English

Find the Coordinates of the Centre of the Circle Inscribed in a Triangle Whose Vertices Are (−36, 7), (20, 7) and (0, −8). - Mathematics

Advertisements
Advertisements

Question

Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8).

Sum

Solution

The coordinates of the in-centre of a triangle whose vertices are \[A\left( x_1 , y_1 \right), B\left( x_2 , y_2 \right)\text{ and }C\left( x_3 , y_3 \right)\] are \[\left( \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c} \right)\], where a = BC, b = AC and c = AB.
Let A(−36, 7), B(20, 7) and C(0, −8) be the coordinates of the vertices of the given triangle.
Now,
\[a = BC = \sqrt{\left( 20 - 0 \right)^2 + \left( 7 + 8 \right)^2} = 25\]

\[b = AC = \sqrt{\left( 0 + 36 \right)^2 + \left( - 8 - 7 \right)^2} = 39\]
\[c = AB = \sqrt{\left( 20 + 36 \right)^2 + \left( 7 - 7 \right)^2} = 56\]

Thus, the coordinates of the in-centre of the given triangle are:
\[\left( \frac{25 \times \left( - 36 \right) + 39 \times 20 + 0}{25 + 39 + 56}, \frac{25 \times 7 + 39 \times 7 + 56\left( - 8 \right)}{25 + 39 + 56} \right)\]= \[\left( \frac{- 120}{120}, 0 \right)\]

\[\left( - 1, 0 \right)\]

Hence, the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8) is \[\left( - 1, 0 \right)\]

shaalaa.com
Brief Review of Cartesian System of Rectanglar Co-ordinates
  Is there an error in this question or solution?
Chapter 22: Brief review of cartesian system of rectangular co-ordinates - Exercise 22.1 [Page 13]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 22 Brief review of cartesian system of rectangular co-ordinates
Exercise 22.1 | Q 5 | Page 13

RELATED QUESTIONS

If the line segment joining the points P (x1, y1) and Q (x2, y2) subtends an angle α at the origin O, prove that
OP · OQ cos α = x1 x2 + y1, y2


The vertices of a triangle ABC are A (0, 0), B (2, −1) and C (9, 2). Find cos B.


Four points A (6, 3), B (−3, 5), C (4, −2) and D (x, 3x) are given in such a way that \[\frac{\Delta DBC}{\Delta ABC} = \frac{1}{2}\]. Find x.


The base of an equilateral triangle with side 2a lies along the y-axis, such that the mid-point of the base is at the origin. Find the vertices of the triangle.


Find the equation of the locus of a point which moves such that the ratio of its distances from (2, 0) and (1, 3) is 5 : 4.

 

Find the locus of a point such that the sum of its distances from (0, 2) and (0, −2) is 6.

 

Find the locus of a point which moves such that its distance from the origin is three times its distance from the x-axis.

 

A (5, 3), B (3, −2) are two fixed points; find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.


If A (−1, 1) and B (2, 3) are two fixed points, find the locus of a point P, so that the area of ∆PAB = 8 sq. units.


A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.


If O is the origin and Q is a variable point on y2 = x, find the locus of the mid-point of OQ.

 

What does the equation (x − a)2 + (y − b)2 = r2 become when the axes are transferred to parallel axes through the point (a − c, b)?

 

Find what the following equation become when the origin is shifted to the point (1, 1).
x2 + xy − 3x − y + 2 = 0


Find what the following equation become when the origin is shifted to the point (1, 1).
xy − x − y + 1 = 0


Find what the following equation become when the origin is shifted to the point (1, 1).
xy − y2 − x + y = 0


To what point should the origin be shifted so that the equation x2 + xy − 3x − y + 2 = 0 does not contain any first degree term and constant term?


Verify that the area of the triangle with vertices (2, 3), (5, 7) and (− 3 − 1) remains invariant under the translation of axes when the origin is shifted to the point (−1, 3).


Find what the following equation become when the origin is shifted to the point (1, 1).
x2 + xy − 3y2 − y + 2 = 0


Find what the following equation become when the origin is shifted to the point (1, 1).
xy − y2 − x + y = 0


Find what the following equation become when the origin is shifted to the point (1, 1).
x2 − y2 − 2x + 2y = 0


Find the point to which the origin should be shifted after a translation of axes so that the following equation will have no first degree terms: x2 + y2 − 5x + 2y − 5 = 0


Find the point to which the origin should be shifted after a translation of axes so that the following equation will have no first degree terms: x2 − 12x + 4 = 0


The vertices of a triangle are O (0, 0), A (a, 0) and B (0, b). Write the coordinates of its circumcentre.


If the points (a, 0), (at12, 2at1) and (at22, 2at2) are collinear, write the value of t1 t2.

 

If the coordinates of the sides AB and AC of  ∆ABC are (3, 5) and (−3, −3), respectively, then write the length of side BC.

 

Write the coordinates of the circumcentre of a triangle whose centroid and orthocentre are at (3, 3) and (−3, 5), respectively.

 

Write the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12).


Write the area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×