S Chand CLASS 10 Chapter 6 Ration and proportion Exercise 6B

  Exercise 6B

Question 1 

Ans: 1 

Given, 

$\frac{x}{y}=\frac{p}{q}$

According to question, $\frac{x}{y}=\frac{p}{q}$

$\frac{x}{y}=\frac{p}{q}$

Multiply both sides by $\frac{5}{7}$

$\frac{5 x}{7 y}=\frac{5 p}{7 q} .$

$\frac{5 x+7 y}{5 x-y}=\frac{5 p+7 q}{5 p-7 q}$ (by compound and divided)

hence proved 

Question 2

Ans:2 Given,

$4 a+9 b: 4 a-9 b=4 c+9 d: 4 c-9 d$.

$\frac{4 a+9 b}{4 a-9 b}=\frac{4 c+9 d}{4 c-9 d} \text {}$

$\frac{4 a+9 b+4 a-9 b}{4 a+9 b-(4 a-9 b)}=\frac{4 c+9 d+4 c-9 d}{4 c+9 d-(4 c-9 d)}$ (by compound and dividends)

$\frac{4 a+9 b+4 a-9b}{4 a+9 b-4 a+9 b}=\frac{4 c+9 d-(4 c-9 d}{4 c+9 d-4 c+9 d}$ 

$\frac{8 a}{18b}=\frac{8 c}{18 d} \text {. }$

Multiply both sides by $\frac{18}{8}$

$\frac{8a}{18b} \times \frac{18}{8}=\frac{8 c}{18 d} \times \frac{18}{8}$◘

$\frac{a}{b}=\frac{c}{d}$

$\therefore a: b:: c \mathrm{~d}$

Hence, Proved

Question 3

Ans:3 $(5 a+11 b):(5 c+11 d)::(5 a-11 b):(5 c-11 d)$

$\frac{5 a+11 b}{5 c+11 d}=\frac{5 a-11 b}{5 c-11 d}$

$\frac{5 a+11 b}{5 a-11 \cdot b}=\frac{5 c+11 d}{5 c-11 . d}$

using comp. 

$\frac{5 a+11 b+5 a-11 b}{5 a+11 b-5 a+11 b}$ = $\frac{5 c-11d+5 c-11d}{5 c+11 d-5 c+11 d}$

$\frac{10 a}{22 b}=\frac{10 c}{22 d}$ proved 

Question 4

Ans:4

 (i) Given, $m a^{2}+h b^{2}: m c^{2}+h d^{2}: m a^{2}-h b^{2}: m c^{2}-h d^{2}$

$\frac{m a^{2}+h b^{2}}{m c^{2}+h d^{2}}=\frac{m a^{2}-h b^{2}}{m c^{2}-h d^{2}}$

$\frac{m n^{2}+h b^{2}}{m a^{2}-h b^{2}}=\frac{m c^{2}+h d^{2}}{m c^{2}-h d^{2}}$ (by alternate)

$\frac{m a^{2}+h b^{2}+m a^{2}-h b^{2}}{m a^{2}+h b^{2}-\left(m a^{2}-h b^{2}\right)}=\frac{m c^{2}+h d^{2}+m c^{2}-h d^{2}}{m c^{2}+h d^{2}-\left(m c^{2}-h d^{2}\right)}$

$\frac{m a^{2}+h b^{2}+m a^{2}-h b^{2}}{m a^{2}+h b^{2}-ma^{2}+h b^{2}}=\frac{m c^{2}+n d^{2}+m c^{2}-b d^{2}}{mc^{2}+n d^{2}-m c^{2}+h d^{2}}$

$\frac{m a^{2}}{h b^{2}}=\frac{m c^{2}}{h d^{2}}$

multiply both sides by $\frac{h}{m}$

$\frac{m  a^{2}}{h b^{2}} \times \frac{h}{m}$= $\frac{m  c^{2}}{h d^{2}} \times \frac{h}{m}$

