P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triange:
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
Why did the triangle go to the circus? It wanted to join the "Pythagorean Theorem Show" and show off its area! Let's find it together.
We know that P and Q are at a distance of 5 units from the origin.
So, the coordinates of P and Q can be represented as (x₁, y₁) and (x₂, y₂) respectively.
Since the line x - y + 1 = 0 passes through the origin (0, 0), we can substitute the values into the equation:
x₁ - y₁ + 1 = 0 ----(1) for Point P
x₂ - y₂ + 1 = 0 ----(2) for Point Q
Let's solve these two equations simultaneously:
From equation (1), we have:
x₁ = y₁ - 1 ----(3)
Substituting equation (3) in equation (2), we get:
(y₁ - 1) - y₂ + 1 = 0
y₁ - y₂ = 0
y₁ = y₂ ----(4)
Substituting equation (4) in equation (1), we get:
x₁ = y₁ - 1
So, for convenience, let's assume both y₁ and y₂ as y.
Thus, the coordinates of P and Q are (y - 1, y) and (y + 1, y) respectively.
Now, let's find the distance between P and Q using the distance formula:
Distance PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(y + 1 - (y - 1))² + (y - y)²]
= √[2² + 0²]
= √4
= 2
So, the length of side PQ is 2 units.
Now to find the height of the triangle, we need to find the distance between the origin and the line x - y + 1 = 0.
We can plug in the coordinates (0, 0) into the equation:
0 - 0 + 1 = 1
So, the distance between the origin and the line is 1 unit.
Now, we can calculate the area of triangle POQ using the formula:
Area = (base × height) / 2
Area = (2 × 1) / 2
Area = 1 square unit
So, the area of triangle POQ is 1 square unit. Now that's a triangle with a sense of balance!
To find the area of triangle POQ, we first need to find the coordinates of points P and Q.
Given that P and Q lie on the line x - y + 1 = 0 and are at a distance of 5 units from the origin, we can use the distance formula to find the coordinates.
The distance formula states that the square of the distance between two points A(x1, y1) and B(x2, y2) is given by:
Distance^2 = (x2 - x1)^2 + (y2 - y1)^2
Applying this formula to the points P and the origin (0,0), we have:
5^2 = (x - 0)^2 + (y - 0)^2
25 = x^2 + y^2
We also know that the point (x, y) lies on the line x - y + 1 = 0. Substituting x = y - 1, we get:
25 = (y - 1)^2 + y^2
25 = y^2 - 2y + 1 + y^2
25 = 2y^2 - 2y + 1
2y^2 - 2y - 24 = 0
Dividing the equation by 2, we get:
y^2 - y - 12 = 0
Factoring the quadratic equation, we have:
(y - 4)(y + 3) = 0
So, y can be either 4 or -3.
If y = 4:
x = y - 1 = 4 - 1 = 3
If y = -3:
x = y - 1 = -3 - 1 = -4
Therefore, the coordinates of points P and Q are (3, 4) and (-4, -3) respectively.
Now, we can find the length of sides PO and OQ using the distance formula:
PO = √[(x2 - x1)^2 + (y2 - y1)^2]
PO = √[(3 - 0)^2 + (4 - 0)^2]
PO = √(3^2 + 4^2)
PO = √(9 + 16)
PO = √25
PO = 5
OQ = √[(-4 - 0)^2 + (-3 - 0)^2]
OQ = √((-4)^2 + (-3)^2)
OQ = √(16 + 9)
OQ = √25
OQ = 5
Area of triangle POQ = 0.5 * base * height
Since PO and OQ are equal and the base and height are the same, we have:
Area = 0.5 * 5 * 5
Area = 0.5 * 25
Area = 12.5 square units
Therefore, the area of triangle POQ is 12.5 square units.
To find the area of triangle POQ, we first need to find the coordinates of points P and Q.
Given that P and Q are equidistant from the origin, we can use the distance formula to find the coordinates.
Let's start by finding the equation of the line x-y+1=0 in the slope-intercept form.
Rearrange the given equation:
x - y + 1 = 0
x + 1 = y
y = x + 1
Now we have the equation in the form y = mx + c, where m is the slope and c is the y-intercept.
From the equation, we can see that the slope is 1 and the y-intercept is 1.
