Question

answer it. class 11 . ch straight line . question no. 4 ​

Answer 1

Answer:

Step-by-step explanation:

Given, equation of the straight line

$\tt{y=mx+c}$

It passes through (-2,3) and (4,-3)

So,

$\sf{3=m(-2)+c}$

And,

$\sf{-3=m(4)+c}$

These equations may be rewritten as,

$\sf{2m-c+3=0\,\,\,\,\,\,\,\,\,...(1)}$

$\sf{4m+c+3=0\,\,\,\,\,\,\,\,\,\,...(2)}$

Add (1) and (2),

$\sf{6m+6=0}$

$\sf{\implies\,\boxed{m=-1}}$

Put this value in (1),

$\sf{2(-1)-c+3=0}$

$\sf{\implies-2+3=c}$

$\sf{\implies\,\boxed{c=1}}$