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 equat

Question

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 equat

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 radius of the in circle of triangle PQR is
The radius of the circle of triangle PQR is
A. 4
B. 5
C. 6
D. 8

Monjur Alahi

9 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - B
The coordinates of the incenter are given as follows:
`bar(x) = (20(-8)+15(-15)+25(1))/(20+15+25) = -6`
`bar(y) = (20(5)+15(-19)+25(-7))/(60) = -6`
Hence, incenter I is (-6, -6).
Now, the equation of side PR is
`y+7 = (12)/(-9)(x-1)`
or 4x-4 = -3y-21
or 4x+3y+17 = 0
Inradius is given as the distance of I from side PR, i.e.,
`(|-24-18+17|)/(5) = 5`