Question

if the line x/a+y/b=1passes through the point of intersection of the lines x=3 and 2x +3y=12 and is perpendicular to the line 2x +3y =8 find value of a and b.

Answer 1

Answer:

here,

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

let, these lines intersect at point M

now,

$2x + 3y = 12 \\ > 2 \times 3 + 3y = 12 \\ > 6 + 3y = 12 \\ > 3y = 6 \\ > y = 2$

so, the coordinates of point M is: (3,2)

now,

$\frac{x}{a} + \frac{y}{b} = 1 \: passes \: through \: point \: < m >$

therefore,

X=3, y=2

$\frac{x}{a} + \frac{y}{b} = 1 \\ > \frac{3}{a} + \frac{2}{b} = 1 \\ > \frac{3b + 2a}{ab} = 1 \\ > 3b + 2a = ab \\ > 2a = ab - 3b \\ > a = \frac{ab - 3b}{2}$

now, this line is perpendicular to the line 2x+3y=8

here,

2x+3y=8

>3y=8-2x

>y=(8-2x)/3

so, slope of the line is -(2/3)

here,

$\tan( \alpha ) = - \frac{2}{3} \\ > \tan( \alpha ) = \tan( - 33.69) \\ > \alpha = - 33.69$

now

the line X/a+y/B= is perpendicular to the line 2x+3y=8

so, angle of line X/a+y/B is (-33.69+90)°=56.31°

here,

slope of the line

tan(56.31)=1.5

now,

$\frac{x}{a } + \frac{y}{b} = 1 \\ > \frac{y}{b} = 1 - \frac{x}{a} \\ > \frac{y}{b} = \frac{a - x}{a} \\ > y = \frac{b(a - x)}{a} = b - \frac{x}{a}$

here slope=-(X/a)

also, slope=1.5=3/2

so,

$- \frac{x}{a} = \frac{3}{2} \\ > \frac{x}{a} = - \frac{3}{2}$

putting this in X/a+y/B=1, we get,

$\frac{x}{a} + \frac{y}{b} = 1 \\ > - \frac{3}{2} + \frac{y}{b} = 1 \\ > \frac{y}{b} = 1 + \frac{3}{2} = \frac{5}{2} \\ > \frac{2}{b} = \frac{5}{2} \\ > b = \frac{4}{5}$

again, putting value of b in the same equation, we get,

$\frac{x}{a} + \frac{y}{b} = 1 \\ > \frac{3}{a} + \frac{2}{ \frac{4}{5} } = 1 \\ >$