Answer to Question #133926 in Analytic Geometry for katie jones

Question

Line L has the equation 4y-6x=33. Line M goes through the point A (5,6) and the point B (-4,k). Line L is perpendicular to M. Work out the value of k.

Answer 1

So the straight line is given by the equation: $y = k * x + b$

Let write line L as: $y_1(x) = k_1 * x + b_1$

Line L: $4 * y - 6 * x = 33$

Let write line L in general form: $y = \dfrac{33 + 6 * x}{4} = 1.5 * x + 8.25$

From L line's equation: $k_1 = 1.5$ , $b_1 = 8.25$

Let write line M as: $y_2(x) = k_2 * x + b_2$

Perpendicularity condition of two lines is: $k_2 = - \dfrac{1}{k_1}$

$k_2 = -\dfrac{1}{1.5} = -\dfrac{2}{3}$

Substitute the numbers in M line's equation:

$M: y_2(x) = -\dfrac{2}{3} * x + b_2$

Let put in this equation values from point $A(x_a = 5, y_a = 6)$

$M: 6 = -\dfrac{2}{3} * 5 + b_2$

$b_2=6 + \dfrac{2}{3} * 5=\dfrac{28}{3}$

Substitute $b_2$ in M line's equation:

$M: y_2(x) = -\dfrac{2}{3} * x + \dfrac{28}{3}$

Let put in this equation values from point $B(x_b = -4, y_b = k)$

$M: k = -\dfrac{2}{3} * (-4) + \dfrac{28}{3} = \dfrac{8+28}{3}=\dfrac{36}{3}=12$

$k = 12$