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Question: If the vertices of a triangle are \[\left( {3,2} \right),\] \[\left( { - 4,1} \right)\] and \[\left(...

If the vertices of a triangle are (3,2),\left( {3,2} \right), (−4,1)\left( { - 4,1} \right) and (−5,8)\left( { - 5,8} \right) then, which of the following is not the orthocentre?
A) (4,1)\left( {4,1} \right)
B) (4,−1)\left( {4, - 1} \right)
C) (−1,4)\left( { - 1,4} \right)
D) (−4,1)\left( { - 4,1} \right)

Explanation

Solution

Hint : In the question, we are asked to find which are not the orthocentre of the given vertices of the triangle. So we will start with finding the orthocentre of the triangle, for that first we will assume that our orthocentre be (h,k) then on applying the slope formula we get our orthocentre, after that on comparing with the options given, we will get the required answer.

Complete step-by-step answer :
We have been given the vertices of a triangle which are A(3,2),\left( {3,2} \right), B(−4,1)\left( { - 4,1} \right) and C(−5,8),\left( { - 5,8} \right), we need to find which are not the orthocentre from the given options. So, we will find the orthocentre, and check which are not the orthocentre of the triangle.
Let us understand first with the help of a diagram.

Here, in the diagram O(h,k) is the orthocentre, which is the intersection point of perpendiculars.
To find the orthocentre of the triangle, we will use the formula of slope given below.
Slope =y2−y1x2−x1= \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
So, slope of line BC =8−1−5+4=−71= \dfrac{{8 - 1}}{{ - 5 + 4}} = \dfrac{{ - 7}}{1}
Slope of line AC =8−2−5−3=−68=−34= \dfrac{{8 - 2}}{{ - 5 - 3}} = \dfrac{{ - 6}}{8} = \dfrac{{ - 3}}{4}
Slope of line AB =1−2−4−3=−1−7=17= \dfrac{{1 - 2}}{{ - 4 - 3}} = \dfrac{{ - 1}}{{ - 7}} = \dfrac{1}{7}
Now, OA ⊥\bot BC
Then, (slope of OA)(slope of BC) = −1 = {\text{ }} - 1
(k−2h−3)(−71)=−1(\dfrac{{k - 2}}{{h - 3}})(\dfrac{{ - 7}}{1}) = - 1
k−2h−3=17 ⇒7k−14=h−3 ⇒7k−h=11....eq.(1)  \dfrac{{k - 2}}{{h - 3}} = \dfrac{1}{7} \\\ \Rightarrow 7k - 14 = h - 3 \\\ \Rightarrow 7k - h = 11....eq.(1) \\\
Now, OB ⊥\bot AC
Then, (slope of OB)(slope of AC) = −1 = {\text{ }} - 1
(k−1h+4)(−34)=−1(\dfrac{{k - 1}}{{h + 4}})(\dfrac{{ - 3}}{4}) = - 1
k−1h+4=43 ⇒3k−3=4h+16 ⇒3k−4h=19....eq.(2)  \dfrac{{k - 1}}{{h + 4}} = \dfrac{4}{3} \\\ \Rightarrow 3k - 3 = 4h + 16 \\\ \Rightarrow 3k - 4h = 19....eq.(2) \\\
So, we get two equations, 7k−h=117k - h = 11 and 3k−4h=193k - 4h = 19 , now on solving the equations, we get
⇒(7k−h=11)4 ⇒3k−4h=19  \Rightarrow (7k - h = 11)4 \\\ \Rightarrow 3k - 4h = 19 \\\
⇒28k−4h=44 ⇒3k−4h=19 \-−−−−−−−− ⇒25k=25 ⇒k=1  \Rightarrow 28k - 4h = 44 \\\ \Rightarrow 3k - 4h = 19 \\\ \- - - - - - - - - \\\ \Rightarrow 25k = 25 \\\ \Rightarrow k = 1 \\\
On substituting the value of k in eq.(2),eq.\left( 2 \right),we get
⇒3(1)−4h=19 ⇒3−19=4h ⇒h=−164 ⇒h=−4  \Rightarrow 3(1) - 4h = 19 \\\ \Rightarrow 3 - 19 = 4h \\\ \Rightarrow h = \dfrac{{ - 16}}{4} \\\ \Rightarrow h = - 4 \\\
So, we get our orthocentre (−4,1).\left( { - 4,1} \right).
Now on checking with the options, option A), B) and C) are not the orthocentre of the given vertices of the triangle.

Note : There is no direct formula of finding the orthocentre of the triangle, So, we do it by first calculating the slopes of sides of the triangle. Then we take the slope of the perpendicular. We get three equations from it, so, by solving any two of those equations, we can get the orthocentre