Three points have coordinates A(-8, 6), B(4, 2) and C(-1, 7). The line through C perpendicular to AB intersects AB at the point P. Find the equations of the line AB and CP.

To find the equation of a line, we need; the gradient of that line and a point on that line. To find the gradient of a line, we require two points on the line which have been provided in the question. Equation of the line ABUsing points A & B, the gradient(m) of line AB is: m = 6 - 2 / (-8) - 4 m = -1/3 Using this gradient and one point (either A or B) the equation of a line can be found.Using y = mx + c method andUsing point A:where y = 6, x = -8 and m = -1/36 = -1/3(-8) + c6 = 8/3 + cc = 6 - 8/3c = 10/ 3 therefore the equation of the line of AB : ANS: y = -x/3 + 10/3 or simplified to 3y = -x + 10 The equation of a line of CPThe same two conditions still apply, however it should be known, in addition, that the gradient of a line (CP) perpendicular to a line (AB) has the negative reciprocal of that lines gradient. In other words, when the two gradients of perpendicular lines are multiplied the answer should always be -1.
This implies that the gradient of line CP is 3.We know the point from the question so using a different formula [just for knowing purposes (NB: consistency in formulas is usually better)] : y - y₁ = m (x - x₁)where y₁ = 7 and x₁= -1 y - 7 = 3 (x - (-1))y - 7 = 3x + 3 ANS: y = 3x + 10