Question

the points (a, b) and (c, d). line p contains the points (-b, a) and (-d, c). prove: the slopes of perpendicular lines are negative reciprocals. given that the lines are perpendicular and we have shown that the slopes of the lines are negative reciprocals. (d - b)/(c - a) (c - a)/(-d + b)=-1 (b - a)/(a + b) (c - d)/(-d - c)=-1 (d - b)/(c - a) (c - a)/(-d + b)=1 (b - a)/(a + b) (c - d)/(-d - c)=1

Explanation:

Step1: Recall slope - formula

The slope formula for a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). The slope of the line passing through \(P(a,b)\) and \(Q(c,d)\) is \(m_1=\frac{d - b}{c - a}\). The slope of the line passing through \(P'(-b,a)\) and \(Q'(-d,c)\) is \(m_2=\frac{c - a}{-d + b}\).

Step2: Use perpendicular - line property

If two lines are perpendicular, the product of their slopes is \(- 1\). So \(m_1\times m_2=\frac{d - b}{c - a}\times\frac{c - a}{-d + b}=-1\).

Answer:

\((\frac{d - b}{c - a})(\frac{c - a}{-d + b})=-1\)