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Mars2501 [29]
3 years ago
6

Part B

Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0

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})

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