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Sagot :
To determine the point on the line that is perpendicular to the given line and passing through a given point on the [tex]\( y \)[/tex]-axis, we should first identify the characteristics of the given line and then apply these characteristics to find the required point.
1. Identify the given line: The given line passes through the points [tex]\((-2, 0)\)[/tex] and [tex]\((0, -2)\)[/tex].
2. Calculate the slope of the given line:
- The slope formula is given by:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- Plugging in the coordinates of the given points:
[tex]\[ \text{slope} = \frac{-2 - 0}{0 - (-2)} = \frac{-2}{2} = -1 \][/tex]
3. Determine the slope of the line perpendicular to the given line:
- The slope of a line perpendicular to another line is the negative reciprocal of the original slope.
- The negative reciprocal of [tex]\(-1\)[/tex] is [tex]\(1\)[/tex].
4. Find the equation of the line perpendicular to the given line that passes through a point on the [tex]\( y \)[/tex]-axis:
- Since this perpendicular line passes through the [tex]\( y \)[/tex]-axis, its [tex]\( x \)[/tex]-coordinate is [tex]\( 0 \)[/tex], making the point [tex]\((0, y)\)[/tex].
- The general equation of a line with a slope [tex]\( m \)[/tex] passing through the point [tex]\((x_1, y_1)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
- Substituting [tex]\( m = 1 \)[/tex] and [tex]\( (x_1, y_1) = (0, y) \)[/tex]:
[tex]\[ y - y = 1(x - 0) \implies y = x + y \][/tex]
5. Check the points provided as options to determine which point lies on both the perpendicular line and the [tex]\( y \)[/tex]-axis:
- Points given: [tex]\((-3.6, 0)\)[/tex], [tex]\((-2, 0)\)[/tex], [tex]\((0, -3.6)\)[/tex], [tex]\((0, -2)\)[/tex].
6. Identify the point on the [tex]\( y \)[/tex]-axis that also satisfies the equation of the line perpendicular to the given line:
- Clearly, among the given options, the point [tex]\((0, -2)\)[/tex] is on the [tex]\( y \)[/tex]-axis and is also where the perpendicular line intersects the [tex]\( y \)[/tex]-axis.
Therefore, the point on the line perpendicular to the given line, passing through the given point on the [tex]\( y \)[/tex]-axis, is:
[tex]\[ (0, -2) \][/tex]
1. Identify the given line: The given line passes through the points [tex]\((-2, 0)\)[/tex] and [tex]\((0, -2)\)[/tex].
2. Calculate the slope of the given line:
- The slope formula is given by:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- Plugging in the coordinates of the given points:
[tex]\[ \text{slope} = \frac{-2 - 0}{0 - (-2)} = \frac{-2}{2} = -1 \][/tex]
3. Determine the slope of the line perpendicular to the given line:
- The slope of a line perpendicular to another line is the negative reciprocal of the original slope.
- The negative reciprocal of [tex]\(-1\)[/tex] is [tex]\(1\)[/tex].
4. Find the equation of the line perpendicular to the given line that passes through a point on the [tex]\( y \)[/tex]-axis:
- Since this perpendicular line passes through the [tex]\( y \)[/tex]-axis, its [tex]\( x \)[/tex]-coordinate is [tex]\( 0 \)[/tex], making the point [tex]\((0, y)\)[/tex].
- The general equation of a line with a slope [tex]\( m \)[/tex] passing through the point [tex]\((x_1, y_1)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
- Substituting [tex]\( m = 1 \)[/tex] and [tex]\( (x_1, y_1) = (0, y) \)[/tex]:
[tex]\[ y - y = 1(x - 0) \implies y = x + y \][/tex]
5. Check the points provided as options to determine which point lies on both the perpendicular line and the [tex]\( y \)[/tex]-axis:
- Points given: [tex]\((-3.6, 0)\)[/tex], [tex]\((-2, 0)\)[/tex], [tex]\((0, -3.6)\)[/tex], [tex]\((0, -2)\)[/tex].
6. Identify the point on the [tex]\( y \)[/tex]-axis that also satisfies the equation of the line perpendicular to the given line:
- Clearly, among the given options, the point [tex]\((0, -2)\)[/tex] is on the [tex]\( y \)[/tex]-axis and is also where the perpendicular line intersects the [tex]\( y \)[/tex]-axis.
Therefore, the point on the line perpendicular to the given line, passing through the given point on the [tex]\( y \)[/tex]-axis, is:
[tex]\[ (0, -2) \][/tex]
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