Question

For $a,b,c,d\in N$, the area of the parallelogram bounded by the lines $y=ax+c,y=ax+d,y=bx+c$ and $y=bx+d$ is $18$ while the area of the parallelogram bounded by the lines $y=ax+c,y=ax-d,y=bx+c$ and $y=bx-d$ is $72$. The least possible value of $a+b+c+d$ will be:

• A. $0$
• B. $12$
• C. $15$
• D. $16$

Answer 1

The correct option is D $16$

So when the equation of parallel line of parallelogram is given then the area of parallelogram will be given by

Area of a parallelogram $=\frac{\left|(c_1-d_1)\right|\left|(c_2-d_2)\right|}{\left|(a_1{b}_2-a_2{b}_1)\right|}$

where $c_1,$ $d_1$ are the intercepts of the two parallel lines and $c_2,$ $d_2$ are the intercept of the other lines

$a_1,$ $a_2$ are the coefficient of $x$ in non-parallel lines and $b_1$ $b_2$ are the coefficient of $y$ in non-parallel lines

Now we have the area of parallelogram which is bounded by

$y=ax+c,y=ax+d,y=bx+c$ and $y=bx+d$

so by the formulae which we have mentioned above we got

$\frac{{|c-d|}^2}{|a-b|}=18$ ....... $\left(1\right)$

When the lines are $y=ax+c,y=ax-d,y=bx+c$ and $y=bx-d$

Then the area will be $72$

So again we will apply the formulae and after that we get

$\frac{{|c+d|}^2}{|a-b|}=72$ ........ $\left(2\right)$

Now from equation $1$ and $2$, we get the relation between $a$ , $b$ , $c$ , $d$ that

$c=3\times d$ and $d^2$ = $\frac{9}{2}\times (a-b)$

So now because we know $a,b,c,d$ are the least natural number which satisfy these relation and it is only possible when

$(a-b)=2$ then $d$ will be $3$ and $c=3\times d$ so $c$ will be $9$

$a$ and $b$ will be $3$ and $1$ orderly

So, now the minimum value of

$a+b+c+d=3+1+9+3$ = $16$

Hence, option D is correct.