Question

class 10th students solve this.
Q21
15 points will be given.
​

Answer 1

Answer:

Step-by-step explanation:

Given that,

A(x₁,y₁) = (2,1)

B(x₂,y₂) = (3,-2)

C(x₃,y₃) = (7/2,y)

Area of the triangle = 5 sq. units

y = ?

Solution:

Area of triangle = 1/2 I x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) I

5 = 1/2 I 2(-2-y) + 3(y-1) + 7/2(1+2) I

5 = 1/2 I -4-2y+3y-3+7/2+7 I

5 = 1/2 I -7+7+y+7/2 I

5 = 1/2 I y+7/2 I

10 = I y+7/2 I

10 = y+7/2      ,      -10 = y+7/2

y = 10 - 7/2     ,        y = -10-7/2

y = (20-7)/2     ,        y = (-20-7)/2

y = 13/2          ,         y = -27/2

These are the required values

Answer 2

Let the points be A , B and C

Slope of A and B

Slope = ( b² - a² ) / ( b - a )

= ( b + a )( b - a )/(b - a )

= b + a

Slope of B and C

Slope = ( b² - 0 ) / ( b - 0 )

= b²/b

= b

Slope of b + a is not equal to b .

This can be possible only if a = b = 0

So line can never be collinear if a ≠ b ≠ 0 .

Hence proved.