Home » » Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between t...

# Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between t...

### Question

Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.

### Options

A) $$\begin{pmatrix} 3 & -3 \\ 8 & 2 \end{pmatrix}$$

B) $$\begin{pmatrix} 3 & 3 \\ 8 & 2 \end{pmatrix}$$

C) $$\begin{pmatrix} -2 & 2 \\ 2 & 2 \end{pmatrix}$$

D) $$\begin{pmatrix} -2 & 3 \\ 3 & 2 \end{pmatrix}$$

### Explanation:

y - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, 3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;
$$[\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]$$
$$[\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]$$
$$[\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]$$
$$[\frac{3}{8}, \frac{-3}{2}]$$

## Dicussion (1)

• y - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, 3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;
$$[\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)]$$
$$[\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)]$$
$$[\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)]$$
$$[\frac{3}{8}, \frac{-3}{2}]$$