Consider a triangle P Q R with coordinates of its vertices as P(-8,5), Q(-15,-19), and R(1,-7). The bisector of the interior angle of P has the equation which can be written in the form ax +2 y+c=0. The distance between the orthocenter and the circumcenter of triangle P Q R is

Question

Consider a triangle P Q R with coordinates of its vertices as P(-8,5), Q(-15,-19), and R(1,-7). The bisector of the interior angle of P has the equation which can be written in the form ax +2 y+c=0. The distance between the orthocenter and the circumcenter of triangle P Q R is (A) 25 / 2 (B) 29 / 2 (C) 37 / 2 (D) 51 / 2

Tardigrade Admin · Accepted Answer

Since the triangle is right-angled, the circumcenter is the midpoint of P Q and the orthocenter is R(1,-7). Hence, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/56b7ec4bd6c16574bad6b969e3fa4c2a-.png" /> R M=|√((23/2)+1)2|=12 (1/2)

$25/2$

$29/2$

$37/2$

$51/2$

Since the triangle is right-angled, the circumcenter is the midpoint of $PQ$ and the orthocenter is $R (1, - 7)$ . Hence,

$RM = ∣ ∣ (\frac{23}{2} + 1)^{2} ∣ ∣ = 12 \frac{1}{2}$

Q. Consider a triangle PQR with coordinates of its vertices as P(−8,5),Q(−15,−19), and R(1,−7). The bisector of the interior angle of P has the equation which can be written in the form ax +2y+c=0. The distance between the orthocenter and the circumcenter of triangle PQR is

25/2

29/2

37/2

51/2

Since the triangle is right-angled, the circumcenter is the midpoint of PQ and the orthocenter is R(1,−7). Hence, RM=∣∣​(223​+1)2​∣∣​=1221​

$25/2$

$29/2$

$37/2$

$51/2$

Since the triangle is right-angled, the circumcenter is the midpoint of $PQ$ and the orthocenter is $R (1, - 7)$ . Hence,

$RM = ∣ ∣ (\frac{23}{2} + 1)^{2} ∣ ∣ = 12 \frac{1}{2}$