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Questions  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the line which passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point

$\begin {array} {1 1} (A)\;9x+20y+96=0 & \quad (B)\;9x-20y+96=0 \\ (C)\;9x-20y-96=0 & \quad (D)\;9x+20y-96=0 \end {array}$

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A)
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  • Section formula : The coordinates of the point which divides the line joining the points $(x_1, y_1) $ and $ (x_2, y_2)$ in the ratio $m : n $ is $ \bigg( \large\frac{mx_2+nx_1}{m+n}$$, \large\frac{my_2+ny_1}{m+n} \bigg)$
Step 1:
Let the equation of the line be $\large\frac{x}{a}$$+\large\frac{y}{b}$$=1$. This line meets the coordinates axes at A(a,0) and B(0,b) respectively. The coordinates of the point which divides the line joining A(a,0) and B(0, b) in the ratio 5:3 are
$ \bigg[ \large\frac{5 \times (0)+3(a)}{5+3}$$, \large\frac{5 \times (b) +3 (0)}{5+3} \bigg]$
(i.e) The coordinates are
$ \Rightarrow \bigg( \large\frac{3a}{8}$$, \large\frac{5b}{8} \bigg)$
Step 2 :
It is given that the point (-4, 3) divides AB in the ratio 5 : 3
$ \therefore \large\frac{3a}{8}$$ = -4 \Rightarrow a = -\large\frac{33}{3}$
and $ \large\frac{5b}{8}$$=3 \Rightarrow = \large\frac{24}{5}$
Hence the equation of the line is
$ \large\frac{x}{\Large\frac{-32}{3}}$$+\large\frac{y}{\Large\frac{24}{5}}$$=1$
(i.e) $9x-20y=-96$
$ \quad 9x-20y+96=0$
Hence the equation of the required line is $ 9x-20y+96=0$
Thank you verymuch@likedit
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