Straight Lines MCQ Questions & Answers in Geometry | Maths

Point $$\left( {0,\,\alpha } \right)$$  lies on the $$y$$-axis. In such questions, it is easier to see the location of point if lines are drawn on the axis.
From the diagram, it is clear that $$\frac{5}{3} \leqslant \alpha \leqslant \frac{7}{2}$$

Clearly for point $$P,$$

$${a^2} - 3a < 0\,{\text{ and }}{a^2} - \frac{a}{2} > 0\,\, \Rightarrow \frac{1}{2} < a < 3$$

Clearly from the figure, the origin is contained in the acute angle. Writing the equations of the lines as $$2x-y+4=0$$    and $$-x+2y+1=0,$$    the required bisector is $$\frac{{2x - y + 4}}{{\sqrt 5 }} = \frac{{ - x + 2y + 1}}{{\sqrt 5 }}.$$

One vertex of square is $$\left( { - 4,\,5} \right)$$  and equation of one diagonal is $$7x - y + 8 = 0$$
Diagonal of a square are perpendicular and bisect each other.
Let the equation of the other diagonal be $$y = mx + c$$   where $$m$$ is the slope of the line and $$c$$ is the $$y$$-intercept.
Since this line passes through $$\left( { - 4,\,5} \right)$$
$$\therefore \,5 = - 4m + c......\left( {\text{i}} \right)$$
Since this line is at right angle to the line $$7x - y + 8 = 0$$    or $$y = 7x + 8,$$   having slope $$ = 7,$$
$$\therefore \,7 \times m = - 1{\text{ or }}\,m = \frac{{ - 1}}{7}$$
Putting this value of m in equation $$\left( {\text{i}} \right)$$ we get $$C = \frac{{31}}{7}$$
Hence, equation of the other diagonal is $$y = - \frac{1}{7}x + \frac{{31}}{7}{\text{ or }}x + 7y = 31$$

$$\eqalign{ & {\text{We have, }}PA = PB \cr & \Rightarrow {\left( {PA} \right)^2} = {\left( {PB} \right)^2} \cr & \Rightarrow {\left[ {x - \left( {a + b} \right)} \right]^2} + {\left[ {y - \left( {b - a} \right)} \right]^2} = {\left[ {x - \left( {a - b} \right)} \right]^2} + {\left[ {y - \left( {a + b} \right)} \right]^2} \cr & \Rightarrow {\left[ {\left( {x - a} \right) - b} \right]^2} + {\left[ {\left( {y - b} \right) + a} \right]^2} = {\left[ {\left( {x - a} \right) + b} \right]^2} + {\left[ {\left( {y - b} \right) - a} \right]^2} \cr & \Rightarrow {\left[ {\left( {x - a} \right) + b} \right]^2} - {\left[ {\left( {x - a} \right) - b} \right]^2} = {\left[ {\left( {y - b} \right) + a} \right]^2} - {\left[ {\left( {y - b} \right) - a} \right]^2} \cr & \Rightarrow 4b\left( {x - a} \right) = 4a\left( {y - b} \right) \cr & \Rightarrow bx = ay......\left( {\text{i}} \right) \cr} $$
Also, $$P\left( {a,\,b} \right)$$  satisfies the condition $$\left( {\text{i}} \right)$$ so that $$P$$ can be $$\left( {a,\,b} \right).$$

The sides are $$y=1$$  and the pair $${x^2} + 7xy + 2{y^2} = 0.$$     Clearly, one vertex is $$\left( {0,\,0} \right)$$  and the y-coordinate of each of the other two vertices is 1. Putting $$y=1$$  in the second equation, we get $${x^2} + 7xy + 2 = 0.$$
If $${x_1},\,{x_2}$$  are the roots then $${x_1} + {x_2} = - 7.$$
$$\therefore $$ the centroid $$ = \left( {\frac{{0 + {x_1} + {x_2}}}{3},\,\frac{{0 + 1 + 1}}{3}} \right) = \left( { - \frac{7}{3},\,\frac{2}{3}} \right).$$

The lines by which $$\Delta $$ is formed are $$x = 0, \,y = 0$$    and $$x+y=1.$$
Clearly, it is right $$\Delta $$ and we know that in a right $$\Delta $$ orthocenter coincides with the vertex at which right $$\angle $$ is formed.
$$\therefore $$ Orthocenter is $$\left( {0,\,0} \right).$$

Let the points be $$A,\,B,\,C$$   and $$D.$$ Clearly
$$\eqalign{ & 'm'\,{\text{of }}AD = 0 = 'm'\,{\text{of }}BC \cr & \therefore AD\left\| {BC} \right\|{\text{ }}x{\text{ - axis}}{\text{. Also }}A{D^2} = B{C^2} \cr & 'm'\,{\text{of }}AC = \infty = 'm'\,{\text{of }}BD \cr & \therefore AC\left\| {BD} \right\|{\text{ }}y{\text{ - axis}}{\text{. Also }}A{C^2} = B{D^2} \cr} $$

The lines are $$y=x-1$$  and $$y=-x+1$$   and $$\left( {\cos \,\theta ,\,\sin \,\theta } \right)$$   is any point on the circle $${x^2} + {y^2} = 1$$   where centre $$ = \left( {0,\,0} \right)$$  and radius $$=1$$
Clearly from the figure, $$\theta $$ can vary from $${ - \frac{\pi }{2}}$$  to $${\frac{\pi }{2}}.$$

Two of the given lines $$y + 3x - 5 = 0$$    and $$3y - x + 10 = 0$$    are $$ \bot $$ to each other, so the triangle is right angled and its circumradius is half the hypotenuse, $$y = x$$  intersects other two lines in $$\left( {\frac{5}{4},\,\frac{5}{4}} \right){\text{ and }}\left( { - 5,\, - 5} \right)$$
$$\therefore $$  Circumradius
$$\eqalign{ & = \frac{1}{2}\left[ {\sqrt {{{\left( { - 5 - \frac{5}{4}} \right)}^2} + {{\left( { - 5 - \frac{5}{4}} \right)}^2}} } \right] \cr & = \frac{1}{2}\left[ {\sqrt {\frac{{625}}{{16}} + \frac{{625}}{{16}}} } \right] \cr & = \frac{1}{2}\left[ {\frac{{25\sqrt 2 }}{4}} \right] \cr & = \frac{{25}}{{4\sqrt 2 }} \cr} $$