# 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

Verified

330.6k+ views

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,

\[\begin{align}

& 4x+5y-20=0 \\

& 5y=20-4x \\

& y=-\dfrac{4}{5}x+4 \\

\end{align}\]

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{align}

& y-1=\dfrac{5}{4}(x-1) \\

& 4(y-1)=5(x-1) \\

& 4y-4=5x-5 \\

& 5x-4y-1=0 \\

\end{align}\]

Next, we have to consider line BC. The slope of the line can be obtained by rearranging the equation

as,

\[\begin{align}

& 3x-2y+6=0 \\

& -2y=-6-3x \\

& y=\dfrac{3}{2}x+3 \\

\end{align}\]

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{align}

& y-1=-\dfrac{2}{3}(x-1) \\

& 3(y-1)=-2(x-1) \\

& 3y-3=-2x+2 \\

& 2x+3y-5=0 \\

\end{align}\]

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{align}

& 4x+5y-20=0 \\

& -4x-6y+10=0 \\

\end{align}}{\begin{align}

& -y-10=0 \\

& y=-10 \\

\end{align}}\]

Substituting the value of \[y\] in \[2x+3y-5=0\], we get,

\[\begin{align}

& 2x+(3\times -10)-5=0 \\

& 2x-30-5=0 \\

& 2x=35 \\

& x=\dfrac{35}{2} \\

\end{align}\]

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{align}

& 6x-4y+12=0 \\

& -5x+4y+1=0 \\

\end{align}}{\begin{align}

& x+13=0 \\

& x=-13 \\

\end{align}}\]

Substituting the value of \[x\] in \[3x-2y+6=0\], we get,

\[\begin{align}

& (3\times -13)-2y+6=0 \\

& -39-2y+6=0 \\

& -2y=33 \\

& y=-\dfrac{33}{2} \\

\end{align}\]

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{align}

& 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{align}\]

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{align}

& 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{align}\]

Multiplying both sides by 2,

\[\begin{align}

& 122y+1220=26x-455 \\

& 26x-122y-1675=0 \\

\end{align}\]

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.

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,

\[\begin{align}

& 4x+5y-20=0 \\

& 5y=20-4x \\

& y=-\dfrac{4}{5}x+4 \\

\end{align}\]

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{align}

& y-1=\dfrac{5}{4}(x-1) \\

& 4(y-1)=5(x-1) \\

& 4y-4=5x-5 \\

& 5x-4y-1=0 \\

\end{align}\]

Next, we have to consider line BC. The slope of the line can be obtained by rearranging the equation

as,

\[\begin{align}

& 3x-2y+6=0 \\

& -2y=-6-3x \\

& y=\dfrac{3}{2}x+3 \\

\end{align}\]

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{align}

& y-1=-\dfrac{2}{3}(x-1) \\

& 3(y-1)=-2(x-1) \\

& 3y-3=-2x+2 \\

& 2x+3y-5=0 \\

\end{align}\]

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{align}

& 4x+5y-20=0 \\

& -4x-6y+10=0 \\

\end{align}}{\begin{align}

& -y-10=0 \\

& y=-10 \\

\end{align}}\]

Substituting the value of \[y\] in \[2x+3y-5=0\], we get,

\[\begin{align}

& 2x+(3\times -10)-5=0 \\

& 2x-30-5=0 \\

& 2x=35 \\

& x=\dfrac{35}{2} \\

\end{align}\]

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{align}

& 6x-4y+12=0 \\

& -5x+4y+1=0 \\

\end{align}}{\begin{align}

& x+13=0 \\

& x=-13 \\

\end{align}}\]

Substituting the value of \[x\] in \[3x-2y+6=0\], we get,

\[\begin{align}

& (3\times -13)-2y+6=0 \\

& -39-2y+6=0 \\

& -2y=33 \\

& y=-\dfrac{33}{2} \\

\end{align}\]

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{align}

& 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{align}\]

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{align}

& 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{align}\]

Multiplying both sides by 2,

\[\begin{align}

& 122y+1220=26x-455 \\

& 26x-122y-1675=0 \\

\end{align}\]

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.

Last updated date: 24th May 2023

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