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Y−√3∣x∣=2

p1 and p2 are points on either of the two lines at a distance of 5 units from their point of point of intersection. Find the coordinates of the foot of perpendiculars drawn from p1 , p2 on the bisector of the Angle between the given lines


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The intersection point of the given lines is (0, 2). The bisector's equation is y - 2 = (√3/3)x. However, there appears to be an issue, as points p1 and p2 cannot be determined at a distance of 5 units from the intersection point.

The given equation is y - √3|x| = 2. To find the points of intersection, we'll consider the two cases separately for x ≥ 0 and x < 0:

For x ≥ 0, the equation becomes y - √3x - 2 = 0. Solving this equation, we get y = √3x + 2.

For x < 0, the equation becomes y + √3x - 2 = 0. Solving this equation, we get y = -√3x + 2.

Now, let's find the intersection point by equating y in both equations:

√3x + 2 = -√3x + 2

√3x = -√3x

x = 0

Now, substitute x = 0 back into either equation to find y:

y = 2

So, the point of intersection is (0, 2).

Next, we need the equation of the bisector. The given line can be written as y = √3|x| + 2, which has a slope of √3. The bisector's slope can be found using the formula tan(θ/2) = (m1 - m2) / (1 + m1m2), where m1 = √3 and m2 is the slope of the bisector.

Solving for m2:

tan(θ/2) = (√3 - m2) / (1 + √3m2)

The solution for m2 is found to be m2 = 1/√3 = √3/3.

Now, we can use the point-slope form to find the equation of the bisector. Let (x1, y1) = (0, 2) be the point of intersection:

y - y1 = m(x - x1)

y - 2 = (√3/3)x

Now, let's find the points p1 and p2 at a distance of 5 units from the intersection point (0, 2). The distance formula is given by d = √((x2 - x1)^2 + (y2 - y1)^2).

Let p1 have coordinates (x1, y1) and p2 have coordinates (x2, y2). The distance between p1 and the intersection point is 5 units:

√((x1 - 0)^2 + (y1 - 2)^2) = 5

Substitute x1 = 0 and y1 = 2 into the equation:

√(0 + (2 - 2)^2) = 5

0 = 5 (This is not possible)

Therefore, there is an issue, and it seems there is no solution for points p1 and p2 at a distance of 5 units from the intersection point.

Y−√3∣x∣=2 p1 and p2 are points on either of the two lines at a distance of 5 units-example-1
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