Straight Lines MCQ Questions & Answers in Geometry | Maths

Here $$p = 7$$  and $$\alpha = {30^ \circ }$$

$$\therefore $$  Equation of the required line is
$$\eqalign{ & x\,\cos \,{30^ \circ } + y\,\sin \,{30^ \circ } = 7 \cr & {\text{or }}x \times \frac{{\sqrt 3 }}{2} + y \times \frac{1}{2} = 7 \cr & {\text{or }}\sqrt 3 \,x + y = 14 \cr} $$

Let $$m$$ be the slope of $$PQ$$  then
$$\eqalign{ & \tan \,{45^ \circ } = \left| {\frac{{m - \left( { - 2} \right)}}{{1 + m\left( { - 2} \right)}}} \right| \cr & \Rightarrow 1 = \left| {\frac{{m + 2}}{{1 - 2m}}} \right| \cr & \Rightarrow \pm 1 = \frac{{m + 2}}{{1 - 2m}} \cr & \Rightarrow m + 2 = 1 - 2m\,\,\,{\text{or}}\,\, - 1 + 2m = m + 2 \cr & \Rightarrow m = - \frac{1}{3}\,\,\,\,\,\,{\text{or}}\,\,\,\,\,m = 3 \cr} $$

As $$PR$$  also makes $$\angle {45^ \circ }$$   with $$RQ.$$
$$\therefore $$ The above two values of $$m$$ are for $$PQ$$  and $$PR.$$
$$\therefore $$ Equation of $$PQ,$$
$$\eqalign{ & y - 1 = - \frac{1}{3}\left( {x - 2} \right) \cr & \Rightarrow 3y - 3 = - x + 2 \cr & \Rightarrow x + 3y - 5 = 0 \cr} $$
and equation of $$PR$$  is $$ \Rightarrow 3x - y - 5 = 0$$
$$\therefore $$ Combined equation of $$PQ$$  and $$PR$$  is
$$\eqalign{ & \left( {x - 3y - 5} \right)\left( {3x - y - 5} \right) = 0 \cr & \Rightarrow 3{x^2} - 3{y^2} + 8xy - 20x - 10y + 25 = 0 \cr} $$

$$x=1,\, y=1$$   make both $$x+y-3$$   and $$2x-y-2$$   negative. So the required bisector is $$\frac{{x + y - 3}}{{\sqrt 2 }} = \frac{{2x - y - 2}}{{\sqrt 5 }}.$$

The given equations of the set of all lines
$$px+qy+r=0 .....(1)$$
and given condition is :
$$\eqalign{ & 3p + 2q + 4r = 0 \cr & \Rightarrow \frac{3}{4}p + \frac{2}{4}q + r = 0.....(2) \cr} $$
From (1) & (2) we get :
$$\therefore x = \frac{3}{4},\,\,\,\,\,y = \frac{1}{2}$$
Hence the set of lines are concurrent and passing through the fixed point $$\left( {\frac{3}{4},\,\frac{1}{2}} \right).$$

Slope of $$x + y = 1$$   is $$ - 1$$
Slope of $$x - y = 1$$   is $$1$$
$$\eqalign{ & {\text{Let }}\tan \,A = - 1,\,\tan \,B = 1 \cr & A = \frac{{3\pi }}{4},\,B = \frac{\pi }{4} \cr & A - B = \frac{\pi }{2} \cr} $$

From the figure, we have
$$\eqalign{ & a = 2,\,\,b = 2\sqrt 2 ,\,\,c = 2 \cr & {x_1} = 0,\,{x_2} = 0,\,{x_3} = 2 \cr} $$

Now, $$x$$-co-ordinate of incentre is given as $$\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}}$$
$$ \Rightarrow \,x$$ -co-ordinate of incentre
$$\eqalign{ & = \frac{{2 \times 0 + 2\sqrt 2 .0 + 2.2}}{{2 + 2 + 2\sqrt 2 }} \cr & = \frac{2}{{2 + \sqrt 2 }} \cr & = 2 - \sqrt 2 \cr} $$

The line passing through the intersection of lines
$$ax+ 2by =3b=0$$     and $$bx-2ay-3a=0$$     is
$$\eqalign{ & ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0 \cr & \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0 \cr} $$
As this line is parallel to x-axis.
$$\eqalign{ & \therefore a + b\lambda = 0\,\, \Rightarrow \lambda = - \frac{a}{b} \cr & \Rightarrow ax + 2by + 3b - \frac{a}{b}\left( {bx - 2ay - 3a} \right) = 0 \cr & \Rightarrow ax + 2by + 3b - ax + \frac{{2{a^2}}}{b}y + \frac{{3{a^2}}}{b} = 0 \cr & y\left( {2b + \frac{{2{a^2}}}{b}} \right) + 3b + \frac{{3{a^2}}}{b} = 0 \cr & y\left( {\frac{{2{b^2} + 2{a^2}}}{b}} \right) = - \left( {\frac{{3{b^2} + 3{a^2}}}{b}} \right) \cr & y = \frac{{ - 3\left( {{a^2} + {b^2}} \right)}}{{2\left( {{b^2} + {a^2}} \right)}} = \frac{{ - 3}}{2} \cr} $$
So it is $$\frac{3}{2}$$ units below $$x$$-axis.

If $$a,\,b,\,c$$  are the $${p^{th}},{q^{th}}$$  and $${r^{th}}$$ terms of an $$AP$$  whose first term $$ = \lambda $$  and the common difference $$ = \mu $$  then the line is
$$\eqalign{ & \left\{ {\lambda + \left( {p - 1} \right)\mu } \right\}x + \left\{ {\lambda + \left( {q - 1} \right)\mu } \right\}y + \lambda + \left( {r - 1} \right)\mu = 0 \cr & {\text{or }}\lambda \left\{ {x + y + 1} \right\} + \mu \left\{ {\left( {p - 1} \right)x + \left( {q - 1} \right)y + r - 1} \right\} = 0, \cr & {\text{which is of the form }}{L_1} + k{L_2} = 0 \cr} $$

Since, the line $$BD$$  divides the triangle into two of equal area. $$BD$$  is a median and $$D$$ is $$\left( { - 1,\, - 2} \right).$$
Slope of $$BD = - \frac{1}{2}$$

So, the required line is
$$\eqalign{ & y + 3 = 2\left( {x - 1} \right) \cr & \Rightarrow y - 2x + 5 = 0 \cr} $$

The diagonals are perpendicular to each other.