Answer :
Let's solve the given problem step by step.
### Problem Restatement
We have a line given by the equation [tex]\(\frac{x}{a} + \frac{y}{b} = 1\)[/tex], which intersects the [tex]\(x\)[/tex]-axis at point [tex]\(P\)[/tex] and the [tex]\(y\)[/tex]-axis at point [tex]\(Q\)[/tex]. The distances [tex]\(PQ\)[/tex] is [tex]\(\sqrt{45}\)[/tex], and the gradient of the line [tex]\(PQ\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex]. We need to find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
### Step-by-Step Solution
1. Determine the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
- The line meets the [tex]\(x\)[/tex]-axis at [tex]\(P\)[/tex], so set [tex]\(y = 0\)[/tex] in the equation: [tex]\(\frac{x}{a} + \frac{0}{b} = 1 \implies x = a\)[/tex]. Hence, [tex]\(P = (a, 0)\)[/tex].
- The line meets the [tex]\(y\)[/tex]-axis at [tex]\(Q\)[/tex], so set [tex]\(x = 0\)[/tex] in the equation: [tex]\(\frac{0}{a} + \frac{y}{b} = 1 \implies y = b\)[/tex]. Hence, [tex]\(Q = (0, b)\)[/tex].
2. Calculate the distance [tex]\(PQ\)[/tex]:
The distance between [tex]\(P (a, 0)\)[/tex] and [tex]\(Q (0, b)\)[/tex] is given by the distance formula:
[tex]\[ PQ = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \][/tex]
Given that [tex]\(PQ = \sqrt{45}\)[/tex], we have:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{45} \][/tex]
Squaring both sides, we get:
[tex]\[ a^2 + b^2 = 45 \][/tex]
3. Determine the gradient of the line [tex]\(PQ\)[/tex]:
The gradient (or slope) of the line passing through [tex]\(P (a, 0)\)[/tex] and [tex]\(Q (0, b)\)[/tex] is:
[tex]\[ \text{Gradient} = \frac{b - 0}{0 - a} = \frac{b}{-a} = -\frac{b}{a} \][/tex]
Given that the gradient is [tex]\(-\frac{1}{2}\)[/tex], we have:
[tex]\[ -\frac{b}{a} = -\frac{1}{2} \][/tex]
Removing the negative signs, this simplifies to:
[tex]\[ \frac{b}{a} = \frac{1}{2} \][/tex]
Solving for [tex]\(b\)[/tex], we get:
[tex]\[ b = \frac{a}{2} \][/tex]
4. Solve the System of Equations:
Now we have two equations:
[tex]\[ a^2 + b^2 = 45 \][/tex]
[tex]\[ b = \frac{a}{2} \][/tex]
Substitute [tex]\(b = \frac{a}{2}\)[/tex] into the first equation:
[tex]\[ a^2 + \left(\frac{a}{2}\right)^2 = 45 \][/tex]
Simplify inside the parentheses:
[tex]\[ a^2 + \frac{a^2}{4} = 45 \][/tex]
Multiply through by 4 to clear the fraction:
[tex]\[ 4a^2 + a^2 = 180 \][/tex]
Combine like terms:
[tex]\[ 5a^2 = 180 \][/tex]
Solve for [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = 36 \][/tex]
Taking the positive square root (since [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are positive):
[tex]\[ a = 6 \][/tex]
Substitute [tex]\(a = 6\)[/tex] back into [tex]\(b = \frac{a}{2}\)[/tex]:
[tex]\[ b = \frac{6}{2} = 3 \][/tex]
### Conclusion
The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ a = 6 \quad \text{and} \quad b = 3 \][/tex]
### Problem Restatement
We have a line given by the equation [tex]\(\frac{x}{a} + \frac{y}{b} = 1\)[/tex], which intersects the [tex]\(x\)[/tex]-axis at point [tex]\(P\)[/tex] and the [tex]\(y\)[/tex]-axis at point [tex]\(Q\)[/tex]. The distances [tex]\(PQ\)[/tex] is [tex]\(\sqrt{45}\)[/tex], and the gradient of the line [tex]\(PQ\)[/tex] is [tex]\(-\frac{1}{2}\)[/tex]. We need to find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
### Step-by-Step Solution
1. Determine the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
- The line meets the [tex]\(x\)[/tex]-axis at [tex]\(P\)[/tex], so set [tex]\(y = 0\)[/tex] in the equation: [tex]\(\frac{x}{a} + \frac{0}{b} = 1 \implies x = a\)[/tex]. Hence, [tex]\(P = (a, 0)\)[/tex].
- The line meets the [tex]\(y\)[/tex]-axis at [tex]\(Q\)[/tex], so set [tex]\(x = 0\)[/tex] in the equation: [tex]\(\frac{0}{a} + \frac{y}{b} = 1 \implies y = b\)[/tex]. Hence, [tex]\(Q = (0, b)\)[/tex].
2. Calculate the distance [tex]\(PQ\)[/tex]:
The distance between [tex]\(P (a, 0)\)[/tex] and [tex]\(Q (0, b)\)[/tex] is given by the distance formula:
[tex]\[ PQ = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \][/tex]
Given that [tex]\(PQ = \sqrt{45}\)[/tex], we have:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{45} \][/tex]
Squaring both sides, we get:
[tex]\[ a^2 + b^2 = 45 \][/tex]
3. Determine the gradient of the line [tex]\(PQ\)[/tex]:
The gradient (or slope) of the line passing through [tex]\(P (a, 0)\)[/tex] and [tex]\(Q (0, b)\)[/tex] is:
[tex]\[ \text{Gradient} = \frac{b - 0}{0 - a} = \frac{b}{-a} = -\frac{b}{a} \][/tex]
Given that the gradient is [tex]\(-\frac{1}{2}\)[/tex], we have:
[tex]\[ -\frac{b}{a} = -\frac{1}{2} \][/tex]
Removing the negative signs, this simplifies to:
[tex]\[ \frac{b}{a} = \frac{1}{2} \][/tex]
Solving for [tex]\(b\)[/tex], we get:
[tex]\[ b = \frac{a}{2} \][/tex]
4. Solve the System of Equations:
Now we have two equations:
[tex]\[ a^2 + b^2 = 45 \][/tex]
[tex]\[ b = \frac{a}{2} \][/tex]
Substitute [tex]\(b = \frac{a}{2}\)[/tex] into the first equation:
[tex]\[ a^2 + \left(\frac{a}{2}\right)^2 = 45 \][/tex]
Simplify inside the parentheses:
[tex]\[ a^2 + \frac{a^2}{4} = 45 \][/tex]
Multiply through by 4 to clear the fraction:
[tex]\[ 4a^2 + a^2 = 180 \][/tex]
Combine like terms:
[tex]\[ 5a^2 = 180 \][/tex]
Solve for [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = 36 \][/tex]
Taking the positive square root (since [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are positive):
[tex]\[ a = 6 \][/tex]
Substitute [tex]\(a = 6\)[/tex] back into [tex]\(b = \frac{a}{2}\)[/tex]:
[tex]\[ b = \frac{6}{2} = 3 \][/tex]
### Conclusion
The values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are:
[tex]\[ a = 6 \quad \text{and} \quad b = 3 \][/tex]
