MATH SOLVE

3 months ago

Q:
# 1)Write an equation in point-slope form for the line through (5, 7) and (6, 8). Please show your work! (Hint: Find the slope. Then, use y - y1 = m (x - x1) )2)Write an equation in point-slope form for the line through (-3, 6) and (3, -6). Please show your work!(Hint: Find the slope. Then, use y - y1 = m (x - x1) )

Accepted Solution

A:

β Straight Lines β

Heya !

[tex]1) \: line \: is \: passing \: through \:points \: (5,7) \: and \: (6,8) \\ \\ slope \: of \: line \: = \frac{(y2 - y1)}{(x2 - x1)} \\ \\ slope \: = \: \frac{(8 - 7)}{(6 - 5)} = \: 1 \\ \\ equation \: of \: line \: = \: (y - y1) \: = m (x - x1) \\ \\ equation \: of \: line \: =( y - 7) = 1(x - 5) \\ \\ equation \: of \: line \: - \\ x - y + 2 = 0 \: ans. \\ \\ \\ 2) \: line \: is \: passing \: through \: ( - 3,6) \: and \: ( 3, - 6) \\ \\ slope \: of \: line \: = \frac{(y2 - y1)}{(x2 - x1)} \\ \\ slope \: of \: line \: = \frac{( - 6 - 6)}{(3 - ( - 3))} = - 2 \\ \\ equation \: of \: line = (y - y1) = m(x - x1) \\ \\ equation \: of \: line \: = (y - 6) = - 2(x + 3) \\ \\ equation \: of \: line \: - \\ 2x + y = 0 \: ans. \\ \\ hope \: it \: helps \: you :)[/tex]

βββββββββββββββββββββββ

Heya !

[tex]1) \: line \: is \: passing \: through \:points \: (5,7) \: and \: (6,8) \\ \\ slope \: of \: line \: = \frac{(y2 - y1)}{(x2 - x1)} \\ \\ slope \: = \: \frac{(8 - 7)}{(6 - 5)} = \: 1 \\ \\ equation \: of \: line \: = \: (y - y1) \: = m (x - x1) \\ \\ equation \: of \: line \: =( y - 7) = 1(x - 5) \\ \\ equation \: of \: line \: - \\ x - y + 2 = 0 \: ans. \\ \\ \\ 2) \: line \: is \: passing \: through \: ( - 3,6) \: and \: ( 3, - 6) \\ \\ slope \: of \: line \: = \frac{(y2 - y1)}{(x2 - x1)} \\ \\ slope \: of \: line \: = \frac{( - 6 - 6)}{(3 - ( - 3))} = - 2 \\ \\ equation \: of \: line = (y - y1) = m(x - x1) \\ \\ equation \: of \: line \: = (y - 6) = - 2(x + 3) \\ \\ equation \: of \: line \: - \\ 2x + y = 0 \: ans. \\ \\ hope \: it \: helps \: you :)[/tex]

βββββββββββββββββββββββ