Question

Two vertices of a triangle are $(4, - 3)$ and $( - 2,5)$ . If the orthocentre of the triangle is at $(1,2)$ . Then the third vertex is

Answer 1

Hint : As we know that the orthocentre of a triangle is the point of intersection of altitudes of the triangle. Here in this question we will assume the third vertex as $(x,y)$ . We should know that to find the slope of the line having two points $({x_1},{y_1})$ and $({x_2},{y_2})$ , we use this formula $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ .

Complete step-by-step answer :

Here let us assume the third vertex be $C = (x,y)$ .
Then we have $A = (4, - 3),B = ( - 2,5)$ . We have the orthocentre i.e. $I = (1,2)$ .
So we can say that the slope of AB is $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ .
Here by comparing we have ${y_2} = 5,{y_1} = - 3,{x_2} = - 2,{x_1} = 4$ .
By putting the values in the formula we have AB $= \dfrac{{5 - ( - 3)}}{{ - 2 - 4}} = \dfrac{{ - 4}}{3}$ .
Similarly we will calculate the slope of IC i.e. $\dfrac{{y - 2}}{{x - 1}}$ . We will now equate both the expressions and we have $\dfrac{{y - 2}}{{x - 1}} = \dfrac{3}{4}$ .
By cross multiplication we have $4(y - 2) = 3(x - 1)$ . By breaking the brackets we have $4y - 8 = 3x - 3$ .
We can write this in the form of the linear equation i.e. $3x - 4y + 5 = 0$ .
Again we have to calculate the slope of AC.
Here by comparing we have ${y_2} = y,{y_1} = - 3,{x_2} = x,{x_1} = 4$ .
By putting the values in the formula we have AC $= \dfrac{{y - ( - 3)}}{{x - 4}} = \dfrac{{y + 3}}{{x - 4}}$ .
For IB we have points $B = ( - 2,5)$ and $I = (1,2)$ . By comparing we have ${y_2} = 5,{y_1} = 2,{x_2} = - 1,{x_1} = 1$ .
Similarly we will calculate the slope of IB i.e.
$\dfrac{{5 - 2}}{{ - 2 - 1}} = - 1$ .
We will now equate both the expressions and we have $\dfrac{{y + 3}}{{x - 4}} = 1$ .
By cross multiplication we have
$y + 3 = x - 4 \Rightarrow x - y - 7 = 0$ .
So we have two equations i.e.
$x - y - 7 = 0$ and $3x - 4y + 5 = 0$ .
WE can solve this by elimination method i.e. by multiplying the first equation with $3$ , and then we subtract the second equation, we can write
$3x - 3y - 21 - 3x + 4y - 5 = 0$ .
On further solving we have
$4y - 3y = 21 + 5 \Rightarrow y = 26$ .
Putting this value in the first equation we have
$x - 26 - 7 = 0 \Rightarrow x = 33$ .
Hence the third vertex is $(26,33)$ .
So, the correct answer is “ $(26,33)$ .”.

Note : Before solving this kind of question we should have the full knowledge of the orthocentre, slope of the lines and their formulas. We should note that the equation of the ;line passing through a point $(x,y)$ with slope $m$ can be written as $y - y' = m(x - x')$ . We should avoid calculation mistakes while solving the equations.