Answer :
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].