If the points A (2,9), B (a,5) and C (5,5) are the vertices of a `triangleABC`. Right-angled at B, then find the values of a and hence the area of `triangleABC`.

Explanation

Given that , the points A (2,9), B(a,5) and C(5,5) are the vertices of a `triangleABC` right angled at B. By Pythagoras theorem, `AC^(2)=AB^(2)+BC^(2)` …(i)
Now , by distance formula, AB `=sqrt((a-2)^(2)+(5-9)^(2))`
[`:.` distance between two points `(x_(1),y_(2))and (x_(2),y_(2))=sqrt((x_(2)-x_(1)^(2))+(y_(2)-y_(1))^(2))]`
`=sqrt(a^(2)+4-4a+16)=sqrt(a^(2)-4a+20)`
BC`=sqrt((5-a)^(2)+(5-5)^(2))`
`sqrt((5-a)^(2))=5-a`
and AC `=sqrt((2-5)^(2)+(9-5)^(2))`
`=sqrt((-3)^(2)+(4)^(2))=sqrt(9+16)=sqrt(25)=5`
Put the values of AB,BC and AC in Eq . (i), we get
`(5)^(2)=(sqrt(a^(2)-4a+20))^(2)+(5-a)^(2)`
`rArr25=a^(2)-4a+20+a^(2)-10a`
`rArr2a^(2)=14a+20=0`
`rArra^(2)-2a-5a+10=0` [by factorisation method]
`rArra(a-2)-5(a-2)=0`
`rArr (a-2)(a-5)=0`
`:. a=2,5`
Here , `ane5,` , since at `a=5` , the lenght of BC`=0`. It is not possible because the sides AB,BC and CA form a right angled triangle.
So, `a=2`
Now , the coordinate of A,B and C becomes (2,9), (2,5) and (5,5) , respectively.
`:'` Area of `triangle` ABC `=(1)/(2)[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(2)(y_(1)-y_(2))]`
`:. triangle=(1)/(2)[2(5-5)+2(5-9)+5(9-5)]`
`(1)/(2)[2xx0+2(-4)+5(4)]`
`(1)/(2)(0=8+20)=(1)/(2)xx12=6`
Hence , the required area of `triangleABC` is sq units.