Question

Let $d_1$ and $d_2$ be the lengths of the perpendiculars drawn from any point of the line $7x-9y+10=0$ upon the lines $3x+4y=5$ and $12x+5y=7$ respectively. Then

• A. $d_1>d_2$
• B. $d_1=d_2$
• C. $d_1<d_2$
• D. $d_1=2d_2$

Answer 1

Answer: (B)

$d_1=d_2$

Perpendicular drawn from any point of $7x-9y+10=0$ ......... $\left(i\right)$ upon the lines

$3x+4y=5$ ...... $\left(ii\right)$ and

$12x+5y=7$ ......... $\left(iii\right)$

Any point on the line $\left(i\right)$ will be of the form $(x,\frac{7x+10}{9})$

$d_1$ and $d_2$ are the lengths of perpendicular drawn from any point of $\left(i\right)$ upon $\left(ii\right)$ and $\left(iii\right)$

Distance from $(a,\frac{7a+10}{9})$ to $\left(ii\right)$ is

$d_1=\frac{|3a+4\left(\frac{7a+10}{9}\right)-5|}{\sqrt{3^2+4^2}}$

$=\frac{\left|\frac{27a+28a+40-45}{9}\right|}{\sqrt{25}}$

$=\frac{|55a-5|}{9\times 5}$

$=\frac{|11a-1|}{9}$

Distance from $(a,\frac{7a+10}{9})$ to $\left(iii\right)$ is

$d_2=\frac{|12a+5\left(\frac{7a+10}{9}\right)-7|}{\sqrt{{12}^2+5^2}}$

$=\frac{\left|\frac{108a+35a+50-63}{9}\right|}{\sqrt{169}}$

$=\frac{|143a-13|}{9\times 13}$

$=\frac{|11a-1|}{9}=d_1$

$\therefore d_1=d_2$