Since P and Q are equidistant from the origin, they lie on a line perpendicular to the given line. The slope of this perpendicular line can be found by taking the negative reciprocal of the slope of the given line.
The slope of the given line is 1, so the slope of the perpendicular line is -1.
Now we can find the equation of the perpendicular line passing through the origin.
Using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Substituting (0,0) for (x1, y1) and -1 for m, we get:
y - 0 = -1(x - 0)
y = -x
Now we have the equation of the line passing through the origin perpendicular to the given line.
To find points P and Q, we need to find the intersection of the two lines.
Solving the system of equations:
x + 1 = y
y = -x
Substituting the second equation into the first equation:
x + 1 = -x
Combining like terms:
2x + 1 = 0
Solving for x:
2x = -1
x = -1/2
Substituting the value of x back into the equation y = -x:
y = -( -1/2)
y = 1/2
Point P: (-1/2, 1/2)
To find point Q, we need to use the distance formula.
The distance between the origin (0, 0) and point P (-1/2, 1/2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the values:
5 = sqrt((-1/2 - 0)^2 + (1/2 - 0)^2)
Simplify:
5 = sqrt((-1/2)^2 + (1/2)^2)
5 = sqrt(1/4 + 1/4)
5 = sqrt(1/2)
5 = 1/sqrt(2)
To rationalize the denominator, multiply the numerator and denominator by sqrt(2):
5 = (1/sqrt(2)) * (sqrt(2)/sqrt(2))
5 = sqrt(2)/2
We can now find the y-coordinate of Q by subtracting the y-coordinate of P from 0:
0 - 1/2 = -1/2
Point Q: (sqrt(2)/2, -1/2)
Now that we have the coordinates of points P and Q, we can calculate the area of triangle POQ using the shoelace formula or the distance formula.
To calculate the area using the distance formula, we can consider the coordinates of points P, O (origin), and Q as the vertices of a triangle.
Using the distance formula to find the lengths of the sides of the triangle:
Side PO: sqrt((x2 - x1)^2 + (y2 - y1)^2) = sqrt((0 - (-1/2))^2 + (0 - 1/2)^2) = sqrt((-1/2)^2 + (-1/2)^2) = sqrt(1/4 + 1/4) = sqrt(1/2) = 1/sqrt(2) = sqrt(2)/2
Side OQ: sqrt((x2 - x1)^2 + (y2 - y1)^2) = sqrt((sqrt(2)/2 - 0)^2 + (-1/2 - 0)^2) = sqrt((sqrt(2)/2)^2 + (-1/2)^2) = sqrt(1/2 + 1/4) = sqrt(3/4) = sqrt(3)/2
Side PQ: sqrt((x2 - x1)^2 + (y2 - y1)^2) = sqrt((sqrt(2)/2 - (-1/2))^2 + (-1/2 - 1/2)^2) = sqrt((sqrt(2)/2 + 1/2)^2 + (-1)^2) = sqrt((2/4 + sqrt(2)/2 + sqrt(2)/2 + 1/4) + 1) = sqrt(1 + sqrt(2) + sqrt(2) + 1) = sqrt(2 + 2sqrt(2))
Now that we have the lengths of all three sides of the triangle, we can use Heron's formula to calculate the area:
Area = sqrt(s * (s - Side PO) * (s - Side OQ) * (s - Side PQ)), where s is the semi-perimeter of the triangle.
Semi-perimeter, s = (Side PO + Side OQ + Side PQ) / 2 = (sqrt(2)/2 + sqrt(3)/2 + sqrt(2 + 2sqrt(2))) / 2
Substitute the values into the formula:
Area = sqrt(((sqrt(2)/2 + sqrt(3)/2 + sqrt(2 + 2sqrt(2))) / 2) * ((sqrt(2)/2 + sqrt(3)/2 + sqrt(2 + 2sqrt(2))) / 2 - sqrt(2)/2) * ((sqrt(2)/2 + sqrt(3)/2 + sqrt(2 + 2sqrt(2))) / 2 - sqrt(3)/2) * ((sqrt(2)/2 + sqrt(3)/2 + sqrt(2 + 2sqrt(2))) / 2 - sqrt(2 + 2sqrt(2)))))
Simplify the expression to get the final answer for the area of triangle POQ.