The given points are A = (-4,-2) B = (4,4) C = (18,-8)
First we'll use points A and C to find the slope of line AC m = (y2-y1)/(x2-x1) m = (-8-(-2))/(18-(-4)) m = (-8+2)/(18+4) m = -6/22 m = -3/11 The slope of AC is -3/11
Take the negative reciprocal of this slope Flip the fraction: -3/11 -----> -11/3 Flip the sign: -11/3 ----> +11/3 = 11/3
The slope of AC is -3/11 while the slope of any line perpendicular to AC is 11/3 Let m = 11/3 and (x,y) = (4,4) which are the coordinates of point B Plug these values into slope intercept form and then solve for b y = mx+b 4 = (11/3)*4+b 4 = 44/3+b 4 - 44/3 = 44/3+b-44/3 b = 4 - 44/3 b = 12/3 - 44/3 b = (12 - 44)/3 b = -32/3
Since m = 11/3 and b = -32/3, we go from this y = mx+b to this y = (11/3)x-32/3
Now clear out the fractions and get the x and y variables to one side y = (11/3)x-32/3 3y = 3*[ (11/3)x-32/3 ] 3y = 11x - 32 3y-11x = 11x-32-11x -11x+3y = -32 -1*(-11x+3y) = -1*(-32) 11x-3y = 32
The equation of the through B that is perpendicular to AC is 11x-3y = 32 (this equation is in Ax+By = C form which is called standard form) This is better known as the altitude through B