Answer :
To determine which point is on the line that is perpendicular to line MN and passes through point K, we need to follow these steps.
1. Identify the Slope of Line MN:
Assume that line MN passes through points [tex]\( M \)[/tex] and [tex]\( N \)[/tex], with coordinates [tex]\( M(x_1, y_1) \)[/tex] and [tex]\( N(x_2, y_2) \)[/tex]. For simplicity, let's use example points [tex]\( M(1, 1) \)[/tex] and [tex]\( N(2, 3) \)[/tex].
The slope [tex]\( m_{MN} \)[/tex] of line MN is calculated using the formula:
[tex]\[ m_{MN} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Using the example points, the slope is:
[tex]\[ m_{MN} = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \][/tex]
2. Determine the Slope of the Perpendicular Line:
The slope of the line perpendicular to MN, [tex]\( m_{perpendicular} \)[/tex], is the negative reciprocal of [tex]\( m_{MN} \)[/tex]:
[tex]\[ m_{perpendicular} = -\frac{1}{m_{MN}} = -\frac{1}{2} \][/tex]
3. Use the Point Through Which Perpendicular Line Passes:
Let's assume point K through which the perpendicular line passes is [tex]\( K(x_3, y_3) \)[/tex]. Using the given example, [tex]\( K \)[/tex] is [tex]\( (2,2) \)[/tex].
4. Form the Equation of the Perpendicular Line:
Using the slope-point form of the equation of a line, [tex]\( y - y_3 = m_{perpendicular} \cdot (x - x_3) \)[/tex]:
[tex]\[ y - 2 = -\frac{1}{2}(x - 2) \][/tex]
Simplifying the equation:
[tex]\[ y - 2 = -\frac{1}{2}x + 1 \][/tex]
[tex]\[ y = -\frac{1}{2}x + 1 + 2 \][/tex]
[tex]\[ y = -\frac{1}{2}x + 3 \][/tex]
5. Check Each Given Point:
Verify which of the given points satisfy the line equation [tex]\( y = -\frac{1}{2}x + 3 \)[/tex].
- For point [tex]\((0, -12)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(0) + 3 = 3 \quad \text{not } -12 \][/tex]
- For point [tex]\((2, 2)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(2) + 3 = -1 + 3 = 2 \quad \text{(Valid Point)} \][/tex]
- For point [tex]\((4,8)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(4) + 3 = -2 + 3 = 1 \quad \text{not } 8 \][/tex]
- For point [tex]\((5,13)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(5) + 3 = -2.5 + 3 = 0.5 \quad \text{not } 13 \][/tex]
Hence, the point [tex]\((2, 2)\)[/tex] is on the line that is perpendicular to MN and passes through point K.
1. Identify the Slope of Line MN:
Assume that line MN passes through points [tex]\( M \)[/tex] and [tex]\( N \)[/tex], with coordinates [tex]\( M(x_1, y_1) \)[/tex] and [tex]\( N(x_2, y_2) \)[/tex]. For simplicity, let's use example points [tex]\( M(1, 1) \)[/tex] and [tex]\( N(2, 3) \)[/tex].
The slope [tex]\( m_{MN} \)[/tex] of line MN is calculated using the formula:
[tex]\[ m_{MN} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Using the example points, the slope is:
[tex]\[ m_{MN} = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \][/tex]
2. Determine the Slope of the Perpendicular Line:
The slope of the line perpendicular to MN, [tex]\( m_{perpendicular} \)[/tex], is the negative reciprocal of [tex]\( m_{MN} \)[/tex]:
[tex]\[ m_{perpendicular} = -\frac{1}{m_{MN}} = -\frac{1}{2} \][/tex]
3. Use the Point Through Which Perpendicular Line Passes:
Let's assume point K through which the perpendicular line passes is [tex]\( K(x_3, y_3) \)[/tex]. Using the given example, [tex]\( K \)[/tex] is [tex]\( (2,2) \)[/tex].
4. Form the Equation of the Perpendicular Line:
Using the slope-point form of the equation of a line, [tex]\( y - y_3 = m_{perpendicular} \cdot (x - x_3) \)[/tex]:
[tex]\[ y - 2 = -\frac{1}{2}(x - 2) \][/tex]
Simplifying the equation:
[tex]\[ y - 2 = -\frac{1}{2}x + 1 \][/tex]
[tex]\[ y = -\frac{1}{2}x + 1 + 2 \][/tex]
[tex]\[ y = -\frac{1}{2}x + 3 \][/tex]
5. Check Each Given Point:
Verify which of the given points satisfy the line equation [tex]\( y = -\frac{1}{2}x + 3 \)[/tex].
- For point [tex]\((0, -12)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(0) + 3 = 3 \quad \text{not } -12 \][/tex]
- For point [tex]\((2, 2)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(2) + 3 = -1 + 3 = 2 \quad \text{(Valid Point)} \][/tex]
- For point [tex]\((4,8)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(4) + 3 = -2 + 3 = 1 \quad \text{not } 8 \][/tex]
- For point [tex]\((5,13)\)[/tex]:
[tex]\[ y = -\frac{1}{2}(5) + 3 = -2.5 + 3 = 0.5 \quad \text{not } 13 \][/tex]
Hence, the point [tex]\((2, 2)\)[/tex] is on the line that is perpendicular to MN and passes through point K.
