Question

ABC is variable triangle whose centroid is fixed at (5,5) if BC=13 such that B and C moves on x and y axis respectively.the locus of A is​

Answer 1

Answer:

Let centroid Q(h,k).

Clearly, centroid = (x1 +x2 +x3/3 , y1 + y2+ y3/3) where x1/2/3, y1/2/3 are coordinates of vertices.

So we get h=(1+sint +cost/3) and k=(sint -cost+2/3)

On rearranging and adding, we get

3/2(h+k-1) = sint

And

3/2(h-k+1/3) = cost

Also, to eliminate t, put formula sin^2t + cos^2t = 1

Rearrange with above values of sint and cost, and in final expression replace h by x and k by y.

Thus you get the required locus.

Step-by-step explanation:

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Answer 2

Answer: The equation of locus of point A is $x^{2} + y^{2} - 30x - 30y + 281 = 0$

Step-by-step explanation:

Given:ABC is variable triangle whose centroid is fixed at (5,5) if BC=13 such that B and C moves on x and y axis respectively.

To find: We have to find the equation of locus of A.

Explanation:

Step 1: On applying the pythagoras theorem in Δ OBC we have,

                                 $a^{2} + b^{2} = (BC)^{2}$

                                 $a^{2} + b^{2} = (13)^{2}$

                                 $a^{2} + b^{2} = 169$         (1)

Step 2:As we know the centroid of ΔABC is fixed at (5,5), so we have,

                                $(\frac{x_{1}+x_{2}+x_{3} }{3} ) (\frac{y_{1}+y_{2}+y_{3} }{y} ) = (5,5)$

Step 3: On comparison we get,

                                 $\frac{a+x}{3} = 5$  

                              $a+x = 5$ × $3$

                              $a + x = 15$

                                    $a = 15 - h$

and                            $\frac{b + y}{3} = 5$

                               $b + y = 5$ × $3$

                               $b + y = 15$

                                     $b = 15 - y$

Step 4:Substituting the value of a and b in equation (1), we have,

                                             $a^{2} + b^{2} = 169$

⇒                     $(15 - x)^{2} + (15 - y)^{2} = 169$

⇒   $225 - 30x + x^{2} + 225 -30y + y^{2} = 169$

⇒             $x^{2} + y^{2} - 30x - 30y +450 = 169$                                                              

⇒   $x^{2} + y^{2} - 30x - 30y +450 - 169 = 0$

⇒             $x^{2} + y^{2} - 30x - 30y + 281 = 0$

Hence, the equation of locus of point A is $x^{2} + y^{2} - 30x - 30y + 281 = 0$.

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