Question

If the points p(-3,9),q(a,b) and r(4,-5) are collinear and a+b=1 . Find the values of a and



b.

Answer 1

Solution:

It is given that the points p(-3,9),q(a,b) and r(4,-5) are collinear and a+b=1 .

Find area of Triangle:

[x1(y2 - y3) + x2( y3 - y1)+ x3(y1 - y2)] = 0.

⇒ [-3(b +5) + a(-5 -9) + -4(9 - b)] = 0

⇒ [ -3b - 15 + a(-14) - 36 + 4b ] = 0

⇒ [ -3b - 15 - 14a - 36 + 4b ] = 0

⇒ [ -14a + b - 51 ] = 0

⇒ -14a + b = 51 .....(i)

We have another equation i.e a + b = 1

Solve both Equations:

• a + b = 1
• -14a + b = 51

Subtract (i) from (ii):

-14a + b - ( a + b) = 51 - 1

-14a + b - a - b = 50

-15a = 50

a = 50/-15

a = 10/-3

Substitute value of x in Equation:

a + b = 1

-10/3 + b = 1

b = 1 + 10/3

b = 3+10/3

b = 13/3

Therefore, Value of x and y are -10/3 and 13/3 respectively.

Answer 2

Solution:

We have been given that the points p(-3,9),q(a,b) and r(4,-5) are collinear and a+b=1 .

Here,

• X1 = -3
• x2 = a
• X3 = -4
• y1 = 9
• y2 = b
• y3 = -5

Colinear:

[x1(y2 - y3) + x2( y3 - y1)+ x3(y1 - y2)] = 0.

[-3(b +5) + a(-5 -9) + -4(9 - b)] = 0

[ -3b - 15 + a(-14) - 36 + 4b ] = 0

[ -3b - 15 - 14a - 36 + 4b ] = 0

-14a + b - 51 = 0

Equation: -14a + b = 51 ..(i)

Subtracting Equation (i) from (ii):

-14a + b - ( a + b) = 51 - 1

-14a + b - a - b = 50

-15a = 50

a = -50/15

a = -10/3..

Now,

a + b = 1

-10/3 + b = 1

b = 1 + 10/3

b = 13/3

Hence, value of x and y are -10/3 and 13/3 respectively.