Answer :
### Question 9
To determine which of the points lie on the line given by the equation [tex]\( y = -5x + 4 \)[/tex], we will substitute the x-values of each point into the equation and verify if the resulting y-value matches the y-value of the point.
#### Point A: [tex]\((-1, 9)\)[/tex]
1. Substitute [tex]\( x = -1 \)[/tex] into the equation [tex]\( y = -5x + 4 \)[/tex]:
[tex]\[ y = -5(-1) + 4 = 5 + 4 = 9 \][/tex]
2. The calculated y-value is 9, which matches the y-coordinate of the point [tex]\((-1, 9)\)[/tex].
Therefore, point [tex]\( (-1, 9) \)[/tex] lies on the line.
#### Point B: [tex]\((0, 4)\)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex] into the equation [tex]\( y = -5x + 4 \)[/tex]:
[tex]\[ y = -5(0) + 4 = 0 + 4 = 4 \][/tex]
2. The calculated y-value is 4, which matches the y-coordinate of the point [tex]\((0, 4)\)[/tex].
Therefore, point [tex]\( (0, 4) \)[/tex] lies on the line.
#### Point C: [tex]\(\left(\frac{2}{5}, 2\right)\)[/tex]
1. Substitute [tex]\( x = \frac{2}{5} \)[/tex] into the equation [tex]\( y = -5x + 4 \)[/tex]:
[tex]\[ y = -5 \left( \frac{2}{5} \right) + 4 = -2 + 4 = 2 \][/tex]
2. The calculated y-value is 2, which matches the y-coordinate of the point [tex]\(\left(\frac{2}{5}, 2\right)\)[/tex].
Therefore, point [tex]\(\left(\frac{2}{5}, 2\right)\)[/tex] lies on the line.
### Question 10
To find the equations of the lines passing through the given points, we first determine the slope and then use one point to find the y-intercept.
#### Part i: Line through [tex]\((1, 1)\)[/tex] and [tex]\((1 + \sqrt{2}, 1 - \sqrt{2})\)[/tex]
1. Calculate the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{(1 - \sqrt{2}) - 1}{(1 + \sqrt{2}) - 1} = \frac{1 - \sqrt{2} - 1}{\sqrt{2}} = - \frac{\sqrt{2}}{\sqrt{2}} = -1 \][/tex]
2. Use point [tex]\((1, 1)\)[/tex] to compute the y-intercept [tex]\( b \)[/tex]:
[tex]\[ y = mx + b \implies 1 = -1(1) + b \implies b = 1 + 1 = 2 \][/tex]
Therefore, the equation of the line is:
[tex]\[ y = -1x + 2 \quad \text{or} \quad y = -x + 2 \][/tex]
#### Part ii: Line through [tex]\(\left(1 \frac{1}{2}, -\frac{5}{2}\right)\)[/tex] and [tex]\(\left(-\frac{3}{2}, 1\right)\)[/tex]
1. Express the points as improper fractions to make calculations easier:
[tex]\[ \left(\frac{3}{2}, -\frac{5}{2} \right) \text{ and } \left(-\frac{3}{2}, 1 \right) \][/tex]
2. Calculate the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{1 - (-\frac{5}{2})}{(-\frac{3}{2}) - \frac{3}{2}} = \frac{1 + \frac{5}{2}}{-3} = \frac{\frac{2}{2} + \frac{5}{2}}{-3} = \frac{\frac{7}{2}}{-3} = \frac{7}{-6} = - \frac{7}{6} \][/tex]
3. Use point [tex]\(\left(\frac{3}{2}, -\frac{5}{2}\right)\)[/tex] to compute the y-intercept [tex]\( b \)[/tex]:
[tex]\[ y = mx + b \implies -\frac{5}{2} = - \frac{7}{6} \left( \frac{3}{2} \right) + b \][/tex]
Simplify:
[tex]\[ -\frac{5}{2} = - \frac{7}{6} \times \frac{3}{2} + b \implies -\frac{5}{2} = - \frac{7 \times 3}{6 \times 2} + b \implies -\frac{5}{2} = - \frac{7}{4} + b \][/tex]
4. Solve for [tex]\( b \)[/tex]:
[tex]\[ b = -\frac{5}{2} + \frac{7}{4} = -\frac{10}{4} + \frac{7}{4} = -\frac{3}{4} \][/tex]
Therefore, the equation of the line is:
[tex]\[ y = - \frac{7}{6} x - \frac{3}{4} \][/tex]
Both forms can be expressed in straightforward linear equations respectively:
1. [tex]\( y = -x + 2 \)[/tex]
2. [tex]\( y = - \frac{7}{6} x - \frac{3}{4} \)[/tex]
To determine which of the points lie on the line given by the equation [tex]\( y = -5x + 4 \)[/tex], we will substitute the x-values of each point into the equation and verify if the resulting y-value matches the y-value of the point.
