Straight Lines MCQ Questions & Answers in Geometry | Maths

The points $$A,\,B,\,C,\,D$$    form a trapezium whose line of symmetry is the $$y$$-axis.

Let $$BE$$  be the required line where $$ED = \lambda .$$
$$\eqalign{ & {\text{ar}}\left( {BCDE} \right) = \frac{1}{2}{\text{ar}}\left( {ABCD} \right) \cr & \Rightarrow \frac{1}{2}.\left( {2 + \lambda } \right).2 = \frac{1}{2}.\frac{1}{2}\left( {4 + 2} \right).2 \cr & \Rightarrow \lambda = 1 \cr & \therefore {\text{ the point }}E = \left( {1,\,4} \right) \cr} $$
So, the required line $$BE$$  is $$y - 2 = \frac{{4 - 2}}{{1 + 1}}\left( {x + 1} \right).$$

$$2x+11y=5$$   is one of the bisectors of the angles between the given lines.

The combined equation of bisectors of angles between the lines of the first pair is $$\frac{{{x^2} - {y^2}}}{{2 - 1}} = \frac{{xy}}{3},$$    and that of the other pair is $$\frac{{{x^2} - {y^2}}}{{4 - 1}} = \frac{{xy}}{9}$$
As these equations are the same, the two pairs are equally inclined to each other.

The lines are $$y=x,\, y=-x$$    and $$x+2y=2$$   as shown in the figure.
Solving $$y=x,\,x+2y=2,$$    the point $$A = \left( {\frac{2}{3},\frac{2}{3}} \right)$$
$$\eqalign{ & \therefore \,\,OA = \sqrt {\frac{4}{9} + \frac{4}{9}} = \frac{{2\sqrt 2 }}{3} \cr & {\text{Solving }}y = - x,\,x + 2y = 2,{\text{ the point}} \cr & B = \left( { - 2,\,2} \right) \cr & \therefore \,\,OB = \sqrt {4 + 4} = 2\sqrt 2 \cr & \therefore \,\,{\text{ar}}\left( {\Delta OAB} \right) = \frac{1}{2}.OA.OB = \frac{1}{2}.\frac{{2\sqrt 2 }}{3}.2\sqrt 2 = \frac{4}{3} \cr} $$

$$\eqalign{ & {\text{If }}m,{\text{ }}2m{\text{ are slopes then}} \cr & m + 2m = - \frac{{\frac{2}{h}}}{{\frac{1}{b}}} = - \frac{{2b}}{h}{\text{ and }}m.2m = \frac{{\frac{1}{a}}}{{\frac{1}{b}}} = \frac{b}{a} \cr & {\text{Eliminating }}m,\,2{\left( { - \frac{{2b}}{{3h}}} \right)^2} = \frac{b}{a}\,\,\,\, \Rightarrow \frac{{ab}}{{{h^2}}} = \frac{9}{8} \cr} $$

Let the line make intercept $$'a'$$ on $$x$$-axis.
Then, it makes intercept $$'2a'$$  on $$y$$-axis.
Therefore, the equation of the line is given by $$\frac{x}{a} + \frac{y}{{2a}} = 1$$
It passes through $$\left( {1,\,2} \right),$$  so, we have $$\frac{1}{a} + \frac{2}{{2a}} = 1{\text{ or }}a = 2$$
Therefore, the required equation of the line is given by $$\frac{x}{2} + \frac{y}{4} = 1{\text{ or }}2x + y = 4$$

Any point on the line through $$\left( {1,\,1} \right)$$  and parallel to $$x+y=1$$   is $$P\left( {1 + r\cos \,\theta ,\,1 + r\sin \,\theta } \right)$$      where $$\tan \,\theta = - 1.$$   Use the fact that $$P$$ is on the line $$2x-3y=4$$    and find $$\left| r \right|.$$

Equation of $$AB$$  is
$$\eqalign{ & x\,\cos \,\alpha + y\,\sin \,\alpha = p\,; \cr & \Rightarrow \frac{{x\,\cos \,\alpha }}{p} + \frac{{y\,\sin \,\alpha }}{p} = 1\,; \cr & \Rightarrow \frac{x}{{\frac{p}{{\cos \,\alpha }}}} + \frac{y}{{\frac{p}{{\sin \,\alpha }}}} = 1 \cr} $$
So, co-ordinates of $$A$$ and $$B$$ are $$\left( {\frac{p}{{\cos \,\alpha }},\,0} \right){\text{ and }}\left( {0,\,\frac{p}{{\sin \,\alpha }}} \right)$$
So, co-ordinates of midpoint of $$AB$$  are
$$\eqalign{ & \left( {\frac{p}{{2\,\cos \,\alpha }},\,\frac{p}{{2\,\sin \,\alpha }}} \right) = \left( {{x_1},\,{y_1}} \right)\left( {{\text{say}}} \right)\,; \cr & {x_1} = \frac{p}{{2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = \frac{p}{{2\,\sin \,\alpha }}\,; \cr & \Rightarrow \cos \,\alpha = \frac{p}{{2{x_1}}}{\text{ and }}\sin \,\alpha = \frac{p}{{2{y_1}}}\,; \cr} $$
Consider $${\cos ^2}\alpha + {\sin ^2}\alpha = 1$$
$$ \Rightarrow \frac{{{p^2}}}{4}\left( {\frac{1}{{x_1^2}} + \frac{1}{{y_1^2}}} \right) = 1$$
$$\therefore \,{\text{Locus }}\left( {{x_1},\,{y_1}} \right){\text{ is}}\,\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} = \frac{4}{{{p^2}}}$$

Clearly $$OP=OQ=1$$    and $$\angle QOP = \alpha - \theta - \theta = \alpha - 2\theta .$$

The bisector of $$\angle QOP$$   will be a perpendicular to $$PQ$$  and also bisect it. Hence $$Q$$ is reflection of $$P$$ in the line $$OM$$  which makes an angle $$\angle MOP + \angle POX$$     with $$x$$-axis, i.e., $$\frac{1}{2}\left( {\alpha - 2\theta } \right) + \theta = \frac{\alpha }{2}$$
So that slope of $$OM$$  is tan $$\frac{\alpha }{2}.$$

Distance of all the points from $$\left( {0,\,0} \right)$$  are $$5$$ unit. That means circumcentre of the triangle formed by the given point is $$\left( {0,\,0} \right).$$  If $$G\left( {h,\,k} \right)$$   be the centroid of triangle, then
$$3h = 3 + 5\left( {\cos \,\theta + \sin \,\theta } \right),\,3k = 4 + 5\left( {\sin \,\theta - \cos \,\theta } \right)$$
If $$H\left( {\alpha ,\,\beta } \right)$$   be the orthocentre, then
$$\eqalign{ & OG:GH = 1:2\, \Rightarrow \alpha = 3h,\,\,\beta = 3k \cr & \cos \,\theta + \sin \,\theta = \frac{{\alpha - 3}}{5},\,\sin \theta - \cos \,\theta = \frac{{\beta - 4}}{5} \cr & \Rightarrow \sin \,\theta = \frac{{\alpha + \beta - 7}}{{10}},\,\,\cos \,\theta = \frac{{\alpha - \beta + 1}}{{10}} \cr} $$
Thus, locus of $$\left( {\alpha ,\,\beta } \right)$$  is $${\left( {x + y - 7} \right)^2} + {\left( {x - y + 1} \right)^{^2}} = 100.$$