Question

The area of a triangle, whose vertices are (3, 4), (5, 2) and the point of intersection of the lines x=a and y=5 is 3 square unit​

Answer 1

Step-by-step explanation:

A (3,2) B (5,2) intersection of x = a ,y =5 (a,5)

$\frac{1}{2} |triangle| = 3 \\ traingle \: = 6$

$111 \\ 35a \\ 225$

=25−2a−15+2a+6−10

=6

For any value of a area =3sq.units

Answer 2

Answer: a=5 or -1

Step-by-step explanation:

A (3,2) B (5,2) intersection of x = a ,y =5 (a,5)

area==3

area=1/2|[3*(2-5) +5(5-4) +a(4-2)]|2==3

==|(-9+5+2a)|=3*2=6

==|-4+2a|=6

-4+2a=+-6

-4+2a=+6               Or     - 4+2a = -6

a= 5 or -1