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# Prove that the three straight lines whose equations are $15x - 18y + 1 = 0$, $12x + 10y - 3 = 0$ and $6x + 66y - 11 = 0$ all meet at a point. Also, show that the third line bisects the angle between the other two.  Verified
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Hint- Here, we will be using a substitution method to find the point of intersection and also the general form of the equation of angle bisector of two straight lines will be used.
Given equations of the straight lines are
$15x - 18y + 1 = 0{\text{ }} \to {\text{(1)}} \\ 12x + 10y - 3 = 0{\text{ }} \to {\text{(2) }} \\ 6x + 66y - 11 = 0{\text{ }} \to {\text{(3)}} \\$
For the point of intersection of the first two straight lines given by equations (1) and (2),
Using equation (2), $12x + 10y - 3 = 0 \Rightarrow x = \dfrac{{3 - 10y}}{{12}}{\text{ }} \to {\text{(4)}}$
Put the above value of $x$ in equation (1), we get
${\text{15}}\left( {\dfrac{{3 - 10y}}{{12}}} \right) - 18y + 1 = 0 \Rightarrow 5\left( {\dfrac{{3 - 10y}}{4}} \right) - 18y + 1 = 0 \Rightarrow \dfrac{{15 - 50y}}{4} - 18y + 1 = 0 \\ \Rightarrow 15 - 50y - 72y + 4 = 0 \Rightarrow 15 - 122y + 4 = 0 \Rightarrow 122y = 19 \Rightarrow y = \dfrac{{19}}{{122}} \\$
Using equation (4), value of $x$ is given by $x = \dfrac{{3 - 10 \times \dfrac{{19}}{{122}}}}{{12}} = \dfrac{{\dfrac{{3 \times 122 - 10 \times 19}}{{122}}}}{{12}} = \dfrac{{366 - 190}}{{122 \times 12}} = \dfrac{{22}}{{183}}$
Therefore, the point of intersection of the lines given by equations (1) and (2) is ${\text{P}}\left( {\dfrac{{22}}{{183}},\dfrac{{19}}{{122}}} \right)$.
For all the given three straight lines to meet at a point ${\text{P}}\left( {\dfrac{{22}}{{183}},\dfrac{{19}}{{122}}} \right)$. This point must also satisfy the third equation of straight line given by equation (3). Let’s check this condition.
Put $x = \dfrac{{22}}{{183}}$ and $y = \dfrac{{19}}{{122}}$ in equation (3) and check whether it gets satisfied or not
$6 \times \dfrac{{22}}{{183}} + 66 \times \dfrac{{19}}{{122}} - 11 = 0 \Rightarrow \dfrac{{44}}{{61}} + \dfrac{{627}}{{61}} - 11 = 0 \Rightarrow \dfrac{{44 + 627}}{{61}} - 11 = 0 \Rightarrow 11 - 11 = 0 \Rightarrow 0 = 0$
This shows that the equation (3) is satisfied by the point ${\text{P}}\left( {\dfrac{{22}}{{183}},\dfrac{{19}}{{122}}} \right)$. Hence, all the given three straight lines meet at a point ${\text{P}}\left( {\dfrac{{22}}{{183}},\dfrac{{19}}{{122}}} \right)$.
Since, the equations of the angle bisector of two straight lines whose Cartesian equations are ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ is given by $\dfrac{{{a_1}x + {b_1}y + {c_1}}}{{\sqrt {{a_1}^2 + {b_1}^2} }} = \pm \dfrac{{{a_2}x + {b_2}y + {c_2}}}{{\sqrt {{a_2}^2 + {b_2}^2} }}$
Therefore, equations of the angle bisector of the first two straight lines whose Cartesian equations are given by equations (1) and (2) are
$\dfrac{{15x - 18y + 1}}{{\sqrt {{{15}^2} + {{18}^2}} }} = \pm \dfrac{{12x + 10y - 3}}{{\sqrt {{{12}^2} + {{10}^2}} }} \Rightarrow \dfrac{{15x - 18y + 1}}{{3\sqrt {61} }} = \pm \dfrac{{12x + 10y - 3}}{{2\sqrt {61} }} \\ \Rightarrow \dfrac{{15x - 18y + 1}}{3} = \pm \dfrac{{12x + 10y - 3}}{2} \Rightarrow 2\left( {15x - 18y + 1} \right) = \pm 3\left( {12x + 10y - 3} \right) \\ \Rightarrow 30x - 36y + 2 = \pm \left( {36x + 30y - 9} \right) \\$
$\Rightarrow 30x - 36y + 2 = 36x + 30y - 9 \Rightarrow 6x + 66y - 11 = 0$ and ${\text{3}}0x - 36y + 2 = - \left( {36x + 30y - 9} \right) \Rightarrow 66x - 6y - 7 = 0$
Therefore, the equation of the third straight line given by equation (3) is one of the angle bisector’s equation which means that the given third straight line is the angle bisector of the first two straight lines.

Note- These types of problems are solved by solving any two equations for the point of intersection and then substituting this point in the third equation to check whether that point is the common point of intersection between all the three given lines. For the second part, the concept of the equation of the angle bisector of straight lines needs to be clear.