#### Point A: [tex]\((-1, 9)\)[/tex]
1. Substitute [tex]\( x = -1 \)[/tex] into the equation [tex]\( y = -5x + 4 \)[/tex]:
[tex]\[ y = -5(-1) + 4 = 5 + 4 = 9 \][/tex]
2. The calculated y-value is 9, which matches the y-coordinate of the point [tex]\((-1, 9)\)[/tex].
Therefore, point [tex]\( (-1, 9) \)[/tex] lies on the line.
#### Point B: [tex]\((0, 4)\)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex] into the equation [tex]\( y = -5x + 4 \)[/tex]:
[tex]\[ y = -5(0) + 4 = 0 + 4 = 4 \][/tex]
2. The calculated y-value is 4, which matches the y-coordinate of the point [tex]\((0, 4)\)[/tex].
Therefore, point [tex]\( (0, 4) \)[/tex] lies on the line.
#### Point C: [tex]\(\left(\frac{2}{5}, 2\right)\)[/tex]
1. Substitute [tex]\( x = \frac{2}{5} \)[/tex] into the equation [tex]\( y = -5x + 4 \)[/tex]:
[tex]\[ y = -5 \left( \frac{2}{5} \right) + 4 = -2 + 4 = 2 \][/tex]
2. The calculated y-value is 2, which matches the y-coordinate of the point [tex]\(\left(\frac{2}{5}, 2\right)\)[/tex].
Therefore, point [tex]\(\left(\frac{2}{5}, 2\right)\)[/tex] lies on the line.
### Question 10
To find the equations of the lines passing through the given points, we first determine the slope and then use one point to find the y-intercept.
#### Part i: Line through [tex]\((1, 1)\)[/tex] and [tex]\((1 + \sqrt{2}, 1 - \sqrt{2})\)[/tex]
1. Calculate the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{(1 - \sqrt{2}) - 1}{(1 + \sqrt{2}) - 1} = \frac{1 - \sqrt{2} - 1}{\sqrt{2}} = - \frac{\sqrt{2}}{\sqrt{2}} = -1 \][/tex]
2. Use point [tex]\((1, 1)\)[/tex] to compute the y-intercept [tex]\( b \)[/tex]:
[tex]\[ y = mx + b \implies 1 = -1(1) + b \implies b = 1 + 1 = 2 \][/tex]
Therefore, the equation of the line is:
[tex]\[ y = -1x + 2 \quad \text{or} \quad y = -x + 2 \][/tex]
#### Part ii: Line through [tex]\(\left(1 \frac{1}{2}, -\frac{5}{2}\right)\)[/tex] and [tex]\(\left(-\frac{3}{2}, 1\right)\)[/tex]
1. Express the points as improper fractions to make calculations easier:
[tex]\[ \left(\frac{3}{2}, -\frac{5}{2} \right) \text{ and } \left(-\frac{3}{2}, 1 \right) \][/tex]
2. Calculate the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{1 - (-\frac{5}{2})}{(-\frac{3}{2}) - \frac{3}{2}} = \frac{1 + \frac{5}{2}}{-3} = \frac{\frac{2}{2} + \frac{5}{2}}{-3} = \frac{\frac{7}{2}}{-3} = \frac{7}{-6} = - \frac{7}{6} \][/tex]
3. Use point [tex]\(\left(\frac{3}{2}, -\frac{5}{2}\right)\)[/tex] to compute the y-intercept [tex]\( b \)[/tex]:
[tex]\[ y = mx + b \implies -\frac{5}{2} = - \frac{7}{6} \left( \frac{3}{2} \right) + b \][/tex]
Simplify:
[tex]\[ -\frac{5}{2} = - \frac{7}{6} \times \frac{3}{2} + b \implies -\frac{5}{2} = - \frac{7 \times 3}{6 \times 2} + b \implies -\frac{5}{2} = - \frac{7}{4} + b \][/tex]
4. Solve for [tex]\( b \)[/tex]:
[tex]\[ b = -\frac{5}{2} + \frac{7}{4} = -\frac{10}{4} + \frac{7}{4} = -\frac{3}{4} \][/tex]
Therefore, the equation of the line is:
[tex]\[ y = - \frac{7}{6} x - \frac{3}{4} \][/tex]
Both forms can be expressed in straightforward linear equations respectively:
1. [tex]\( y = -x + 2 \)[/tex]
2. [tex]\( y = - \frac{7}{6} x - \frac{3}{4} \)[/tex]
