Question

A(2, -1), B(3,0) and C(1,k) are the vertices of a right triangle with right angle at B.
Find the value of k.
Find the area of the triangle.

Answer 1

To determine the value of k, we can use the fact that the slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line.

The slope of the line AB is (change in y) / (change in x) = (0 - (-1)) / (3 - 2) = 1/1 = 1.

Therefore, the slope of the line BC is -1/1 = -1.

The line BC passes through points B(3,0) and C(1,k). Using the slope-intercept form of a linear equation (y = mx + b) and substituting the values of B and C, we can find the equation of the line BC.

0 = -1(3) + b
0 = -3 + b
b = 3

The equation of the line BC is y = -x + 3.

Since B is a right angle, the product of the slopes of BA and BC is -1.

The slope of BA is (change in y) / (change in x) = (0 - (-1)) / (3 - 2) = 1.

Thus, 1 * -1 = -1, confirming that the slopes are perpendicular.

To find the value of k, we substitute the x-coordinate of C (1) into the equation of the line BC.

k = -1(1) + 3
k = 2

Therefore, the value of k is 2.

To find the area of the triangle, we can use the formula for the area of a triangle A = (1/2) * base * height.

The base of the triangle is the length of AB, which can be found using the distance formula:

AB = sqrt((3 - 2)^2 + (0 - (-1))^2)
AB = sqrt(1^2 + 1^2)
AB = sqrt(2)

The height of the triangle is the distance from point C to the line AB. We can find this distance using the formula for the distance between a point and a line:

Distance = (|Ax + By + C|) / sqrt(A^2 + B^2)

In this case, we take the equation of the line AB (y = x - 1) and use the point C(1,2) to find the distance:

Distance = (|1(1) + 1(2) + (-1)|) / sqrt(1^2 + 1^2)
Distance = (|1 + 2 - 1|) / sqrt(2)
Distance = 2 / sqrt(2)
Distance = sqrt(2)

Now, we can calculate the area of the triangle:

Area = (1/2) * AB * Distance
Area = (1/2) * sqrt(2) * sqrt(2)
Area = 1

Therefore, the area of the triangle is 1 square unit.