Question

Let the equations of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0$. If the
orthocentre of this triangle is at $(1,1)$, then the equation of its third side is
(a) $122y-26x-1675=0$
(b) $26x+61y+1675=0$
(c) $122y+26x+1675=0$
(d) $26x-122y-1675=0$

Answer 1

Hint: Find the slope of the given two lines and use this slope and given orthocentre to find the coordinates of the triangle.

The equations of the sides of a triangle given in the questions are,
$3x-2y+6=0$ and $4x+5y-20=0$
The orthocentre of a triangle is the point where all the altitudes intersect. Its coordinates are given
in the question as $(1,1)$. Draw a triangle ABC with all the given data as shown below,
Consider the line AC. The slope of the line can be obtained by rearranging the equation as,

& 4x+5y-20=0 \\\ & 5y=20-4x \\\ & y=-\dfrac{4}{5}x+4 \\\ \end{aligned}$$ Comparing it with the general equation, $$y=mx+c$$, we have the slope as $$-\dfrac{4}{5}$$. The slope of perpendicular lines is related as $$m\times n=-1$$. Now since the line BR is perpendicular to the line AC, the slope of BR would be $$\dfrac{5}{4}$$. The line BR has slope $$\dfrac{5}{4}$$ and passes through the point $$(1,1)$$. The equation of a line passing through point $$({{x}_{1}},{{y}_{1}})$$ and having slope $$m$$ is given by,$$y-{{y}_{1}}=m(x- {{x}_{1}})$$. The equation of the line BR can hence be obtained as, $$\begin{aligned} & y-1=\dfrac{5}{4}(x-1) \\\ & 4(y-1)=5(x-1) \\\ & 4y-4=5x-5 \\\ & 5x-4y-1=0 \\\ \end{aligned}$$ Next, we have to consider line BC. The slope of the line can be obtained by rearranging the equation as, $$\begin{aligned} & 3x-2y+6=0 \\\ & -2y=-6-3x \\\ & y=\dfrac{3}{2}x+3 \\\ \end{aligned}$$ Comparing it with the general equation, $$y=mx+c$$, we have the slope as $$\dfrac{3}{2}$$. The slope of perpendicular lines is related as $$m\times n=-1$$. Now since the line AQ is perpendicular to the line BC, the slope of AQ would be $$-\dfrac{2}{3}$$. The line AQ has slope $$-\dfrac{2}{3}$$ and passes through the point $$(1,1)$$. The equation of a line passing through point $$({{x}_{1}},{{y}_{1}})$$ and having slope $$m$$ is given by,$$y-{{y}_{1}}=m(x- {{x}_{1}})$$. The equation of the line AQ can hence be obtained as, $$\begin{aligned} & y-1=-\dfrac{2}{3}(x-1) \\\ & 3(y-1)=-2(x-1) \\\ & 3y-3=-2x+2 \\\ & 2x+3y-5=0 \\\ \end{aligned}$$ The vertex A is passing through both lines AC and AQ. So, we can get the coordinates of vertex A by solving the equations $$4x+5y-20=0$$ and $$2x+3y-5=0$$. Multiplying the equation $$2x+3y-5=0$$ by 2 and subtracting from the equation $$4x+5y-20=0$$, we get, $$\dfrac{\begin{aligned} & 4x+5y-20=0 \\\ & -4x-6y+10=0 \\\ \end{aligned}}{\begin{aligned} & -y-10=0 \\\ & y=-10 \\\ \end{aligned}}$$ Substituting the value of $$y$$ in $$2x+3y-5=0$$, we get, $$\begin{aligned} & 2x+(3\times -10)-5=0 \\\ & 2x-30-5=0 \\\ & 2x=35 \\\ & x=\dfrac{35}{2} \\\ \end{aligned}$$ Therefore, the coordinates of A are $$\left( \dfrac{35}{2},-10 \right)$$. The vertex B is passing through both lines BC and BR. So, we can get the coordinates of vertex B by solving the equations $$3x-2y+6=0$$ and $$5x-4y-1=0$$ . Multiplying the equation $$3x-2y+6=0$$ by 2 and subtracting the equation $$5x-4y-1=0$$ from it, we get, $$\dfrac{\begin{aligned} & 6x-4y+12=0 \\\ & -5x+4y+1=0 \\\ \end{aligned}}{\begin{aligned} & x+13=0 \\\ & x=-13 \\\ \end{aligned}}$$ Substituting the value of $$x$$ in $$3x-2y+6=0$$, we get, $$\begin{aligned} & (3\times -13)-2y+6=0 \\\ & -39-2y+6=0 \\\ & -2y=33 \\\ & y=-\dfrac{33}{2} \\\ \end{aligned}$$ Therefore, the coordinates of B are $$\left( -13,-\dfrac{33}{2} \right)$$. The equation of a line passing through two points $$({{x}_{1}},{{y}_{1}})$$ and $$({{x}_{2}},{{y}_{2}})$$ is given by, $$\begin{aligned} & y-{{y}_{1}}=m(x-{{x}_{1}}) \\\ & \Rightarrow y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\times (x-{{x}_{1}}) \\\ \end{aligned}$$ The equation of line AB passing through points $$\left( \dfrac{35}{2},-10 \right)$$ and $$\left( -13,- \dfrac{33}{2} \right)$$ can hence be found as, $$\left( y+10 \right)=\dfrac{\left( -\dfrac{33}{2}+10 \right)}{\left( -13-\dfrac{35}{2} \right)}\times \left( x-\dfrac{35}{2} \right)$$ Taking the LCM of the terms in the numerator and denominator, $$\begin{aligned} & y+10=\dfrac{-\dfrac{13}{2}}{-\dfrac{61}{2}}\times \left( x-\dfrac{35}{2} \right) \\\ & y+10=\dfrac{13}{61}\times \left( x-\dfrac{35}{2} \right) \\\ & 61(y+10)=13\left( x-\dfrac{35}{2} \right) \\\ & 61y+610=13x-\dfrac{455}{2} \\\ \end{aligned}$$ Multiplying both sides by 2, $$\begin{aligned} & 122y+1220=26x-455 \\\ & 26x-122y-1675=0 \\\ \end{aligned}$$ Therefore, the required equation of the straight line is $$26x-122y-1675=0$$. We get option (d) as the correct answer. Note: There is one more way to approach this problem. The coordinates of the vertices A and B can be taken as $$\left( p,\dfrac{20-4p}{5} \right)$$ and $$\left( q,\dfrac{3+6q}{2} \right)$$ by assuming $$x$$ as $$p$$ and $$q$$ respectively in the equations $$4x+5y-20=0$$and $$3x-2y+6=0$$. Then we can formulate the equations using the slope of the lines and solve for $$p$$ and $$q$$. The equation of AB can then be obtained.