Q.1 Determine the values of m and n so that the following system of linear equations have infinite number of solutions: 

$(2𝑚 - 1)𝑥 + 3𝑦 - 5 = 0 𝑎𝑛𝑑 3𝑥 + (𝑛 - 1)𝑦 - 2 = 0$

 

(OR)

 

Q.2 If ΔABC ≃ ΔPQR and AD, PS are the bisectors of the corresponding angles A & P then prove that,  $\frac{Area\left(\Delta ABC\right) }{Area\left(\Delta PQR\right)}= \frac{AD²}{PS²}$

Given that,  

 The following system of linear equations have infinite number of solutions: 

$(2𝑚 - 1)𝑥 + 3𝑦 - 5 = 0 \dots .(𝑖)$

$3𝑥 + (𝑛 - 1)𝑦 - 2 = 0 \dots .(𝑖𝑖)$

We have to find the values of m and n

The above equations are of the form,  

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

${𝑎}_1 = (2𝑚 - 1),{ 𝑏}_1 = 3, {𝑐}_1 =- 5$

${𝑎}_2 =3,{ 𝑏}_2 = (n-1), {𝑐}_2 =- 2$

For infinitely many solutions,  

$$\frac{a₁}{a₂}=\frac{b₁}{b₂}=\frac{c₁}{c₂}$$

$$\Rightarrow \frac{(2m-1)}{3}=\frac{3}{(n-1)}=\frac{-5}{-2}$$

$\Rightarrow \frac{(2m-1)}{3}=\frac{-5}{-2} and \frac{ 3 }{ ( n - 1 )} = \frac{ - 5 }{- 2}$

 $\Rightarrow (- 2)(2𝑚 - 1) =- 5\left(3\right)$

$\Rightarrow - 4𝑚 + 2 =- 15$

$\Rightarrow - 4𝑚 =- 17$

$\Rightarrow 𝑚 = \frac{17 }{4}$

$Similarly, \frac{3}{(n-1)}=\frac{-5}{-2 }$

$\Rightarrow (-5)(n-1)=-2\left(3\right)$

$\Rightarrow -5n+5 =-6$

$\Rightarrow -5n =-11$

$\Rightarrow n= \frac{11}{5}$

Hence, the values of m and n are$\frac{17}{4},\frac{11 }{5}$

(OR)

Given that, 

ΔABC ≃ ΔPQR and AD, PS  are the bisectors of the corresponding angles A & P

Now we have to prove that, 

$\frac{Area\left(\Delta ABC\right) }{Area\left(\Delta PQR\right)}= \frac{AD²}{PS²}$

Since,

ΔABC ≃ ΔPQR

Then

∠A= ∠P

∠B= ∠Q

$$\angle C= \angle R$$

We know that ratio of area of two similar triangles is equal to the ratio of the  square of the corresponding sides 

$Since, \angle A= \angle P$

$\Rightarrow \frac{1}{2}\angle A=\frac{1}{2} \angle P$

$$\Rightarrow \angle BAD=\angle PQS$$ 

By AA similarity,

ΔBAD ≃ ΔQPS

Then the corresponding sides are in proportional  

$$\frac{AB}{PQ} =\frac{AD}{PS}$$......(i)

From

From (i) and (ii)we get,  

$\Rightarrow \frac{ A r e a ( \Delta A B C )}{ A r e a ( \Delta P Q R )} = \frac{A B ²}{P{S}^2}$

hence proved