Question

A(1,2), B(2,3) and C(-2,5) are vertices of A ABC. Find the slope of the altitude drawn
from A.

​

Answer 1

Answer:

2

Step-by-step explanation:

A line from 'A' must be perpendicular to BC, in order to be 'altitude'.

     Therefore, altitude and BC are perpendicular to each other.

If two lines are perpendicular to each other, then the product of their slopes is -1.

   ∴ slope(BC) * slope(altitude) = -1

Using m = (y₂ - y₁)/(x₂ - x₁)

    Slope(BC) = (5 - 3)/(-2 -2)

                      = 2/(-4)

                      = -1/2

Hence,

      Slope(BC)*slope(altitude) =-1

      (-1/2) * slope(altitude) = -1

       slope(altitude) = 2

∴ slope of the altitude drawn  from A is 2.

Answer 2

Given:-

• A(1,2), B(2,3) and C(-2,5) are vertices of A ABC.

To Find:-

• The slope altitude drawn from A.

Solution:-

• We know that if two lines are perpendicular to each other then the product of their slopes is -1

By using m, $\; \; \large\sf \to\frac{y_2-y_1}{x_2-x_1}$

Slope (BC) $\large \sf \: \: \: \: \to \frac{(5-3)}{(-2-2)}$

$\large\qquad \qquad\quad\to \sf \frac{(5-3)}{(-2-2)}$

$\large \qquad\qquad \quad\to\sf \frac{(2)}{(-4)} = \sf \cancel \frac{ - 2}{ - 4} = \frac{ - 1}{ \: \: \: 2}$

Hence,

➪Slope(BC) × Slope(altitude) = -1

➪(-1/2) × slope(altitude) = -1

➪Slope(altitude) = 2

Hence the altitude drawn from A is 2.

$\underline{\rule{200pt}{2pt}}$