Triangle ABC is an isosceles triangle such that AB = AC. Given A (4,3) and B (1,-1) ,gradient of the straight line BC is 1/2 and AD is perpendicular to BC.
(a) Find the equation of BC and AD.
(b) Find the coordinates:
(i) C
(ii) D
Answers
Given that,
↝ Triangle ABC is an isosceles triangle such that AB = AC.
↝ Coordinates of A is (4,3) and B is (1,-1)
↝ Gradient of the straight line BC is 1/2.
↝ AD is perpendicular to BC.
Part (a) Equation of BC
As,
Line BC passes through the point (1, - 1) and having gradient or slope 1/2.
We know,
Equation of line which passes through the point (a, b) and having gradient or Slope m is
So, using this,
The equation of BC is
Part (a) Equation of AD
Now, it is given that AD is perpendicular to BC.
We know,
Two lines having slope m and M are perpendicular iff Mm = - 1.
So, given that,
↝ Slope or gradient of BC = 1/2.
↝ So, gradient of AD = - 2.
Now, Line AD passes through the point (4, 3) and having gradient - 2.
So, Equation of AD is
Part (b) Coordinates of D
Now, D is the point of intersection of line AD and BC.
So, Solve the system of linear equations.
From equation (1), we have
Substituting this value in equation (2), we get
On substituting y = 1 in equation (1), we get
Hence, Coordinates of D is (5, 1).
Part (b) Coordinates of C
As we have with us
↝ Coordinates of B (1, - 1)
↝ Coordinates of D (5, 1)
↝ Also, AB = AC
↝ So, AD is perpendicular bisector of BC.
It means, D is the midpoint of BC.
Let assume that coordinates of C be (a, b).
So, using Midpoint Formula, we have
So, on comparing,
Hence, Coordinates of C is (9, 3).
Answer:
Given that,
↝ Triangle ABC is an isosceles triangle such that AB = AC.
↝ Coordinates of A is (4,3) and B is (1,-1)
↝ Gradient of the straight line BC is 1/2.
↝ AD is perpendicular to BC.
Part (a) Equation of BC
As,
Line BC passes through the point (1, - 1) and having gradient or slope 1/2.
We know,
Equation of line which passes through the point (a, b) and having gradient or Slope m is
So, using this,
The equation of BC is
Part (a) Equation of AD
Now, it is given that AD is perpendicular to BC.
We know,
Two lines having slope m and M are perpendicular iff Mm = - 1.
So, given that,
↝ Slope or gradient of BC = 1/2.
↝ So, gradient of AD = - 2.
Now, Line AD passes through the point (4, 3) and having gradient - 2.
So, Equation of AD is
Part (b) Coordinates of D
Now, D is the point of intersection of line AD and BC.
So, Solve the system of linear equations.
From equation (1), we have
Substituting this value in equation (2), we get
On substituting y = 1 in equation (1), we get
Hence, Coordinates of D is (5, 1).
Part (b) Coordinates of C
As we have with us
↝ Coordinates of B (1, - 1)
↝ Coordinates of D (5, 1)
↝ Also, AB = AC
↝ So, AD is perpendicular bisector of BC.
It means, D is the midpoint of BC.
Let assume that coordinates of C be (a, b).
So, using Midpoint Formula, we have
So, on comparing,
Hence, Coordinates of C is (9, 3).