Home
Class 12
MATHS
In triangle A B C , the equation of the ...

In triangle `A B C` , the equation of the right bisectors of the sides `A B` and `A C` are `x+y=0` and `y-x=0` , respectively. If `A-=(5,7)` , then find the equation of side `B Cdot`

Text Solution

Verified by Experts

The correct Answer is:
14y = 10x

B and C will be the images of A on y+x=0 and y-x=0 respectively. Thus,
`B-=(-7,-5),C-=(7,5)`
Hence, the equation of BC is
`y-5 = (-5-5)/(-7-7)(x-7)`
i.e., 14y = 10x
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    CENGAGE|Exercise Exercise 2.6|5 Videos
  • STRAIGHT LINES

    CENGAGE|Exercise Exercise (Single)|82 Videos
  • STRAIGHT LINES

    CENGAGE|Exercise Exercise 2.4|8 Videos
  • STRAIGHT LINE

    CENGAGE|Exercise Multiple Correct Answers Type|8 Videos
  • THEORY OF EQUATIONS

    CENGAGE|Exercise JEE Advanced Previous Year|9 Videos

Similar Questions

Explore conceptually related problems

The equations of the perpendicular bisectors of the sides A Ba n dA C of triangle A B C are x-y+5=0 and x+2y=0 , respectively. If the point A is (1,-2) , then find the equation of the line B Cdot

If the middle points of the sides B C ,C A , and A B of triangle A B C are (1,3),(5,7), and (-5,7), respectively, then find the equation of the side A Bdot

If the coordinates of the vertices of triangle A B C are (-1,6),(-3,-9) and (5,-8) , respectively, then find the equation of the median through Cdot

Find the equation of the bisectors of the anglebetween the lines 4x+3y=5 and x +2y+3=0.

Find the equation of the bisector of the obtuse angle between the lines 3x-4y+7=0 and 12 x+5y-2=0.

If one of the sides of a square is 3x-4y-12=0 and the center is (0,0) , then find the equations of the diagonals of the square.

In triangle A B C , the equation of side B C is x-y=0. The circumcenter and orthocentre of triangle are (2, 3) and (5, 8), respectively. The equation of the circumcirle of the triangle is

Find the equations of the bisector of the acute angle between the lines 3x + 4y + 2 = 0 and 5x + 12y - 5 = 0 .

Let A,B,C be angles of triangles with vertex A -= (4,-1) and internal angular bisectors of angles B and C be x - 1 = 0 and x - y - 1 = 0 respectively. If A,B,C are angles of triangle at vertices A,B,C respectively then cot ((B)/(2))cot .((C)/(2)) =

In a triangle A B C , the bisectors of angles Ba n dC lies along the lines x=ya n dy=0. If A is (1,2) , then the equation of line B C is (a) 2x+y=1 (b) 3x-y=5 (c) x-2y=3 (d) x+3y=1