$\frac{a^{2}}{b^{2}}=\frac{c^{2}}{d^{2}}$

Talking square root both sides 

$\begin{aligned} \therefore \quad \frac{a}{b} &=\frac{c}{d} \\ a : b: & \therefore c=d \end{aligned}$

Hence a , b , c and d are in proportion 

(ii)

$\begin{aligned}&(a+b+c+d)(a-b-c+d)=(a+b-c-d)(a-b+c-d) \\&\frac{a+b+c d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}\end{aligned}$

$\frac{a+b+c d}{a+b-c d}=\frac{a-b+c-d}{a-b-c+d}$

$\frac{a+b+c+d+a+b-c-d}{a+b+c+d-(a+b-c-d)}$ =$\frac{a-b+c-d+a-b-c+d}{a-b+c-d-(a-b-c+d)}$ 

(By component and dividers)

$\frac{2 a+2 b}{2 c+2 d}=\frac{2 a-2 b}{2 c-2 d}$

$\frac{2 a+2 b}{2 a-2 b}=\frac{2 c+2 d}{2 c-2 d}$(By Alternate)

$\frac{2 a+2 b+2 a-2 b}{2 a+2 b-(2 a-2 b)}=\frac{2 c+2 d+2 c-2 d}{2 c+2 d-(2 c-2 d)}$ (By component and dividends)

$\frac{2 a+2 b+2 a-2 b}{2 b+2 b-2 b+2 b}=\frac{2 c+2 b+2 c-2 c}{2 c+2 d-2 c+2 d}$

$\frac{4_{a}}{4_{b}}=\frac{4_{c}}{4_{d}}$

$a=b :: c: d .$

∴ a , b , c and d are in proportion 

Question 5

Ans:5

Given,

a: b=c: d  and c: f=g: h

$\frac{a}{b}=\frac{c}{d}$ and $\frac{c}{f}=\frac{g}{h}$

Multiplying each other 

$\frac{a}{b} \times \frac{e}{f}=\frac{c}{d} \times \frac{g}{h}$

$\frac{a e}{b y}=\frac{c g}{d h}$

$\frac{a e+b f}{a e-b f}=\frac{c g+d h}{c g-d h}$ (By component and dividends)

∴ $a e+b f=$ $a e-b y=$ $c g+d h= c y-d h$

Hence proved, 

Question 6

Ans:6

Given,

$\begin{aligned}x &=\frac{10 p q}{p+q} \\x &=\frac{5 p \times 2 q}{p+q} \\\frac{x}{5 p} &=\frac{2 q}{p+q} \\\frac{x+5 p}{x-5 p} &=\frac{2 q+p+q}{2 q-(p+q)}\end{aligned}$  (By component and dividends)

$\frac{x+5 p}{x-5 p}=\frac{3 q+p}{2 q-p-q}$

$\frac{x+5 p}{x-5 p}=\frac{3 q+p}{q-p}$............(i)

Again , 

$x=\frac{10 p q}{p+q}$

$x=\frac{2 p \times 5 q}{p+q}$

$\frac{x}{5 q}=\frac{2 p}{p+q}$

$\frac{x+5 q}{x-5 q}=\frac{2 p+p+q}{2 p-(p+q)}$   (By component and dividends)

$\frac{x+5 q}{x-5 q}=\frac{3 p+q}{2 p-p-q}$

$\frac{x+5 q}{x-5 q}=\frac{3 p+q}{p-q}$ ..........(ii)

Add the eqn (i) and (ii)

$\begin{aligned} \frac{x+5 p}{x-5 p}+\frac{x+5 q}{x-5 q} &=\frac{3 q+p}{q-p}+\frac{3 p+q}{p-q} \\ &=\frac{3 q+p}{q-p}-\frac{3 p+q}{q-p} \\ &=\frac{3 q+p-(3 p+q)}{q-p} \\ &=\frac{3 q+p-3 p-q}{q-p} \\ &=\frac{2 q-2 p}{q-p} \end{aligned}$

$=\frac{2(q-p)}{q-p}$

=2 

Hence the value of $\frac{x+5 p}{x-5 p}+\frac{x+5 q}{x-5 q}$ is 2 .

