第 11 课 RD Sharma 解决方案-第 23 章直线-练习 23.17

问题 1. 证明由直线组成的平行四边形的面积

a ₁ x + b ₁ y + c ₁ = 0，a ₁ x + b ₁ y + d ₁ = 0，a ₂ x + b ₂ y + c ₂ = 0，a ₂ x + b ₂ y + d ₂ = 0 是$\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|$平方单位。

推导出这些线形成菱形的条件。

解决方案：

Given:

The given lines are

a₁x + b₁y + c₁ = 0 —(equation-1)

a₁x + b₁y + d₁ = 0 —(equation-2)

a₂x + b₂y + c₂ = 0 —(equation-3)

a₂x + b₂y + d₂ = 0 —(equation-4)

Let us prove, the area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0, a₂x + b₂y + d₂ = 0 is  

$\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|$ sq. units.

The area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0 and a₂x + b₂y + d₂ = 0 is given below:

Area  = $\left|\frac{(c_1-d_1)(c_2-d_2)}{\begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}}\right|$

Since, \begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix} = a₁b₂ – a₂b₁

Therefore, Area = $\left|\frac{(c_1-d_1)(c_2-d_2)}{a_1b_2-a_2b_1}\right|=\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|$

If the given parallelogram is a rhombus, then the distance between the pair of parallel lines is equal.

Therefore, $\left|\frac{c_1-d_1}{\sqrt{a_1^2+b_1^2}}\right|=\left|\frac{c_2-d_2}{\sqrt{a_1^2+b_1^2}}\right|$

Hence proved.

问题 2. 证明由线 3x – 4y + a = 0、3x – 4y + 3a = 0、4x – 3y – a = 0 和 4x – 3y – 2a = 0 形成的平行四边形的面积是$\frac{2a^2}{7}$平方单位。

解决方案：

Given:

The given lines are

3x − 4y + a = 0 —(equation-1)

3x − 4y + 3a = 0 —(equation-2)

4x − 3y − a = 0 —(equation-3)

4x − 3y − 2a = 0 —(equation-4)

We have to prove, the area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is $\frac{2a^2}{7}$ sq. units.

From above solution, we know that

Area of the parallelogram = $\left|\frac{(c_1-d_1)(c_2-d_2)}{a_1b_2-a_2b_1}\right|$

Area of the parallelogram = $\left|\frac{(a-3a)(2a-a)}{(-9+16)}\right|=\frac{2a^2}{7}$ sq. units

Hence proved.

问题 3. 证明边为 lx + my + n = 0、lx + my + n' = 0、mx + ly + n = 0 和 mx + ly + n' = 0 的平行四边形的对角线包含一个角π /2。

解决方案：

Given:

The given lines are

lx + my + n = 0 —(equation-1)

mx + ly + n’ = 0 —(equation-2)

lx + my + n’ = 0 —(equation-3)

mx + ly + n = 0 —(equation-4)

We have to prove, the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0 and mx + ly + n’ = 0 include an angle $\frac{\pi}{2}$.

By solving equation (1) and (2), we will get

B = $\left(\frac{mn'-ln}{l^2-m^2}, \frac{mn-ln'}{l^2-m^2}\right)$

Solving equation (2) and (3), we get 

C = $\left(-\frac{n'}{m+1'}, -\frac{n'}{m+1}\right)$

Solving equation (3) and (4), we get 

D = $\left(\frac{mn-ln'}{l^2-m^2}, \frac{mn'-ln}{l^2-m^2}\right)$

Solving equation (1) and (4), we get

A = $\left(-\frac{n}{m+1'}, -\frac{n}{m+1}\right)$ 

Let m₁ and m₂ be the slope of AC and BD.

Now,

m₁ = $\frac{\frac{-n'}{m+1}+\frac{n}{m+1}}{\frac{-n'}{m+1}+\frac{n}{m+1}}=1$

m₂ = $\frac{\frac{mn'-ln}{l^2-m^2}+\frac{mn-ln'}{l^2-m^2}}{\frac{mn-ln'}{l^2-m^2}+\frac{mn'-ln}{l^2-m^2}}=-1$

Therefore, 

m₁m₂ = -1

Hence proved.