Find the equations of the altitudes of a ΔABC whose vertices are A (1, 4), B (-3, 2) and C (-5, -3).

Asked by Sakshi | 1 year ago |  50

##### Solution :-

Given:

The vertices of ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).

Now let us find the slopes of ∆ABC.

Slope of AB = $$\dfrac{(2 – 4) }{ (-3-1)}$$

$$\dfrac{1}{2}$$

Slope of BC = $$\dfrac{(-3 – 2) }{ (-5+3)}$$

$$\dfrac{5}{2}$$

Slope of CA =$$\dfrac{(4 + 3) }{ (1 + 5)}$$

$$\dfrac{7}{6}$$

Thus, we have:

Slope of CF = -2

Slope of AD = $$\dfrac{-2}{5}$$

Slope of BE = $$\dfrac{-6}{7}$$

Hence,

Equation of CF is:

y + 3 = -2(x + 5)

y + 3 = -2x – 10

2x + y + 13 = 0

y – 4 = ($$\dfrac{-2}{5}$$) (x – 1)

5y – 20 = -2x + 2

2x + 5y – 22 = 0

Equation of BE is:

y – 2 = ($$\dfrac{-6}{7}$$) (x + 3)

7y – 14 = -6x – 18

6x + 7y + 4 = 0

The required equations are 2x + y + 13 = 0, 2x + 5y – 22 = 0, 6x + 7y + 4 = 0.

Answered by Aaryan | 1 year ago

### Related Questions

#### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

#### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

#### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.