR = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{3\sqrt 3 }}{3}$$
$$\eqalign{ & \Rightarrow \tan \,\theta = \sqrt 3 \cr & \Rightarrow \theta = \frac{\pi }{3} \cr & \Rightarrow \angle RQX = \frac{\pi }{3} \cr & \therefore \,\angle RQP = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}\,; \cr} $$
Let $$QM$$  bisects the $$\angle PQR,$$
$$\therefore $$  Slope of the line $$QM = \tan \frac{{2\pi }}{3} = - \sqrt 3 $$
$$\therefore $$  Equation of line $$QM$$  is
$$\eqalign{ & \left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right) \cr & \Rightarrow y = - \sqrt 3 x \cr & \Rightarrow \sqrt 3 x + y = 0 \cr} $$

Let $$A\left( {{x_1},\,{y_1}} \right),\,B\left( {{x_2},\,{y_2}} \right)$$     and $$C\left( {{x_3},\,{y_3}} \right)$$   be the vertices of $$\Delta ABC$$   and let $$lx + my + n = 0$$     be the equation of the line. If $$P$$ divides $$BC$$  in the ratio $$\lambda :1,$$  then the coordinates of $$P$$ are $$\left( {\frac{{\lambda {x_3} + {x_2}}}{{\lambda + 1}},\,\frac{{\lambda {y_3} + {y_2}}}{{\lambda + 1}}} \right)$$

Also, as $$P$$ lies on $$L,$$ we have
$$\eqalign{ & l\left( {\frac{{\lambda {x_3} + {x_2}}}{{\lambda + 1}}} \right) + m\left( {\frac{{\lambda {y_3} + {y_2}}}{{\lambda + 1}}} \right) + n = 0 \cr & \Rightarrow - \frac{{l{x_2} + m{y_2} + n}}{{l{x_3} + m{y_3} + n}} = \lambda = \frac{{BP}}{{PC}}......\left( {\text{i}} \right) \cr} $$
Similarly, we obtain
$$\eqalign{ & \frac{{CQ}}{{QA}} = - \frac{{l{x_3} + m{y_3} + n}}{{l{x_1} + m{y_1} + n}}......\left( {{\text{ii}}} \right) \cr & {\text{and }}\frac{{AR}}{{RB}} = - \frac{{l{x_1} + m{y_1} + n}}{{l{x_2} + m{y_2} + n}}......\left( {{\text{iii}}} \right) \cr} $$
On multiplying equations $$\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right)$$  and $$\left( {{\text{iii}}} \right),$$ we get
$$\frac{{BP}}{{PC}}.\frac{{CQ}}{{QA}}.\frac{{AR}}{{RB}} = - 1$$

Let $$\left( {{a^2},\,a} \right)$$   be the point of shortest distance on $$x = {y^2}.$$   Then distance between $$\left( {{a^2},\,a} \right)$$   and line $$x-y+1=0$$   is given by
$$D = \frac{{{a^2} - a + 1}}{{\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\left[ {{{\left( {a - \frac{1}{2}} \right)}^2} + \frac{3}{4}} \right]$$
It is minimum when $$a = \frac{1}{2}$$   and $${D_{\min .}} = \frac{3}{{4\sqrt 2 }} = \frac{{3\sqrt 2 }}{8}$$

$$\eqalign{ & 1.{L_1} - 1.{L_2} - 1.{L_3} \equiv 0 \cr & \Rightarrow {\text{the straight lines are concurrent}}{\text{.}} \cr} $$

We know that orthocenter is the meeting point of altitudes of a $$\Delta $$

Equation of alt. AD
$$ \Rightarrow $$ line parallel to $$y$$-axis through (3, 4)
$$ \Rightarrow x = 3.....(1)$$
Similarly equation of $$OE \bot AB$$   is
$$\eqalign{ & y = - \frac{{3 - 4}}{{4 - 0}}x \cr & \Rightarrow y = \frac{x}{4}.....(2) \cr} $$
Solving (1) and (2), we get orthocenter as $$\left( {3,\,\frac{3}{4}} \right).$$

Slope of $$PQ = \frac{{3 - 4}}{{k - 1}} = \frac{{ - 1}}{{k - 1}}$$
$$\therefore $$ Slope of perpendicular bisector of $$PQ = \left( {k - 1} \right)$$
Also mid point of $$PQ\,\,\,\,\left( {\frac{{k + 1}}{2},\,\frac{7}{2}} \right)$$
$$\therefore $$ Equation of perpendicular bisector is
$$\eqalign{ & y - \frac{7}{2} = \left( {k - 1} \right)\left( {x - \frac{{k + 1}}{2}} \right) \cr & \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right) \cr & \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0 \cr & \therefore {\text{y - intercept}} = - \frac{{8 - {k^2}}}{{ - 2}} = - 4 \cr & \Rightarrow 8 - {k^2} = - 8\,\,{\text{or}}\,\,{k^2} = 16\,\,\,\, \Rightarrow k = \pm 4 \cr} $$

$$\eqalign{ & \sin \,\alpha - \frac{1}{{\sqrt 2 }} > 0{\text{ and }}\frac{1}{{\sqrt 2 }} - \cos \,\alpha > 0.....(1) \cr & {\text{or }}\sin \,\alpha - \frac{1}{{\sqrt 2 }} < 0{\text{ and }}\frac{1}{{\sqrt 2 }} - \cos \,\alpha < 0.....(2) \cr & \left( 1 \right) \Rightarrow \sin \,\alpha > \frac{1}{{\sqrt 2 }}{\text{ and }}\cos \,\alpha < \frac{1}{{\sqrt 2 }}\,\, \Rightarrow \frac{\pi }{4} < \alpha < \frac{{3\pi }}{4} \cr & \left( 2 \right) \Rightarrow \sin \,\alpha < \frac{1}{{\sqrt 2 }}{\text{ and }}\cos \,\alpha > \frac{1}{{\sqrt 2 }}\,\, \Rightarrow - \frac{\pi }{4} < \alpha < \frac{\pi }{4} \cr} $$

Use the fact that the two pairs have the same bisectors of angles.

$$\eqalign{ & {\text{Equation of }}AB = 2x - y + 3 = 0 \cr & \Rightarrow \Delta PAD \cong \Delta PBD \cr & \Rightarrow D{\text{ is foot of perpendicular}} \cr & \Rightarrow {\text{from }}P{\text{ to }}AB \cr & \frac{{\alpha - 1}}{2} = \frac{{\beta - 2}}{{ - 1}} = \frac{{ - \left( {2 \times 1 - 1 \times 2 + 3} \right)}}{{4 + 1}} \cr & \Rightarrow \frac{{\alpha - 1}}{2} = \frac{{\beta - 2}}{{ - 1}} = \frac{{ - 3}}{5} \cr & \Rightarrow \alpha = \frac{{ - 1}}{5},\,\,\beta = \frac{{13}}{5} \cr} $$