Math  /  Algebra

Question56 If 1a,1b,1c\frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in harmonic progression (HP). and the straight-line ax+2by+c=0a x+2 b y+c=0 passes through the points (2,k)(2, k) and (1,3)(-1,3), find the value of kk

Studdy Solution

STEP 1

1. The sequence 1a,1b,1c\frac{1}{a}, \frac{1}{b}, \frac{1}{c} is in harmonic progression (HP).
2. A sequence is in HP if the reciprocals are in arithmetic progression (AP).
3. The line ax+2by+c=0ax + 2by + c = 0 passes through the points (2,k)(2, k) and (1,3)(-1, 3).
4. We need to find the value of kk.

STEP 2

1. Use the definition of harmonic progression to find a relationship between aa, bb, and cc.
2. Use the condition that the line passes through the given points to establish equations.
3. Solve the equations to find the value of kk.

STEP 3

Since 1a,1b,1c\frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in HP, their reciprocals a,b,ca, b, c are in AP. Therefore, we have:
2b=a+c 2b = a + c
This is the relationship between aa, bb, and cc.

STEP 4

Substitute the point (2,k)(2, k) into the line equation ax+2by+c=0ax + 2by + c = 0:
a(2)+2b(k)+c=0 a(2) + 2b(k) + c = 0
This simplifies to:
2a+2bk+c=0 2a + 2bk + c = 0
Equation 1: 2a+2bk+c=0 2a + 2bk + c = 0

STEP 5

Substitute the point (1,3)(-1, 3) into the line equation ax+2by+c=0ax + 2by + c = 0:
a(1)+2b(3)+c=0 a(-1) + 2b(3) + c = 0
This simplifies to:
a+6b+c=0 -a + 6b + c = 0
Equation 2: a+6b+c=0-a + 6b + c = 0

STEP 6

Now, we have three equations:
1. 2b=a+c2b = a + c
2. 2a+2bk+c=02a + 2bk + c = 0
3. a+6b+c=0-a + 6b + c = 0

First, express cc from Equation 1:
c=2ba c = 2b - a
Substitute c=2bac = 2b - a into Equations 2 and 3.

STEP 7

Substitute c=2bac = 2b - a into Equation 2:
2a+2bk+(2ba)=0 2a + 2bk + (2b - a) = 0
Simplify:
a+2bk+2b=0 a + 2bk + 2b = 0
Equation 4: a+2bk+2b=0 a + 2bk + 2b = 0

STEP 8

Substitute c=2bac = 2b - a into Equation 3:
a+6b+(2ba)=0 -a + 6b + (2b - a) = 0
Simplify:
2a+8b=0 -2a + 8b = 0
Equation 5: 2a+8b=0 -2a + 8b = 0

STEP 9

From Equation 5, solve for aa:
2a+8b=0 -2a + 8b = 0
2a=8b 2a = 8b
a=4b a = 4b
Substitute a=4ba = 4b into Equation 4:
4b+2bk+2b=0 4b + 2bk + 2b = 0
6b+2bk=0 6b + 2bk = 0
2b(3+k)=0 2b(3 + k) = 0
Since b0b \neq 0, we have:
3+k=0 3 + k = 0
k=3 k = -3
The value of kk is:
3 \boxed{-3}

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