What is the equation of the line passing through $(88,93)$ and $(-120,3)$ ?

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Answer 1 · 2018-03-03T11:34:06

The equation of the line is $45x-104y= -5712$

The slope of the line passing through $(88,93) and (-120,3)$

is $m= (y_2-y_1)/(x_2-x_1)= (3-93)/(-120-88)= 90/208=45/104$

Let the equation of the line in slope-intercept form be $y=mx+c$

$:. y=45/104x+c$ . The point $(88,93)$ will satisfy the equation .

, $:. 93= 45/104*88+c or 104*93=45*88+104c$ or

$104c=104*93-45*88or c=(104*93-45*88)/104$ or

$c= 5712/104=1428/26=714/13$

Hence the equation of the line is $y= 45/104x+714/13$ or

$104y = 45x+5712 or 45x-104y= -5712$ [Ans]

What is the equation of the line passing through (88,93)(88,93) and (-120,3)(-120,3)?