Question

Find the equation of the line which passes through the point (-4,3) and the portion of the line intercepted between the axis is divided internally in the ratio 5:3 by this point.

Answer 1

Answer:

9x-20y+96 =0

Step-by-step explanation:

Let us assume that the equation of the straight line is

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

[where, A(a,0) and B(0,b) are the points of intersections of the straight line with X-axis and Y-axis respectively]

Now given that the point (-4,3) divides AB in the ratio 5:3 internally.

Hence, ($\frac{5*0+3*a}{5+3}, \frac{5*b+3*0}{5+3}$)≡ (-4,3)

⇒($\frac{3a}{8},\frac{5b}{8}$) ≡ (-4,3)

{Here we have applied the formula that, if $P(x_{1},y_{1}) , Q(x_{2},y_{2})$ are two points and R is another point which divide PQ line in m:n ratio internally, then the coordinates of R are given by

[$\frac{mx_{2}+nx_{1}  }{m+n},\frac{my_{2}+ny_{1}}{m+n}$]}

So, from the above equation we can write that,

$\frac{3a}{8}=-4$  and  $\frac{5b}{8}=3$ ......(2)

⇒ $a=-\frac{32}{3}$  and  $b=\frac{24}{5}$

Hence, from equations (1) and (2), the equation of the straight line will be

$-\frac{3x}{32}+\frac{5y}{24} =1$

⇒-9x+20y=96

⇒9x-20y+96 =0 (Answer)