Question

16. If the points (h, k), (x1, y1) and (x2, y2) are collinear, show that: (h-xi) (y2 - yı) = (k - yı) (x2 – x1) .​

Answer 1

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

Given that, the points

$\rm \: (h,k), \: (x_1,y_1), \: (x_2,y_2) \: \: are \: collinear.$

Let assume that

$\rm \:(h,k), \: (x_1,y_1), \: (x_2,y_2) \: be \: represented \: as \: A,B,C.$

We know,

Three points A, B, C are collinear iff Slope of AB = Slope of BC.

And

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

$\boxed{\rm{  \:Slope \: of \: line \: = \: \frac{d - b}{c - a} \: \: }}$

So, we have

$\rm \: Slope \: of \: AB \: = \: Slope \: of \: BC$

$\rm \: \dfrac{x_1 - h}{y_1 - k} = \dfrac{x_2 - x_1}{y_2 - y_1}$

can be further rewritten as

$\rm \: \dfrac{ - (h - x_1)}{ - (k - y_1)} = \dfrac{x_2 - x_1}{y_2 - y_1}$

$\rm \: \dfrac{h - x_1}{k - y_1} = \dfrac{x_2 - x_1}{y_2 - y_1}$

$\rm \: (h - x_1)(y_2 - y_1) = (k - y_1)(x_2 - x_1)$

Hence,

$\rm\implies\boxed{\rm{\rm \: (h - x_1)(y_2 - y_1) = (k - y_1)(x_2 - x_1) \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

1. Slope of a line is defined as tangent of an angle which a line makes with the positive direction of x axis and is denoted by symbol m.

2. Two lines having slope m and M are parallel, iff M = m

3. Two lines having slope m and M are perpendicular, iff Mm = - 1

4. Slope of a line is 0, if its parallel to x axis or itself x axis.

5. Slope of a line is not defined, if its parallel to y axis or it self y axis.

Answer 2

Answer:

Given :-

• The points A(h , k), B(x₁ , y₁) and C(x₂ , y₂) are collinear.

Show That :-

• (h - x₁) (y₂ - y₁) = (k - y₁) (x₂ - x₁).

Solution :-

Given :

$\mapsto \bf A\: , B\: and\: C\: are\: collinear$

So,

$\longrightarrow \sf\bold{\pink{Slope\: of\: AB =\: Slope\: of\: BC}}$

As we know that :

$\clubsuit$ Slope Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{m =\: \dfrac{y_2 - y_1}{x_2 - x_1}}}}\: \: \: \bigstar$

☆ In case of Slope of AB :-

Given Points :

$\mapsto \sf A =\: (h, k)$

$\mapsto \sf B =\: (x_1 , y_1)$

where,

• x₁ = h
• x₂ = x₁
• y₁ = k
• y₂ = y₁

By putting those values we get,

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

$\implies \sf\bold{\purple{m =\: \dfrac{y_1 - k}{x_1 - h}}}$

☆ In case of Slope of BC :-

Given Points :

$\mapsto \sf B =\: (x_1 , y_1)$

$\mapsto \sf C =\: (x_2 , y_2)$

where,

• x₁ = x₁
• x₂ = x₂
• y₁ = y₁
• y₂ = y₂

By putting those values we get,

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

$\implies \sf\bold{\purple{m =\: \dfrac{y_2 - y_1}{x_2 - x_1}}}$

Now,

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

$\implies \sf \dfrac{y_1 - k}{x_1 - h} =\: \dfrac{y_2 - y_1}{x_2 - x_1}$

We can write as,

$\implies \sf \dfrac{\cancel{-}(k - y_1)}{\cancel{-}(h - x_1)} =\: \dfrac{y_2 - y_1}{x_2 - x_1}$

$\implies \sf \dfrac{k - y_1}{h - x_1} =\: \dfrac{y_2 - y_1}{x_2 - x_1}$

By doing cross multiplication we get,

$\implies \sf (h - x_1)(y_2 - y_1) =\: (k - y_1)(x_2 - x_1)$

$\implies \sf\bold{\red{(h - x_1)(y_2 - y_1) =\: (k - y_1)(x_2 - x_1)}}$

$\dashrightarrow \sf\bold{\underline{\blue{HENCE\: PROVED}}}$

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