Question 7

Ans:7

Given,

$x=\frac{6 p q}{p+q}$

$x=\frac{3 p \cdot x 2 q}{p+q}$

$\frac{x}{3 p}=\frac{2 q}{p+q}$

$\frac{x+3 p}{x-3 p}=\frac{2 q+p+q}{2 q-(p+q)}$  (By component and dividends)

$\frac{x+3 p}{x-3 p}=\frac{3 q+p}{2 q-p-q}$

$\frac{x+3 p}{x-3 p}=\frac{3 q+p}{q-p}$..........(i)

Again,

$\begin{aligned}&x=\frac{6 p q}{p+q} \\&x=\frac{2 p \times 3 q}{p+q} \\&\frac{x}{3 q}=\frac{2 p}{p+q}\end{aligned}$

$\frac{x+3 q}{x-3 q}=\frac{2 p+p+q}{2 p-(p+q)}$ (By component and dividends)

$\frac{x+3 q}{x-3 q}=\frac{3 p+q}{2 p-p-q .}$

$\frac{2 c+3 q}{x-3 q}=\frac{3 p+q}{p-q}$............(ii)

On adding eq (i) and (ii)

$\begin{aligned} \frac{x+3 p}{x-3 p}+& \frac{x+3 q}{x-3 q}=\frac{3 q+p}{q-p}+\frac{3 p+q}{p-q} \\ &=\frac{3 q+p}{q-p}-\frac{3 p+q}{q-p} \\ &=\frac{3 q+p-(3 p+q)}{q-p} \end{aligned}$

$=\frac{3 q+p-3 p-q}{q-p}$

$=\frac{2 q-2 p}{q-p}$

=2

Hence the value of $\frac{x+3 p}{x-3 p}+\frac{x+3 q}{x-3 q}=2$

Question 8

Ans:8

(i) Given

$\frac{3 x+\sqrt{9 x^{2}-5}}{3 x-\sqrt{9 x^{2}-5}}=\frac{5}{1}$

$\frac{3 x+\sqrt{9 x^{2}-5}+3 x-\sqrt{9 x^{2}-5}}{3 x+\sqrt{9 x^{2}-5}-\left(3 x-\sqrt{9 x^{2}-5}\right)}=\frac{5+1}{5-1}$   (By component and dividends)

$\frac{3 x+3 x}{3 x+\sqrt{9 x^{2}-5}-3 x+\sqrt{9 x^{2}-5}}=\frac{6}{4} .$

$\frac{6 x}{2 \sqrt{9 x^{2}-5}}=\frac{6}{4}$

$\frac{3 x}{\sqrt{9 x^{2}-5}}\frac{3}{2}$

$3 x \times 2=3 \sqrt{9 x^{2}-5}$

$6 x=\frac{3 \sqrt{9 x^{2}-5}}{6}$

$x=\frac{3 \sqrt{9 x^{2}-5}}{62}$

$x=\frac{\sqrt{9 x^{2}-5}}{2}$

$2 x=\sqrt{9 x^{2}-5}$

Squaring both side 

$\begin{aligned} \therefore \quad &(2 x)^{2}=\left(\sqrt{9 x^{2}-5}\right)^{2} \\ & 4 x^{2}=9 x^{2}-5 \\ & 4 x^{2}-9 x^{2}=-5 \\ &-5 x^{2}=-5 \end{aligned}$

$x^{2}=1$

$x=\sqrt{1}$

$x=1$

Hence the value of x is 1

(ii)Given,

$\frac{\sqrt{x+12}+\sqrt{x-3}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{5}{1}$

$\frac{\sqrt{x+2}+\sqrt{x-3}+(\sqrt{x+2}-\sqrt{x-3})}{\sqrt{x+2}+\sqrt{x-3}-(\sqrt{x+2}-\sqrt{x-3})}=\frac{5+1}{5-1}$ (By component and dividends)

