Answer:
12
Step-by-step explanation: So the straight line is given by the equation: y = k * x + by=k∗x+b
Let write line L as: y_1(x) = k_1 * x + b_1y
1
(x)=k
1
∗x+b
1
Line L: 4 * y - 6 * x = 334∗y−6∗x=33
Let write line L in general form: y = \dfrac{33 + 6 * x}{4} = 1.5 * x + 8.25y=
4
33+6∗x
=1.5∗x+8.25
From L line's equation: k_1 = 1.5k
1
=1.5 , b_1 = 8.25b
1
=8.25
Let write line M as: y_2(x) = k_2 * x + b_2y
2
(x)=k
2
∗x+b
2
Perpendicularity condition of two lines is: k_2 = - \dfrac{1}{k_1}k
2
=−
k
1
1
k_2 = -\dfrac{1}{1.5} = -\dfrac{2}{3}k
2
=−
1.5
1
=−
3
2
Substitute the numbers in M line's equation:
M: y_2(x) = -\dfrac{2}{3} * x + b_2M:y
2
(x)=−
3
2
∗x+b
2
Let put in this equation values from point A(x_a = 5, y_a = 6)A(x
a
=5,y
a
=6)
M: 6 = -\dfrac{2}{3} * 5 + b_2M:6=−
3
2
∗5+b
2
b_2=6 + \dfrac{2}{3} * 5=\dfrac{28}{3}b
2
=6+
3
2
∗5=
3
28
Substitute b_2b
2
in M line's equation:
M: y_2(x) = -\dfrac{2}{3} * x + \dfrac{28}{3}M:y
2
(x)=−
3
2
∗x+
3
28
Let put in this equation values from point B(x_b = -4, y_b = k)B(x
b
=−4,y
b
=k)
M: k = -\dfrac{2}{3} * (-4) + \dfrac{28}{3} = \dfrac{8+28}{3}=\dfrac{36}{3}=12M:k=−
3
2
∗(−4)+
3
28
=
3
8+28
=
3
36
=12
k = 12k=12
