Math, asked by jhalakshah0106, 5 months ago

a straight line passes through point (2,5) and has a slope of -3 find the equation of straight line and the value of K if line passes theough point(k,k+3)

Answers

Answered by Asterinn
7

Given :

  • A straight line passes through point (2,5) and has a slope of -3.

To find :

  • Equation of straight line

  • value of K if line passes theough point (k,k+3)

Concept used :

 \sf equation \: of \: straight \: line \: when \: it \: passes \: through \: (x_1,y_1) \: and \: its \: slope \: is \:  \bf \: m :

 \bf y - y_1 =( x - x_1)m

Solution :

Equation of straight line passes through point (2,5) and has a slope of -3 :-

 \implies\sf y -5 =( x  - 2)( - 3)

\implies\sf y -5 = - 3x   + 6

\implies\sf y  + 3x - 5 - 6 = 0

\implies\sf y  + 3x  - 11 = 0

If line passes through point (k,k+3) then put x = k and y = k+3 in y+3x-11=0. Then you can easily find the value of k.

\implies\sf( k + 3)  + 3k  - 11 = 0

\implies\sf k + 3  + 3k  - 11 = 0

\implies\sf 4k + 3   - 11 = 0

\implies\sf 4k  - 8 = 0

\implies\sf4k = 8

\implies\sf k  =  \dfrac{8}{4}

\implies\sf k  =  \dfrac{ \cancel8 \:  \: 2}{\cancel4 \:  \: 1} = 2

Answer :

  • Equation of straight line = y +3x-11=0

  • k = 2

Answered by Anonymous
2

Given ,

A straight line passes through point (2,5) , (k,k+3) and has slope of -3

As we know that , the slope of the line is given by

 \boxed{ \tt{m =   \frac{ y_{2} -  y_{1}}{ x_{2} -  x_{1}}}}

∴ The equation of the line will be

-3 = (5 - y)/(2 - x)

-6 + 3x = 5 - y

3x + y - 11 = 0

Therefore , the equation of line is 3x + y - 11= 0

Since , (k , k + 3) lies on the line 3x + y - 11 = 0

∴ Put x = k and y = k + 3 in equation of line

3k + (k + 3) - 11 = 0

4k - 8 = 0

4k = 8

k = 2

Therefore , the value of k is 2

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