$\frac{\sqrt{x+2}+\sqrt{x-3}+\sqrt{x+2}-\sqrt{ x} -3}{\sqrt{x+2}+\sqrt{x-3}-\sqrt{x+2}+\sqrt{x-3}}=\frac{6}{4}$

$\frac{2 \sqrt{x+2}}{2 \sqrt{x-3}}$ $=\frac{6}{4}$

$\frac{\sqrt{x+2}}{\sqrt{x-3}}=\frac{3}{2}$

Squaring both sides 

$\begin{aligned}\left(\frac{\sqrt{x+2}}{\sqrt{x-3}}\right)^{2} &=\left(\frac{3}{2}\right)^{2} \\ \frac{x+2}{x-3} &=\frac{9}{4} \\ 4(x+2) &=9(x-3) \\ 4 x+8 &=9 x-27 \\ 4 x-9 x &=-8-27 \\-5 x &=-35 \end{aligned}$

x=7

Hence, the value of $x$ is 7 .

(iii) $\frac{x^{3}+3 x-341}{3 x^{2}+1}=\frac{31}{91}$

$\frac{x^{3}+3 x+\left(3 x^{2}+1\right)}{x^{3}+3 x-\left(3 x^{2}+1\right)}=\frac{341+91}{341-91}$  (By component and dividends)

$\frac{x^{3}+3 x+3 x^{2}+1}{x^{2}+3 x-3 x^{2}-1}=\frac{432}{250}$

$\frac{(x+1)^{3}}{(x-1)^{3}}$ = $\frac{216}{125}$

$\frac{(x+1)^{3}}{(x-1)^{3}}=\frac{(6)^{3}}{(5)^{3}}$

On taking cube root 

$\frac{x+1}{x-1}=\frac{6}{5}$

$5(x+1)=6(x-1)$

$5 x+5=6 x-6$

$5 x-6 x=-5-6$

$+x=+11$

$x=11$

Hence the value of x is 11

 

(iv) $\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{4 x-1}{2}$

$\frac{\sqrt{x+1}+\sqrt{x-1}+(\sqrt{x+1}-\sqrt{x-1})}{\sqrt{x+1}+\sqrt{x-1}-(\sqrt{x+1}-\sqrt{x-1})}=\frac{4 x-1+2}{4 x-1-2}$  (By component and dividends)

$\frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}+\sqrt{x-1}}=\frac{4 x-1}{4 x-3}$

$\frac{2 \sqrt{x+1}}{2 \sqrt{x-1}}=\frac{4 x+1}{4 x-3}$

$\frac{\sqrt{x+1}}{\sqrt{x-1}}=\frac{4 x+1}{4 x-3}$

Squaring both sides 

$\frac{(\sqrt{x+1})^{2}}{(\sqrt{x-1})^{2}}=\frac{(4 x+1)^{2}}{(4 x-3)^{2}}$

$\frac{x+1}{x-3}=\frac{(4 x)^{2}+(1)^{2}+2 x 4 x x 1}{(4 x)^{2}+(3)^{2}-2 \times 4 x+3}$

$\frac{x+1}{x-1}=\frac{16 x^{2}+1+8 x}{16 x^{2}+9-24 x}$

$(x+1)\left(16 x^{2}+9-24 x\right)=(x-2)\left(16 x^{2}+1+8 x\right)$

$16 x^{3}+9 x-24 x^{2}+16 x^{2}+9-24 x=16 x^{3}+x+8 x^{2}-16 x^{2}-1-8 x$

$\begin{aligned} 16 x^{3}-15 x &-8 x^{2} 9=16 x^{3}-7 x-8 x^{2}-1 \\ 16 x^{3}-15 x &=8 x^{2}+9-16 x^{3}+7 x+8 x^{2}+1=0 \\ &-8 x+10=0 \\ &=8 x=0-10 \\ &-8 x=-10 \end{aligned}$

