Straight Lines MCQ Questions & Answers in Geometry | Maths

Vertex of triangle is (1, 1) and midpoint of sides through this vertex is ($$-$$ 1, 2) and (3, 2)

$$ \Rightarrow $$ vertex $$B$$ and $$C$$ come out to be ($$-$$ 3, 3) and (5, 3)
$$\therefore $$ Centroid is $$\frac{{1 - 3 + 5}}{3},\,\frac{{1 + 3 + 5}}{3} \Rightarrow \left( {1,\,\frac{7}{3}\,} \right)$$

Given equation of straight lines are $$x + 2y - 9 = 0,\,3x + 5y - 5 = 0$$       and $$ax + by - 1 = 0$$
They are concurrent, if
$$\eqalign{ & - 5 + 5b - 2\left( { - 3 + 5a} \right) - 9\left( {3b - 5a} \right) = 0 \cr & \Rightarrow 35a - 22b + 1 = 0 \cr} $$
Thus, given straight lines are concurrent if the straight line $$35x - 22y + 1 = 0$$    passes through $$\left( {a,\,b} \right).$$

Put $$x = 0$$  in the given equation
$$ \Rightarrow b{y^2} + 2fy + c = 0$$
For unique point of intersection $${f^2} - bc = 0$$
$$ \Rightarrow a{f^2} - abc = 0$$
Since $$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$$
$$\eqalign{ & \Rightarrow 2fgh - b{g^2} - c{h^2} = 0 \cr & \Rightarrow 2fgh = b{g^2} + c{h^2} \cr} $$

Given that
$$\eqalign{ & P\left( {1,\,0} \right),\,Q\left( { - 1,\,0} \right)\,\,{\text{and}}\,\,\frac{{AP}}{{AQ}} = \frac{{BP}}{{BQ}} = \frac{{CP}}{{CQ}} = \frac{1}{3} \cr & \Rightarrow 3AP = AQ \cr & {\text{Let}}\,A = \left( {x,\,y} \right)\,\,{\text{then}}\,3AP = AQ \Rightarrow 9\,A{P^2} = A{Q^2} \cr & \Rightarrow 9{\left( {x - 1} \right)^2} + 9{y^2} = {\left( {x + 1} \right)^2} + {y^2} \cr & \Rightarrow 9{x^2} - 18x + 9 + 9{y^2} = {x^2} + 2x + 1 + {y^2} \cr & \Rightarrow 8{x^2} - 20x + 8{y^2} + 8 = 0 \cr & \Rightarrow {x^2} + {y^2} - \frac{5}{3}x + 1 = 0.....(1) \cr} $$
$$\therefore $$ A lies on the circle given by equation (1). As $$B$$ and $$C$$ also follow the same condition, they must lie on the same circle.
$$\therefore $$ Centre of circumcircle of $$\Delta ABC$$
$$=$$ Centre of circle given by $$\left( 1 \right) = \left( {\frac{5}{4},\,0} \right)$$

Total number of points within the square $$OABC =20 × 20=400$$

Points on line $$AB = 20\left( {\left( {1,1} \right),\,\left( {2,\,2} \right),.....\left( {20,\,20} \right)} \right)$$
$$\therefore $$ Points within $$\Delta OBC{\text{ and }}\Delta ABC = 400 - 20 = 380$$
By symmetry points within $$\Delta OAB = \frac{{380}}{2} = 190$$

The equation of the line in parametric form is $$\frac{{x - 3}}{{\cos \,\theta }} = \frac{{y - 4}}{{\sin \,\theta }} = r$$
Any point on this line is $$\left( {3 + r\,\cos \,\theta ,\,4 + r\,\sin \,\theta } \right)$$
It lies on $$x = 6$$  if $$3 + r\,\cos \,\theta = 6 \Rightarrow r = 3\,\sec \,\theta $$
$$\therefore \,PR = 3\,\sec \,\theta $$
Again the point lies on $$y = 8$$  if $$4 + r\,\sin \,\theta = 8$$
$$\therefore \,r = 4\,{\text{cosec}}\,\theta {\text{ or }}PS = 4\,{\text{cosec}}\,\theta $$
other options can be checked easily.

Taking co-ordinates as
$$\left( {\frac{x}{r},\,\frac{y}{r}} \right);\left( {x,\,y} \right)\,\,\& \,\,\left( {xr,\,yr} \right)$$
Then slope of line joining
$$\left( {\frac{x}{r},\,\frac{y}{r}} \right),\left( {x,\,y} \right)\, = \frac{{y\left( {1 - \frac{1}{r}} \right)}}{{x\left( {1 - \frac{1}{r}} \right)}} = \frac{y}{x}$$
and slope of line joining $$\left( {x,\,y} \right)\,\,\& \,\,\left( {xr,\,yr} \right)$$
$$\eqalign{ & = \frac{{y\left( {r - 1} \right)}}{{x\left( {r - 1} \right)}} = \frac{y}{x} \cr & \therefore \,\,\,{m_1} = {m_2} \cr} $$
$$ \Rightarrow $$ Points lie on the straight line.

The family of line through the given lines is
$$L \equiv x - 7y + 5 + \lambda \left( {x + 3y - 2} \right) = 0......\left( {\text{i}} \right)$$
For line $$L = 0$$  in the diagram, the distance of any point say $$\left( {2,\,0} \right)$$  on the line $$x + 3y - 2 = 0$$    from the line $$x - 7y + 5 = 0$$    and the line $$L = 0$$  must be the same.

Therefore,
$$\eqalign{ & \left| {\frac{{2 + 5}}{{\sqrt {50} }}} \right| = \left| {\frac{{2 + 2\lambda + 5 - 2\lambda }}{{\sqrt {{{\left( {1 + \lambda } \right)}^2} + {{\left( {3\lambda - 7} \right)}^2}} }}} \right| \cr & {\text{or }}10{\lambda ^2} - 40\lambda = 0 \cr & {\text{i}}{\text{.e}}{\text{., }}\lambda = 4{\text{ or }}0 \cr} $$
Hence, $$L = 0,\,\lambda = 4$$
Therefore, the required line is $$5x + 5y - 3 = 0.$$

The given pair is $$\left( {2x + y} \right)\left( {x - y} \right) = 0.$$
So, the required pair is $$\left( {2x + y + k} \right)\left( {x - y + k'} \right) = 0,$$       where $$2x+y+k=0$$    and $$x - y + k' = 0$$    pass through (1, 0).
$$\therefore \,\,\,k = - 2,\,\,k' = - 1$$
$$\therefore $$ the required pair is $$\left( {2x + y - 2} \right)\left( {x - y - 1} \right) = 0.$$

Here $$AB=BC=CA=2.$$
So, it is an equilateral triangle and the incentre coincides with centroid.
Therefore,
Incentre $$ = \left( {\frac{{0 + 1 + 2}}{3},\,\frac{{0 + 0 + \sqrt 3 }}{3}} \right) = \left( {1,\,\frac{1}{{\sqrt 3 }}} \right)$$