Question

7) Show that the points (a+5, a-4), (a-2, a+3), and (a,a) do not lie on a
straight line for any value of a.

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume the given coordinates as

Coordinatesof A (a+5, a-4)

Coordinates of B (a-2, a+3)

Coordinates of C (a,a)

Now,

In order to show that points are not collinear, its enough to show that

Slope of AB is not equals to slope of BC.

We know,

Slope of line joining the points ( a, b ) and ( c, d ) is represented as m and given by

$\boxed{ \quad \bf{ \:m \: = \: \dfrac{d \: - \: b}{c \: - \: a} \quad}}$

So,

Slope of AB joining the points (a+5, a-4) and (a-2, a+3)

$\rm :\longmapsto\:Slope \: of \: AB = \dfrac{a + 3 - a + 4}{a - 2 - a - 5}$

$\rm :\longmapsto\:Slope \: of \: AB = \dfrac{7}{ - 7}$

$\bf :\longmapsto\:Slope \: of \: AB = - 1$

Now,

Slope of BC joining the points (a-2, a+3) and (a,a).

$\rm :\longmapsto\:Slope \: of \: BC = \dfrac{a - a - 3}{a - a + 2}$

$\rm :\longmapsto\:Slope \: of \: BC = \dfrac{ - 3}{2}$

$\bf :\longmapsto\:Slope \: of \: BC = - \: \dfrac{3}{2}$

So,

$\bf\implies \:Slope \: of \: AB \: \ne \: Slope \: of \: BC$

Hence, Points A, B, C are not collinear.

So,

It means A, B, C donot lie on a line for any real value of a.

$\red{ \bf \: Remark}$

The other method to show that points don't lie on a line.

1. Show that area of triangle ABC = 0

2. Using line method. Find the equation of line segment joining AB and show that C don't lie on AB.

Additional Information :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of the line passes through h, k) which is parallel to the x-axis is y = k.

and Equation of line which is parallel to y-axis is x = h

2. Point-slope form

Consider a line whose slope is m and passes through the point ( a, b ), then equation of line is given by y - b = m(x - a)

3. Slope-intercept form

Consider a line whose slope is m which cuts the y-axis at a distance ‘a’ from the origin then equation of line is given by y = mx + a.

4. Intercept Form of Line

Consider a line having x– intercept a and y– intercept b, then the equation of line is x/a + y/b = 1.