$x=\frac{5}{4}$

Hence the value of x is  $=\frac{5}{4}$

Question 9

Ans:9

Given,

$16\left[\frac{a-x}{a+x}\right]^{3}=\frac{a+x}{a-x} \text {. }$

Multiply both sides by $\frac{a-x}{a+x}$

$16\left[\frac{a-x}{a+x}\right]^{3} \times \frac{a-x}{a+x}=\frac{a+x}{a-x} \times \frac{a-\pi}{a+x}$

$16\left[\frac{a-x}{a+x}\right]^{4}=1$

$\left[\frac{a-x}{a+x}\right]^{4}=\frac{1}{16}$

$\left[\frac{a-x}{a+x}\right]^{4}=\left[\frac{1}{2}\right]^{4}$

$\begin{aligned} \therefore \quad & \frac{a-x}{a+x}=\frac{1}{2} \\ & 2(a-x)=a+x \\ & 2 a-2 x=a+x \\ & 2 a-a=2 x+x \\ & a=3 x \\ & \therefore \quad x=\frac{a}{3} \end{aligned}$

Hence the value of x is $\frac{a}{3}$

Question 10

Ans-10:

$\begin{gathered}\frac{1+x+x^{2}}{1-x+x^{2}}=\frac{171(1+x)}{172(1-x)} \\\frac{(1-x)\left(1+x+x^{2}\right)}{(1+x)\left(1-x+x^{2}\right)}=\frac{171}{172} \\\frac{1-x^{3}}{1+x^{3}}=\frac{171}{172} \\172\left(1-x^{3}\right)=171\left(1+x^{3}\right) \\172-172 x^{3}=171+171 x^{3} \\172-171=172 x^{3}+171 x^{3} \\1=343 x^{3}\end{gathered}$

$\begin{aligned} \therefore x^{3} &=\frac{1}{343} \\ x &=\sqrt[3]{\frac{1}{343}} \\ x &=\sqrt[3]{\frac{1 \times 1 \times 1}{7 \times 7 \times 7}} \\ x &=\frac{1}{7} \end{aligned}$

Hence the value of x is  $\frac{1}{7}$

Question 11

Ans: 

Given

$\frac{x}{1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b^{2}}

\sqrt{a^{2}b^{2}}}$

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}+\left(\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}\right)}{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}-\left(\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}\right)}$

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b}+\sqrt{a^{2}-b^{2}}-\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}$

$\frac{x+1}{x-1}=\frac{2 \sqrt{a^{2}+b^{2}}}{2\sqrt{a^{2}-b^{2}}}$

$\frac{x+1}{x-1}=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}-b^{2}}}$

Squaring both sides 

$\frac{\left(x+b^{2}\right.}{(x-1)^{2}}=\left(\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}-b^{2}}}\right)^{2}$

$\frac{x^{2}+1+2 x x+1}{x^{2}+1-2 x^{2}(x)}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

$\frac{x^{2}+1+2 x}{x^{2}+1-2 x}=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

$\frac{x^{2}+1+2 x+\left(x^{2}+1-2 x\right)}{x^{2}+1+2 x-\left(x^{2}+1-2 x\right)}=\frac{a^{2}+b^{2}+\left(a^{2}-b^{2}\right)}{x^{2}+b^{2}-\left(a^{2}-b^{2}\right)}$ (By components and dividends)

$\frac{2 x^{2}+2}{4 x}=\frac{2 a^{2}}{2 b^{2}}$

$\frac{2\left(x^{2}+1\right)}{4 x}=\frac{a^{2}}{b^{2}}$

$\frac{x^{2}+1}{2 x}=\frac{a^{2}}{b^{2}}$

$b^{2}\left(x^{2}+1\right)=2 a^{2} x$

$b^{2} x^{2}+b^{2}=2 a^{2} x$

$b^{2} x^{2}-2 a^{2} x+b^{2}=0$

Hence proved .

