The line segment joining the points A(2,1) and B(5,−8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line 2x−y+k=0, find the value of k.
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Step-by-Step Solution
Step 1: Understand Trisection and Section Formula
When a line segment is trisected, it is divided into three equal parts. Since point P is nearer to A, it divides the line segment AB in the ratio 1:2. This means the distance from A to P is one-third of the total length of AB, and the distance from P to B is two-thirds of the total length.
Step 2: Apply Section Formula to find coordinates of P
We use the section formula to find the coordinates of point P. Given A(x1,y1)=(2,1) and B(x2,y2)=(5,−8), and the ratio m:n=1:2. Substitute these values into the formula to find the x and y coordinates of P.
Step 3: Calculate coordinates of P
Perform the arithmetic operations to simplify the expressions for the x and y coordinates of P. This gives us the exact location of point P on the coordinate plane.
Step 4: Substitute P's coordinates into the line equation
The problem states that point P(3,−2) lies on the line 2x−y+k=0. To find the value of k, we substitute the x and y coordinates of P into the equation of the line.
Step 5: Solve for k
Simplify the equation by performing the multiplication and addition. Then, isolate k to find its value. This gives us the final answer for the unknown constant k.