Question

PLEASE HELP WITH MATH GRADIENTS I WILL GIVE BRAINLIEST.

Answer 1

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

Answer 2

Answer: 12

Step-by-step explanation: do not lie to me you better mark me a brainly or going to report you