Question

Hey Mates!


( 6,6 ) ( h,0 ) ( 0, k ) are collinear then find 1/h + 1/k


Disclaimer - No irrelevant answers.

Answer 1

Let the given points be A(6,6), B(h,0) and C(0,k).

Here x1 = 6 and y1 = 6, 

         x2 = h and y2 = 0

         x3 = 0 and y3 = k.


We know that Slope a line m = y2 - y1/x2 - x1.

Slope of AB:

(x1,y1) = (6,6) and (x2,y2) = (h,0).

$= \ \textgreater \ \frac{0 - 6}{h - 6}$

$= \ \textgreater \ \frac{-6}{h - 6}$


Slope of BC:

(x1,y1) = (h,0) and (x2,y2) = (0,k).

$= \ \textgreater \ m = \frac{k - 0}{0 - h}$

$= \ \textgreater \ m = \frac{k}{-h}$

Now,

Slope of AB = slope of BC

$= \ \textgreater \ \frac{-6}{h-6} = \frac{k}{-h}$

$= \ \textgreater \ 6h = kh - 6k$

$= \ \textgreater \ 6h + 6k = kh$

$= \ \textgreater \ 6(h + k) = kh$

$\frac{1}{h} + \frac{1}{k} = \frac{1}{6}$



Hope this helps!

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\therefore let\:the\:givencollinear\:be,$

$\sf\dashrightarrow A(6,6)= x_1=6 , y_1=6$

$\sf\dashrightarrow B(h,0)= x_2=h,y_2=0$

$\sf\dashrightarrow C(0,k)=x_3=0,y_3=k$

✯.FORMULA IN USE ,

$\large{\boxed{\bf{\star\:\:m= \dfrac{y_2-y_1}{x_2-x_1}\:\: \star }}}$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow the\:value\:of\:\dfrac{1}{k} + \dfrac{1}{h}$

$\large\underline\bold{SOLUTION,}$

✯SIDE AB,

$\sf\dashrightarrow A(6,6)= x_1=6 , y_1=6$

$\sf\dashrightarrow B(h,0)= x_2=h,y_2=0$

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

$\sf\implies \dfrac{0 - 6}{h - 6}$

$\sf\implies \dfrac{-6}{h - 6}$

$\rm{\boxed{\bf{\star\:\: \dfrac{-6}{h - 6} \:\: \star }}}$

✯SIDE BC

$\sf\dashrightarrow B(h,0)= x_2=h,y_2=0$

$\sf\dashrightarrow C(0,k)=x_3=0,y_3=k$p

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

$\sf\implies \dfrac{k - 0}{0 - h}$

$\sf\implies \dfrac{k}{-h}$

$\rm{\boxed{\bf{\star\:\: \dfrac{k}{-h}\:\: \star }}}$

✯SIDE AC,

$\sf\dashrightarrow A(6,6)= x_1=6 , y_1=6$

$\sf\dashrightarrow C(0,k)=x_3=0,y_3=k$

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

$\sf\implies \dfrac{0-6}{k-6}$

$\sf\implies \dfrac{-6}{k-6}$

$\rm{\boxed{\bf{\star\:\: \dfrac{-6}{k-6}\:\: \star }}}$

NOW,

SIDE AB = SIDE BC

$\sf\implies \dfrac{-6}{h-6} = \dfrac{k}{-h}$

$\sf\implies (-6) \times (-h) = k \times (h-6)$

$\sf\implies 6h= kh-6k$

$\sf\implies 6h+6k= kh$

$\sf\implies 6(h+k)=kh$

$\sf\implies h+k= \dfrac{kh}{6}$

$\sf\implies h+k= \dfrac{1}{6} \times kh$

$\sf\implies \dfrac{h+k}{kh} = \dfrac{1}{6}$

$\sf\implies \dfrac{h}{kh} + \dfrac{k}{kh}= \dfrac{1}{6}$

$\sf\implies \dfrac{\cancel{h}}{k \cancel{h}} + \dfrac{\cancel{k}}{\cancel{k}h}= \dfrac{1}{6}$

$\sf\implies \dfrac{1}{k} + \dfrac{1}{h} = \dfrac{1}{6}$

$\large{\boxed{\bf{\star\:\: \dfrac{1}{k} + \dfrac{1}{h} = \dfrac{1}{6} \:\: \star }}}$

$\large\underline\bold{the\:value\:of\:\dfrac{1}{k} + \dfrac{1}{h} = \dfrac{1}{6} }$

______________________________