Question

Find the equation of the line perpendicular to 2x+3y=1 and whose distance from the
origin is 7.

Answer 1

Let the meeting point of the perpendicular lines be $(a, b)$

$/OP/=\sqrt{(a^2 +b^2)} =7$ ..... (i)

a²+b²=49

2x+3y=1 ;y=$-(\frac{2}{3}) x+\frac{1} {3}$

Gradient =$-(\frac{2}{3})$

For perpendicular lines,the product of Gradient I and gradient II hence gradient II =$\frac{3}{2}$

$Gradient =\frac{\Delta y} {\Delta x}$

$\frac{3}{2} =\frac{b-0} {a-0}$

2b=3a .... (ii)

Substituting (ii) in (i) ;

$a^2 +(\frac{3a}{2} )^2 =49$

$4a^2+9a^2=196$

a =4 or $-$4; b=6 or $-$ 6.

To get the equation, we take the point (4,6) and the gradient $\frac{3}{2}$

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

2y-12=3x -12

2y=3x

When a=$-$ 4 and b=$-$6

$\frac{3} {2} =\frac{y-(-6)}{x-(-4)}$

3x$+$ 12=2y$+$12

2y=3x