All categories

3 years ago

what is the equation of the line perpendicular to the line y=4/9x-2 that passes through the point (4,3)?

1 answer:

8 0

Y = 4/9x - 2....the slope here is 4/9, but for a perpendicular line, we need the negative reciprocal. All that means is " flip " the slope and change the sign.
So the slope we need to use is -9/4

y = mx + b
slope(m) = - 9/4
(4,3)...x = 4 and y = 3
now we sub and solve for b, the y int
3 = -9/4(4) + b
3 = - 9 + b
3 + 9 = b
12 = b

so ur perpendicular line is :y = -9/4x + 12

You might be interested in

Divide. express your answer in simplest form. 3 5/6 ÷ 3 3/4

Harlamova29_29 [7]

Answer:

$\frac{46}{45}$

Step-by-step explanation:

3 5/6 ÷ 3 3/4 = 46/45

3 0

2 years ago

This is adding and subtracting rational expressions. i don’t know how to do this one

Sindrei [870]

Answer:

Step-by-step explanation:

1-0=1 totally trust me

5 0

2 years ago

Simplify and show your work please. 3x+6y+x+3(2x+3y

kumpel [21]

Look at the attached picture⤴

Hope it will help u...

7 0

3 years ago

Read 2 more answers

Sarah purchased a car that was on sale for 3/4 of the original price .if the original price was 6000, what was the price that Sa

umka21 [38]

So you do 3/4 divided by 6000 which make the thing switches and so you make 4/3 times by 6000 equals to your answers so you gotten 24,000 dividend it by 3 so you get will get.

Answer: 8,000

4 0

3 years ago

Part B

alisha [4.7K]

Question:

Consider ΔABC, whose vertices are A (2, 1), B (3, 3), and C (1, 6); let the line segment AC represent the base of the triangle.

(a) Find the equation of the line passing through B and perpendicular to the line AC

(b) Let the point of intersection of line AC with the line you found in part A be point D. Find the coordinates of point D.

Answer:

$y = \frac{1}{5}x + \frac{12}{5}$

$D = (\frac{43}{26},\frac{71}{26})$

Step-by-step explanation:

Given

$\triangle ABC$

$A = (2,1)$

$B = (3,3)$

$C = (1,6)$

Solving (a): Line that passes through B, perpendicular to AC.

First, calculate the slope of AC

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$C = (1,6)$ --- $(x_2,y_2)$

The slope is:

$m = \frac{6- 1}{1 - 2}$

$m = \frac{5}{-1}$

$m = -5$

The slope of the line that passes through B is calculated as:

$m_2 = -\frac{1}{m}$ --- because it is perpendicular to AC.

So, we have:

$m_2 = -\frac{1}{-5}$

$m_2 = \frac{1}{5}$

The equation of the line is the calculated using:

$m_2 = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

$m_2 = \frac{1}{5}$

$B = (3,3)$ --- $(x_1,y_1)$

$(x_2,y_2) = (x,y)$

So, we have:

$\frac{1}{5} = \frac{y - 3}{x - 3}$

Cross multiply

$5(y-3) = 1(x - 3)$

$5y - 15 = x - 3$

$5y = x - 3 + 15$

$5y = x +12$

Make y the subject

$y = \frac{1}{5}x + \frac{12}{5}$

Solving (b): Point of intersection between AC and $y = \frac{1}{5}x + \frac{12}{5}$

First, calculate the equation of AC using:

$y = m(x - x_1) + y_1$

Where:

$A = (2,1)$ --- $(x_1,y_1)$

$m = -5$

So:

$y=-5(x - 2) + 1$

$y=-5x + 10 + 1$

$y=-5x + 11$

So, we have:

$y=-5x + 11$ and $y = \frac{1}{5}x + \frac{12}{5}$

Equate both to solve for x

i.e.

$y = y$

$-5x + 11 = \frac{1}{5}x + \frac{12}{5}$

Collect like terms

$-5x -\frac{1}{5}x = \frac{12}{5} - 11$

Multiply through by 5

$-25x-x = 12 - 55$

Collect like terms

$-26x = -43$

Solve for x

$x = \frac{-43}{-26}$

$x = \frac{43}{26}$

Substitute $x = \frac{43}{26}$ in $y=-5x + 11$

$y = -5 * \frac{43}{26} + 11$

$y = \frac{-5 *43}{26} + 11$

Take LCM

$y = \frac{-5 *43+11 * 26}{26}$

$y = \frac{71}{26}$

Hence, the coordinates of D is:

$D = (\frac{43}{26},\frac{71}{26})$

3 0

2 years ago