IDNLearn.com provides a collaborative environment for finding and sharing knowledge. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
Let's solve the given problem step-by-step.
First, we need to find the points of intersection between the line [tex]\( y = 2x - 2 \)[/tex] and the curve [tex]\( x^2 - y = 5 \)[/tex].
1. We start by substituting the equation of the line [tex]\( y = 2x - 2 \)[/tex] into the equation of the curve:
[tex]\[ x^2 - (2x - 2) = 5 \][/tex]
2. Simplifying the equation:
[tex]\[ x^2 - 2x + 2 - 5 = 0 \][/tex]
[tex]\[ x^2 - 2x - 3 = 0 \][/tex]
3. Next, we solve this quadratic equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex]. The roots of this equation are given by the formula for solving quadratic equations:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 4}{2} \][/tex]
So the roots are:
[tex]\[ x = 3 \quad \text{and} \quad x = -1 \][/tex]
Thus, the points of intersection have [tex]\( x \)[/tex] coordinates [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex].
4. To find the corresponding [tex]\( y \)[/tex]-coordinates for these [tex]\( x \)[/tex]-values, we use the equation of the line [tex]\( y = 2x - 2 \)[/tex]:
When [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2(3) - 2 = 6 - 2 = 4 \][/tex]
So, the point [tex]\( B \)[/tex] is [tex]\( (3, 4) \)[/tex].
When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 2(-1) - 2 = -2 - 2 = -4 \][/tex]
So, the point [tex]\( A \)[/tex] is [tex]\( (-1, -4) \)[/tex].
5. Now we need to find the coordinates of the point [tex]\( P \)[/tex] that divides the segment [tex]\( AB \)[/tex] in the ratio [tex]\( 3:1 \)[/tex]. Using the section formula:
[tex]\[ P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \][/tex]
Here, [tex]\( A(-1, -4) = (x_1, y_1) \)[/tex] and [tex]\( B(3, 4) = (x_2, y_2) \)[/tex], with [tex]\( m = 3 \)[/tex] and [tex]\( n = 1 \)[/tex]:
[tex]\[ x_P = \frac{3 \cdot 3 + 1 \cdot (-1)}{3+1} = \frac{9 - 1}{4} = \frac{8}{4} = 2 \][/tex]
[tex]\[ y_P = \frac{3 \cdot 4 + 1 \cdot (-4)}{3+1} = \frac{12 - 4}{4} = \frac{8}{4} = 2 \][/tex]
So, the coordinates of point [tex]\( P \)[/tex] are [tex]\( (2, 2) \)[/tex].
As a final answer:
- Points of intersection are [tex]\( A(-1, -4) \)[/tex] and [tex]\( B(3, 4) \)[/tex].
- The coordinates of point [tex]\( P \)[/tex] are [tex]\( (2, 2) \)[/tex].
First, we need to find the points of intersection between the line [tex]\( y = 2x - 2 \)[/tex] and the curve [tex]\( x^2 - y = 5 \)[/tex].
1. We start by substituting the equation of the line [tex]\( y = 2x - 2 \)[/tex] into the equation of the curve:
[tex]\[ x^2 - (2x - 2) = 5 \][/tex]
2. Simplifying the equation:
[tex]\[ x^2 - 2x + 2 - 5 = 0 \][/tex]
[tex]\[ x^2 - 2x - 3 = 0 \][/tex]
3. Next, we solve this quadratic equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex]. The roots of this equation are given by the formula for solving quadratic equations:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 4}{2} \][/tex]
So the roots are:
[tex]\[ x = 3 \quad \text{and} \quad x = -1 \][/tex]
Thus, the points of intersection have [tex]\( x \)[/tex] coordinates [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex].
4. To find the corresponding [tex]\( y \)[/tex]-coordinates for these [tex]\( x \)[/tex]-values, we use the equation of the line [tex]\( y = 2x - 2 \)[/tex]:
When [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2(3) - 2 = 6 - 2 = 4 \][/tex]
So, the point [tex]\( B \)[/tex] is [tex]\( (3, 4) \)[/tex].
When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 2(-1) - 2 = -2 - 2 = -4 \][/tex]
So, the point [tex]\( A \)[/tex] is [tex]\( (-1, -4) \)[/tex].
5. Now we need to find the coordinates of the point [tex]\( P \)[/tex] that divides the segment [tex]\( AB \)[/tex] in the ratio [tex]\( 3:1 \)[/tex]. Using the section formula:
[tex]\[ P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \][/tex]
Here, [tex]\( A(-1, -4) = (x_1, y_1) \)[/tex] and [tex]\( B(3, 4) = (x_2, y_2) \)[/tex], with [tex]\( m = 3 \)[/tex] and [tex]\( n = 1 \)[/tex]:
[tex]\[ x_P = \frac{3 \cdot 3 + 1 \cdot (-1)}{3+1} = \frac{9 - 1}{4} = \frac{8}{4} = 2 \][/tex]
[tex]\[ y_P = \frac{3 \cdot 4 + 1 \cdot (-4)}{3+1} = \frac{12 - 4}{4} = \frac{8}{4} = 2 \][/tex]
So, the coordinates of point [tex]\( P \)[/tex] are [tex]\( (2, 2) \)[/tex].
As a final answer:
- Points of intersection are [tex]\( A(-1, -4) \)[/tex] and [tex]\( B(3, 4) \)[/tex].
- The coordinates of point [tex]\( P \)[/tex] are [tex]\( (2, 2) \)[/tex].
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.