Question

using section formula show that the points (-1,2) (5,0) and (2,1) are collinear? (pls answer with proper explanation)​

Answer 1

Answer:

A-(-1,2).

B-(5.0).

C-(2,1).

Step-by-step explanation:

AB=square root (-1-5)hole sq+(2-0)hole sq

..........."......"......"...=)(-6)hole sq +4

.........."......".....".....=)6+4

........"......."......".....=)10

"""""""""

Answer 2

Given:

Three points :

• (-1,2)

• (5,0)

• (2,1)

To Find:

To show that the three points are colinear using section formula.

Concept:

To show that three points are colinear , the slope of first and second points should be equal to the slope of second and third points.

Solution:

Let the points be A(-1,2) , B(5,0) And C(2,1) respectively.

To find the slope of two points, we will be using the formula:

m = $\dfrac{y_2-y_1}{x_2-x_1}$

where,

m = slope of line

To prove, colinearity, we will prove slope of AB to be equal to slope of BC.

First , calculating the slope of AB.

In this case,

$x_1 = - 1 \\ y_1 = 2 \: \: \: \: \\ x_2 = 5 \: \: \: \: \\ y_2 = 0 \: \: \: \:$

Substituting the values , we get:

Slope of AB = $\dfrac{0-2}{5-(-1)}$

Slope of AB = -$\dfrac{1}{3}$

Now , we will calculate slope of BC.

In this case,

$x_1 = 5 \\ y_ 1 =0 \\ x_2 =2 \\ y_2 = 1$

Substituting the values , we get:

Slope of BC = $\dfrac{1-0}{2-5}$

Slope of BC = -$\dfrac{1}{3}$

Since,

Slope of AB = Slope of BC

-$\dfrac{1}{3}$ = -$\dfrac{1}{3}$

LHS = RHS

Hence Proved.

Therefore, the three points A(-1,2), B(5,0) and C(2,1) are colinear.

Other Formulas:

1) Slope of Line

• Slope of a non-vertical line passing through points A(x_1,y_1) and B(x_1,y_2) is:

$\:\:\:\:\:\:\:\:\:\:\:\:m=\dfrac{y_2-y_1}{x_2-x_1}$

• If a line makes an angle \thetaθ with the positive side of x-axis, then the slope of line is:

$\:\:\:\:\:\:\:\:\:\:\:\:m = \tan\theta$

2) Equation of a Line

• Equation of a line parallel to x-axis at a distance b is:

$\:\:\:\:\:\:\:\:\:\:\:\:$y = by=b (where b is constant)

• Equation of a line parallel to y-axis at a distance is:

$\:\:\:\:\:\:\:\:\:\:\:\:$x = ax=a (where a is constant)

• Equation of a line having a slope and making an intercept with y-axis is:

$\:\:\:\:\:\:\:\:\:\:\:\:$y = mx+c (where m is the slope and c is the y-intercept made by line)

• Equation of a line when the line is passing through one point and slope is given:

$\:\:\:\:\:\:\:\:\:\:\:\:(y-y_1)=m(x-x_1)$(where $x_1,y_1$ are co-ordinates of point through which line passes and m is the slope).

• Equation of a non-vertical line passing through two points is:

$\:\:\:\:\:\:\:\:\:\:\:\:(y-y_1)=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$ (where $(x_1,y_1)\:and\:(x_2,y_2)$ are co-ordinates of two points through which line passes).

3) Conditions for two lines to be:

• Parallel is that the slope of both lines should ve equal.

$\:\:\:\:\:\:\:\:\:\:\:\:$ Let the slope of first line and second line be $m_1\:$and $m_2$ respectively.

Therefore, the two lines are parallel if $m_1=m_2$

• Perpendicular is that the product of the slopes of the two lines should be equal to -1.

$\:\:\:\:\:\:\:\:\:\:\:\:$ Let the slope of first and second line be $m_1$ and $m_2$ respectively.

Therefore, the two lines are perpendicular if $m_1\times m_2=-1$.