Question

The points P(3, 0), Qla, -2), R(4, -1), are the verties of APQR right angled at vertex P. Find the value of a.​

Answer 1

Step-by-step explanation:

Given :-

The points P(3, 0), Q(a, -2), R(4, -1), are the verties of APQR right angled at vertex P.

To find:-

Find the value of a ?

Solution:-

Given that

The points are P(3, 0), Q(a, -2), R(4, -1)

Given that P,Q and R are the vertices of the ∆PQR and right angle at P

=>QR^2 = PQ^2 + PR^2 -----------(1)

by Pythagoras theorem

I) Distance between P and Q:-

Let (x1, y1) = P(3,0)=>x1 =3 and y1 = 0

Let (x2, y2)=Q(a,-2)=>x2 = a and y2 = -2

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2 units

PQ = √[(a-3)^2+(-2-0)^2]

=>PQ = √[(a-3)^2+(-2)^2]

PQ = √[a-3)^2+4] units

Now , PQ^2 = [√[a-3)^2+4]]^2

=>PQ^2 = (a-3)^2+4

=>PQ^2=a^2-6a+9+4

(Since (a-b)^2=a^2-2ab+b^2))

PQ^2=a^2-6a+13 --------------(2)

ii)Distance between Q and R:-

Let (x1, y1) = Q(a,-2)=>x1 = a and y1 = -2

Let (x2, y2)=R(4,-1)=>x2 = 4 and y2 = -1

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2 units

QR = √[(4-a)^2+(-1-(-2))^2]

=>QR = √[(4-a)^2+(-1+2)^2]

=>QR = √[(4-a)^2+(1)^2]

QR = √[(4-a)^2+1] units

Now ,

QR^2 = [√[(4-a)^2+1]]^2

=>QR^2 = (4-a)^2+1

=>QR^2 = 4^2-2(4)(a)+a^2+1

=>QR^2=16-8a+a^2+1

QR^2 = a^2-8a+17 -------------(3)

iii)Distance between P and R:-

Let (x1, y1) = P(3,0)=>x1 =3 and y1 = 0

Let (x2, y2)=R(4,-1)=>x2 = 4 and y2 = -1

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2 units

PR = √[(4-3)^2+(-1-0)^2]

=>PR =√[1^2+(-1)^2]

=>PR = √(1+1)

PR = √2 units

Now,

PR^2 = [√2]^2

PR^2 = 2 -----------------(4)

We have , from (1)

QR^2 = PQ^2 + PR^2

=>a^2-8a+17 = (a^2-6a+13)+2

=>a^2-8a+17 = a^-6a+13+2

=>a^2-8a+17=a^2-6a+15

=>a^2-8a+17-a^2+6a-15 = 0

=>(a^2-a^2)+(-8a+6a)+(17-15) = 0

=>0+(-2a)+2 = 0

=>-2a+2 = 0

=>-2a = -2

=>2a = 2

=>a = 2/2

=>a = 1

Therefore, a = 1

Answer:-

The value of a for the given problem is 1

Used formulae:-

1. The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2 units

2.(a-b)^2=a^2-2ab+b^2

3. Pythagoras Theorem:-

• In a right angled triangle ,The square of the hypotenuse is equal to the sum of the squares of the other two sides .