Question

Three points P (h, k), Q(x, y,) and R (X,, y,) lie on a line. Show that
(h – x,) (2 - y) = (k – y,) (x, - x,).
​

Answer 1

$\bigstar \underline{\underline{\mathfrak{Given:-}}}$

• Three points  P(h,k);Q(x1,y1);R(x2,y2) lie on a line.

$\bigstar \underline{\underline{\mathfrak{To\ Prove:-}}}$

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

$\bigstar \underline{\underline{\mathfrak{Proof:-}}}$

We need to know :

--------------------------------

❥ If 3 points are collinear ,then the slope of any 2 line segments should be same.

❥ Slope of a line passing through (x1,y1)&(x2,y2) is,

$\mapsto \boxed{\sf Slope(m)=\frac{y_2-y_1}{x_2-x_1}}$

---------------------------------

Acc to question,

↳ Slope of QP = Slope of RQ

↬ Slope of QP :-

Line is passing through Q(x1,y1) & P(h,k).

Here,

=> x1 =h &y1=k

=> x2=x1&y2=k

Substituiting these values in the formula we get,

$:\mapsto \sf m=\frac{y_1-k}{x_1-h}$

-------------------------

↬Slope of RQ:-

Line is passing throughR(x2,y2)&Q(x1,y1)

Here,

=> x1 =x1 & y1 = y1

=> x2 = x2& y2=y2

Substituiting the values in the formula we get,

$:\mapsto \sf m=\frac{y_2-y_1 }{x_2-x_1}$

-----------------------

Now,

As Slope of PQ = Slope of RQ ,

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

Crossmultiplying,

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

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

Hence proved !

----------------------------

Answer 2

$\bigstar \underline{\underline{\mathfrak{Given:-}}}$

Three points  P(h,k);Q(x1,y1);R(x2,y2) lie on a line.

$\bigstar \underline{\underline{\mathfrak{To\ Prove:-}}}$

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

$\bigstar \underline{\underline{\mathfrak{Proof:-}}}$

We need to know :

--------------------------------

❥ If 3 points are collinear ,then the slope of any 2 line segments should be same.

❥ Slope of a line passing through (x1,y1)&(x2,y2) is,

$\mapsto \boxed{\sf Slope(m)=\frac{y_2-y_1}{x_2-x_1}}$

---------------------------------

Acc to question,

↳ Slope of QP = Slope of RQ

↬ Slope of QP :-

Line is passing through Q(x1,y1) & P(h,k).

Here,

=> x1 =h &y1=k

=> x2=x1&y2=k

Substituiting these values in the formula we get,

$:\mapsto \sf m=\frac{y_1-k}{x_1-h}$

-------------------------

↬Slope of RQ:-

Line is passing throughR(x2,y2)&Q(x1,y1)

Here,

=> x1 =x1 & y1 = y1

=> x2 = x2& y2=y2

Substituiting the values in the formula we get,

$:\mapsto \sf m=\frac{y_2-y_1 }{x_2-x_1}$

-----------------------

Now,

As Slope of PQ = Slope of RQ ,

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

Crossmultiplying,

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

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

Hence proved !

----------------------------