Question 12

Ans: 

Given,

$\frac{y}{1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}$

$\frac{y+1}{y-1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}+(\sqrt{a+3 b}-\sqrt{a-3 b})}{\sqrt{a+3 b}+\sqrt{a-3 b}-(\sqrt{a+3}-\sqrt{a-3 b})}$  (By components and dividends)

$\frac{y+1}{y-1}=\frac{\sqrt{a+3 b}+\sqrt{a-3 b}+\sqrt{a+3 b}-\sqrt{a-3 b}}{\sqrt{a+3 b}+\sqrt{a-3 b}-\sqrt{a+3 b}+\sqrt{a-3 b}}$

$\frac{y+1}{y-1}=\frac{2 \sqrt{a+3 b}}{2 \sqrt{a-3 b}}$

$\frac{y+1}{y-1}=\frac{\sqrt{a+3 b}}{\sqrt{a-3 b}}$

On squaring both sides 

$\frac{(y+1)^{2}}{(y-1)^{2}}=\frac{(\sqrt{a+3 b})^{2}}{(\sqrt{a-3 b})^{2}}$

$\frac{y^{2}+1+2 x y x 1}{y^{2}+1-2 x y \times 1}=\frac{a+3 b}{a-3 b}$

$\frac{y^{2}+1+2 y}{y^{2}+1-2 y}=\frac{a+2 b}{a-3 b}$

$\frac{y^{2}+1+2 y+\left(y^{2}+1-2 y\right)}{y^{2}+1+2 y-\left(y^{2}+1-2 y\right)}=\frac{a+3 b+(a-3 b)}{a+3b-(a-3 b)}$

$\frac{2 y^{2}+2}{4 y}=\frac{2 a}{6 b}$

$\begin{aligned} \frac{y^{2}+1}{2 y} &=\frac{a}{3 b} \\ 3 b\left(y^{2}+1\right) &=2 a y . \\ 3 b y^{2}+3 b &=2 a y \end{aligned}$

$3 b y^{2}-2 a y+3 b=0$

Hence, proved.

Question 13

Ans: 

given,

$\frac{a^{3}+3 a b^{2}}{3 a^{2} b+b^{3}}=\frac{x^{3}+3 x y^{2}}{3 x^{2} y+y^{3}}$

$\frac{a^{3}+3 a b^{2}+\left(3 a^{2} b+b^{3}\right)}{9^{3}+3 a b^{2}-\left(3 a^{2} b+b^{3}\right)}=\frac{x^{3}+3 x y^{2}+\left(3 x^{2} y+y^{3}\right)^{3}}{x^{3}+3 x y^{2}-\left(3 x^{2} y+y^{3}\right)}$

$\frac{9^{3}+3 a b^{2}+3 a^{2} b+b^{3}}{a^{3}+3 a b^{2}-3 a^{2} b-b^{3}}=\frac{x^{3}+3 x y^{2}+3 x^{2} y+y^{3}}{x^{3}+3 x y^{2}-3 x^{2} y-y^{3}}$

$\frac{a^{3}+b^{3}+3 a b^{2}+3 a^{2} b}{a^{3}-b^{3}+3 a b^{2}-3 a^{2} b}=\frac{x^{3}+y^{3}+3 x^{2} y+3 x y^{2}}{x^{3}-y^{3}+3 x y^{2}-3 x^{2} y}$

$\frac{(a+b)^{3}}{(a-b)^{3}}=\frac{(x+y)^{3}}{(x-y)^{3}}$

On taking cube both sides 

$\sqrt[3]{(a+b)^{3}}=\sqrt[3]{(a-y)^{3}}$

$\frac{a+3}{9-b}=\frac{x+y}{x-y}$

$\frac{a+b+(a-b)}{a+b-(a-b)}=\frac{x+y+(x-y)}{x+y-(x-y)}$

$\frac{a+b+a-b}{c+b-d+b}=\frac{x+y+x-y}{x+y-x+y}$

$\frac{ 2a}{2b}=\frac{2 x}{2 y}$

$\frac{a}{b}=\frac{x}{y}$

$\frac{y}{b}=\frac{x}{a} \quad$

$: \quad \frac{x}{a}=\frac{y}{b}$

Hence proved