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Miscellaneous Examples · Example 15

Q.A line is such that its segment between the lines 5x−y+4=05x - y + 4 = 0 and 3x+4y−4=03x + 4y - 4 = 0 is bisected at the point (1,5)(1, 5). Obtain its equation.

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Letting AA lie on 5x−y+4=05x-y+4=0 and using (1,5)(1,5) as the midpoint to locate the corresponding point BB on 3x+4y−4=03x+4y-4=0 gives slope 1073\dfrac{107}{3}, so the required line is 107x−3y−92=0107x-3y-92=0.

Step 1: Set up the midpoint condition

Let A=(x1,y1)A=(x_1,y_1) lie on 5x−y+4=05x-y+4=0, so y1=5x1+4y_1=5x_1+4.

Since (1,5)(1,5) bisects the segment from AA to a point BB on the second line 3x+4y−4=03x+4y-4=0:

B=(2−x1, 10−y1)=(2−x1, 6−5x1)B=(2-x_1,\ 10-y_1)=(2-x_1,\ 6-5x_1)

Step 2: Force BB onto the second line

3(2−x1)+4(6−5x1)−4=03(2-x_1)+4(6-5x_1)-4=0

6−3x1+24−20x1−4=06-3x_1+24-20x_1-4=0

26−23x1=0  ⟹  x1=262326-23x_1=0 \implies x_1=\frac{26}{23}

y1=5(2623)+4=22223y_1=5\left(\frac{26}{23}\right)+4=\frac{222}{23}

So A=(2623,22223)A=\left(\dfrac{26}{23},\dfrac{222}{23}\right), and

B=(2−2623, 6−13023)=(2023,823)B=\left(2-\frac{26}{23},\ 6-\frac{130}{23}\right)=\left(\frac{20}{23},\frac{8}{23}\right)

Check midpoint: (26/23+20/232,222/23+8/232)=(1,5)\left(\dfrac{26/23+20/23}{2},\dfrac{222/23+8/23}{2}\right)=(1,5) ✓ …

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