p and q are two points on the line x - y + 1 = 0 such that each of them is 5 units far from the origin. Find the co-ordinates of the points
we have
x^2 + y^2 = 25
x^2 + (x+1)^2 = 25
2x^2 + 2x - 24 = 0
x^2+x-12 = 0
(x+4)(x-3) = 0
So, the points are (-4,-3) and (3,4)
To find the coordinates of points p and q on the line x - y + 1 = 0 that are each 5 units away from the origin, we can follow these steps:
Step 1: Find the equation of the line x - y + 1 = 0 in slope-intercept form.
Step 2: Use the distance formula to set up an equation with the given distance of 5 units.
Step 3: Solve the equation to find the coordinates of points p and q.
Let's go through these steps:
Step 1: Find the equation of the line x - y + 1 = 0 in slope-intercept form.
Rearrange the equation to solve for y:
x - y + 1 = 0
-y = -x - 1
y = x + 1
So, the equation of the line is y = x + 1, and its slope is 1.
Step 2: Use the distance formula to set up an equation with the given distance of 5 units.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Since both points p and q are 5 units away from the origin (0, 0), we can set up the distance formula using the coordinates of p and q:
For point p:
d = √((x - 0)^2 + (y - 0)^2) = 5
Simplifying, we have:
√(x^2 + y^2) = 5
For point q:
d = √((x - 0)^2 + (y - 0)^2) = 5
√(x^2 + y^2) = 5
Step 3: Solve the equation to find the coordinates of points p and q.
Since both equations are the same, we can solve one of them:
√(x^2 + y^2) = 5
Square both sides of the equation to eliminate the square root:
x^2 + y^2 = 25
Now, substitute y = x + 1 from the equation of the line:
x^2 + (x + 1)^2 = 25
x^2 + (x^2 + 2x + 1) = 25
2x^2 + 2x + 1 = 25
2x^2 + 2x - 24 = 0
Divide the equation by 2 to simplify it further:
x^2 + x - 12 = 0
Factoring the quadratic equation, we get:
(x + 4)(x - 3) = 0
Setting each factor equal to zero and solving for x, we have:
x + 4 = 0 OR x - 3 = 0
x = -4 OR x = 3
Now, substitute these x-values into the equation of the line to find the corresponding y-values:
For x = -4:
y = x + 1 = -4 + 1 = -3
For x = 3:
y = x + 1 = 3 + 1 = 4
Therefore, the coordinates of point p are (-4, -3) and the coordinates of point q are (3, 4).
To find the coordinates of points P and Q on the line x - y + 1 = 0, we need to use the distance formula between a point and the origin.
Let's start by finding the equation of the line x - y + 1 = 0 in the slope-intercept form, y = mx + b. Rearranging the equation, we have y = x + 1.
Now, we know that the distance between a point (x, y) and the origin (0, 0) is given by the formula √((x - 0)^2 + (y - 0)^2) = √(x^2 + y^2).
Since the points P and Q are 5 units away from the origin, we can set up the following equations:
√(x^2 + y^2) = 5 (for point P)
√(x^2 + y^2) = 5 (for point Q)
We can square both sides of these equations to eliminate the square root:
x^2 + y^2 = 25 (for point P)
x^2 + y^2 = 25 (for point Q)
Now, since both equations are the same, we can conclude that the coordinates of points P and Q are the same.
Substituting y = x + 1 into the equations, we have:
x^2 + (x + 1)^2 = 25
Expanding and simplifying:
x^2 + x^2 + 2x + 1 = 25
2x^2 + 2x - 24 = 0
Dividing the equation by 2 to simplify the coefficient:
x^2 + x - 12 = 0
Now we need to solve this quadratic equation. Factoring or using the quadratic formula will give us the two possible values of x.
Factoring the quadratic equation, we have:
(x + 4)(x - 3) = 0
Setting each factor equal to zero:
x + 4 = 0 --> x = -4
x - 3 = 0 --> x = 3
So we have two possible x-coordinates: -4 and 3.
Now we can substitute these values back into the equation for y = x + 1 to find the y-coordinates:
For x = -4:
y = (-4) + 1 --> y = -3
For x = 3:
y = 3 + 1 --> y = 4
Hence, the coordinates of the points P and Q are (-4, -3) and (3, 4), respectively.