Question

find the ratio in which the line segment joining the points a 3 -3 and -2 7 is divided by x-axis. also find the coordinates of the point of division

Answer 1

Answer: 1) The required ratio is 3 : 7

2) The point of division is (-1.5,0)

Step-by-step explanation:

Since, the point of division lies on the x axis,

Hence, the y-coordinate of the point of division = 0,

Let the ratio in which the x-axis divides the line segment having endpoints (3,-3) and (-2,7) is m:n,

By the section formula,

$\frac{7\times m+(-3)\times n}{m+n}=0$

$\frac{7m-3n}{m+n}=0$

$7m-3n=0$

$7m=3n$

$m:n=3:7$

Hence, the ratio is 3 : 7,

Now, again by the section formula,

The coordinates of the point that divides are,

$( \frac{3\times -2+7\times 3}{3+7}, 0)$

$(\frac{-6+21}{10},0)$

$(\frac{15}{10},0)$

$(1.5,0)$

Answer 2

Answer:

• Hence, the ratio in which the line segment joining the points a 3 -3 and -2 7 is divided by x-axis is:

3:7

• Hence, the coordinate of the point is:

$(\dfrac{3}{2},0)$

Step-by-step explanation:

Let 1:k be the ratio in which the line segment joining the points  A(3,-3) and B(-2,7) is divided by x-axis. The coordinate of a point on the x-axis is:

(x,0).

Now using the ratio formula we have the coordinates as:

$x=\dfrac{1\times (-2)+k\times (3)}{1+k}\\\\x=\dfrac{3k-2}{1+k}----(1)$

and,

$0=\dfrac{1\times (7)+k\times (-3)}{1+k}\\\\0=\dfrac{-3k+7}{1+k}\\\\7-3k=0\\\\3k=7\\\\k=\dfrac{7}{3}$

Hence, the ratio is:

$1:\dfrac{7}{3}$

which could also be written as:

$3:7$

Also on putting the value of k in equation (1) we have:

$x=\dfrac{3}{2}$

Hence, the coordinate of the point is:

$(\dfrac{3}{2},0)$

Hence, the ratio in which the line segment joining the points a 3 -3 and -2 7 is divided by x-axis is:

3:7