Question

The vertex of a variable \[\Delta ABC\] are \[a(3,4)\], \[b(5\cos t,5\sin t)\], and \[c(5\sin t, - 5\cos t)\]. Find the locus of orthocentres of \[\Delta ABC\].

Answer 1

Hint: The given problem has to be solved with distance formula and orthocentre formula which is also known as centroid. Centroid means the midpoint of a triangle. By also circumcenter and basic calculation solved by complete step-by-step explanation.

Formulas used:
Distance between two points \[{\text{ = }}\sqrt {{{({x_2} - {x_1})}^2} - {{({y_2} - {y_1})}^2}} \]
Centroid\[{\text{ = }}\left( {\dfrac{{{x_1} + {x_2} + x{}_3}}{3},\dfrac{{{y_1} + {y_2} + y{}_3}}{3}} \right)\]
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[{(a + b)^2} = {a^2} + {b^2} + 2ab\]

Complete step by step answer:
It is the given vertex of a variable \[a(3,4)\], \[b(5\cos t,5\sin t)\], and \[c(5\sin t, - 5\cos t)\].
We have to find the locus of the orthocentre of \[\Delta ABC\].
First, we need to find the distance of the vertex \[ABC\], and also check the circumcenter of the circle.
Let the origin be \[\left( {0,0} \right)\]
Then finding distance formula for \[OA\], \[OB\], \[OC\]
\[ \Rightarrow {\text{OA = }}\sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}} \]
Simplifying we get,
\[ \Rightarrow {\text{OA = }}\sqrt {{{(3)}^2} + {{(4)}^2}} \]
Now, squaring numbers and taking square root, we get
\[ \Rightarrow {\text{OA = }}\sqrt {9 + 16} \]
Adding the terms we get,
\[ \Rightarrow OA = \sqrt {25} \]
Since, \[25\] is a multiple of \[5\],
\[ \Rightarrow OA = \sqrt {{5^2}} \]
Taking square root we get,
\[ \Rightarrow OA = 5\]
Likewise, we are going to find the other distance \[OB\& OC\], we get,
\[ \Rightarrow {\text{OB = }}\sqrt {{{(5\cos t - 0)}^2} + {{(5\sin t - 0)}^2}} \]
Simplifying we get,
\[ \Rightarrow {\text{OB = }}\sqrt {{{(5\cos t)}^2} + {{(5\sin t)}^2}} \]
Squaring the terms,
\[ \Rightarrow {\text{OB = }}\sqrt {25{{\cos }^2}t + 25{{\sin }^2}t} \]
Taking Commonly \[25\], we get,
\[ \Rightarrow {\text{OB = }}\sqrt {25({{\cos }^2}t + {{\sin }^2}t)} \]
By applying formula mentioned in formula used that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], we get
\[ \Rightarrow {\text{OB = }}\sqrt {25} \]
Since, \[25\] is a multiple of \[5\],
\[ \Rightarrow OB = \sqrt {{5^2}} \]
Taking square root we get,
\[ \Rightarrow OB = 5\]
Likewise we find \[{\text{OC}}\], applying distance formula
\[ \Rightarrow {\text{OC = }}\sqrt {{{(5\sin t - 0)}^2} + {{( - 5\cos t - 0)}^2}} \]
Simplifying we get,
\[ \Rightarrow {\text{OC = }}\sqrt {{{(5\sin t)}^2} + {{( - 5\cos t)}^2}} \]
Squaring the terms,
\[ \Rightarrow {\text{OC = }}\sqrt {25{{\cos }^2}t + 25{{\sin }^2}\theta } \]
Taking Commonly \[25\], we get
\[ \Rightarrow {\text{OC = }}\sqrt {25({{\cos }^2}t + {{\sin }^2}t)} \]
By applying formula mentioned in formula used that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], we get
\[ \Rightarrow {\text{OC = }}\sqrt {25} \]
Since, \[25\] is a multiple of \[5\],
\[ \Rightarrow OC = \sqrt {{5^2}} \]
Taking square root we get,
\[ \Rightarrow OC = 5\]
Thus the distance of \[OA\], \[OB\], \[OC\] is the same so \[\left( {0,0} \right)\] is the circumcenter of \[\Delta ABC\].
Let \[H\] be the orthocentre of \[\Delta ABC\]
Now, Solve using centroid \[G\] divides using section formula in the ratio \[2:1\]
Then, Centroid \[G\] is using vertex,
\[ \Rightarrow G = \left( {\dfrac{{3 + 5\cos t + 5\sin t}}{3},\dfrac{{4 + 5\sin t - 5\cos t}}{3}} \right) - - - - - (1)\]
Now, Centroid \[G\] using Section formula,
We know that the centroid, \[H\left( {h,k} \right)\] and circumcenter \[\left( {0,0} \right)\] in the ratio \[2:3\]
\[ \Rightarrow G = \left( {\dfrac{h}{3},\dfrac{k}{3}} \right) - - - - (2)\]
Now, equating equation (1) & (2), we get
\[ \Rightarrow \dfrac{h}{3} = \dfrac{{3 + 5\cos t + 5\sin t}}{3}\]
\[ \Rightarrow \dfrac{k}{3} = \dfrac{{4 + 5\sin t - 5\cos t}}{3}\]
Now, canceling denominator value because same on both sides,
\[ \Rightarrow h = 3 + 5\cos t + 5\sin t\]
\[ \Rightarrow k = 4 + 5\sin t - 5\cos t\]
Now, changing the constant term towards the left-hand side,
\[ \Rightarrow k - 4 = 5\sin t - 5\cos t - - - - (3)\]
\[ \Rightarrow h - 3 = 5\cos t + 5\sin t - - - - - - (4)\]
Now, squaring on both sides and adding equation (3) & (4), we get
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = {(5\cos t + 5\sin t)^2} + {(5\sin t - 5\cos t)^2}\]
Now, after squaring the RHS side only, and applying ${(a+b)}^2$ formula mention in the formula used, we get
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = 25{\cos ^2}t + 25{\sin ^2}t + 50\cos t\sin t + 25{\cos ^2}t + 25{\sin ^2}t - 50\cos t\sin t\]
Canceling the same positive and negative values, we get
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = 25{\cos ^2}t + 25{\sin ^2}t + 25{\cos ^2}t + 25{\sin ^2}t\]
By adding the right-hand side of the equation, we get
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50{\cos ^2}t + 50{\sin ^2}t\]
By taking 50 commonly, we get
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50({\cos ^2}t + {\sin ^2}t)\]
Once again we applying the formula mentioned in the formula used, we get
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50(1)\]
\[ \Rightarrow {(h - 3)^2} + {(k - 4)^2} = 50\]

$\therefore $ The locus of orthocentre is \[{(h - 3)^2} + {(k - 4)^2} = 50\]

Note:
This kind of problem has special attention to points for the values we substitute to that we get things right in the manner. By knowing formula and calculation more than we want to know about the content of the question. This may be given in the form of orthocentre but no clear explanation regarding centroid. So, we want to create an idea to make a centroid value and section form and compare those we solved. Like this, we must try to find out in many ways.