Question

The sides AB, BC, CD and DA of a quadrilateral have the
equations x + 2y = 3, x= 1, x-3y = 4 and 5x + y + 12 = 0
respectively. Find the angle between the diagonals AC and
BD.​

Answer 1

Answer:

Step-by-step explanation:

Eqn of line AB & BC are x+2y=3 & x=1 respectively, hence

Solving we get coordinates of B as (1,1)

Eqn of line BC & CD are x=1 & x-3y=4 respectively, hence

Solving we get coordinates of C as (1,-1)

Eqn of line CD & DA are x-3y=4 & 5x+y+12=0 respectively, hence

Solving we get coordinates of D as (-2,-2)

Eqn of line DA & AB are 5x+y+12=0 & x+2y=3 respectively, hence

Solving we get coordinates of A as (-3, 3)

So we have:

A(-3,3)

B(1,1)

C(1,-1)

D(-2,-2)

Suppose,

theta1 is the angle which AC makes with the positive direction of X-axis

theta2 is the angle which BD makes with the positive direction of X-axis

theta is the angle between AC and BD, there will be 2 values for theta one acute and the other obtuse angle

Now,

Slope of AC: m1 = tan(theta1) = (-1-3)/(1-(-3)) = -4/4 = -1 => theta1 = (tan^-1) (-1) = 135 deg

Slope of BD: m2 = tan(theta2) = (-2-1)/(-2-1) = 1 => theta2 = (tan^-1) (1) = 45 deg

Hence the angle between AC and BD i.e. theta = theta1 -theta2 = 135–45 = 90 deg

We can also solve it algebraically using trigonometry formulas

theta = theta1 - theta2

=> tan(theta) = tan (theta1 - theta2)

Now as tan(A-B) = (tanA -tanB) / (1- tanA*tanB)

=> tan(theta) = (m1 - m2) / (1-m1*m2) = (-1-1)/(1-(-1)) = -2/2 = -1

=> theta = (tan^-1) (-1), - (tan^-1) (-1) one for the acute angle, other for the obtuse

We no tan^-1 (1) or tan^-1 (-1) = pi/2 or 90 degress.

Hence AC & BD will cut each other at right angles.

Hope it helps.