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Find an equation for the line in slope-intercept form

a)The line that passes through the origin and is parallel to the line containing (2,4) and (4,-4) b)The line that passes through the point (1,7) and is perpendicular to the line x-3y+16=0

To find the equation of line we need two things 1)the point from where line is passes 2)slope of line somehow a) line passes from origin(0,0) given point (2,4) and (4,-4) m=y2-y1 x2-x1 m=-4-4 ⇒-8/2⇒-4 4-2 equation of line is given by m(x-x1)=(y-y1) as x=y=0 -4(x)=y y=-4x b) x-3y+16=0 slope of line=1/3 slope of perpendicular line will be=-1/slope of line=-3=m the point from which line is passes (1,7) equation of line (x-x1)m=(y-y1) (x-1)-3=(y-7) -3x+3=y-7 -3x-y+3+7=0 -3x-y+10=0 3x+y-10=0

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Equation of a straight line in slope intercept form. y = m x + c

slope = m = slope of line passing thru two points = = (y2 - y1) / (x2- x1) = (-4 -4) /(4 -2 ) = -4 the line passes thru origin, so substitute (0, 0) in y = m x + c hence c = 0. so equation: y = - 4 x or y + 4 x = 0 =================== b) slope of the line x - 3 y + 16 = 0 is 1/3 slope of the line perpendicular to it: -3 as product of slopes is -1.

let the line be: y = - 3 x + c it passes thru (1, 7) => 7 = - 3 * 1 + c => c = 10 so the line is : y = -3 x + 10 or, y + 3x - 10 = 0