Question

Determine the ratio in which the point (–6, m) divides the join of A(–3, –1) and B(–8, 9). Also, find the value of m.

Answer 1

Answer:

m=5

Step-by-step explanation:

Given: Point $C(x_3,y_3)=(-6,m)$ divides the join of point $A(x_1,y_1)=(-3,-1)$ and point $B(x_2,y_2)=(8,9)$

Let the line AB divides by Point C in a ratio m:n=k:1

Then, Using section formula   $(x_3,y_3)=\frac{x_1n+x_2m}{m+n},\frac{y_1n+y_2m}{m+n}$

Applying formula,

$x_3,y_3=\frac{x_1n+x_2m}{m+n},\frac{y_1n+y_2m}{m+n}$

$x_3,y_3=\frac{-8k-3}{k+1},\frac{9k-1}{k+1}$

But, $x_3=-6$

Therefore,  $x_3=\frac{-8k-3}{k+1}$

$-6=\frac{-8k-3}{k+1}$

$-6k-6=-8k-3$

$2k-3=0$

$k=\frac{3}{2}$

Therefore, C divides line AB in 3:2

Now,  $m=\frac{9k-1}{k+1}$  where, k=3/2

$m=\frac{9\frac{3}{2}-1}{\frac{3}{2}+1}$

$m=\frac{\frac{25}{2}}{\frac{5}{2}}$

$m=\frac{25\times2}{2\times5}$

$m=5$

Answer 2

Answer:

Ratio = 3 : 2 and value of m = 5.

Step-by-step explanation:

We are given the end points ( -3,-1 ) and ( -8,9 ) of a line and a point P = ( -6,m ) divides this line in a particular ratio.

Let us assume that it cuts the line in k : 1 ratio.

Then, the co-ordinates of P = $( \frac{-8k-3}{k+1},\frac{9k-1}{k+1} )$.

But, $\frac{-8k-3}{k+1}$ = -6

i.e. -8k-3 = -6k-6

i.e. -2k = -3

i.e. $k = \frac{3}{2}$

So, the ratio is k : 1 i.e $\frac{3}{2} : 1$ i.e. 3 : 2.

Hence, the ratio in which P divides the line is 3 : 2.

Also, $\frac{9k-1}{k+1}$ = m where $k = \frac{3}{2}$

i.e. m = $\frac{\frac{9 \times 3}{2}-1}{\frac{3}{2}-1}$

i.e. m = $\frac{27-2}{3+2}$

i.e. m = $\frac{25}{5}$

i.e. m = 5.

Hence, the value of m is 5.