第11类RD Sharma解决方案–第23章直线-练习23.10 |套装2

问题11。找出三角形的等边线，其等式的边为x + y = 1、2x + 3y = 6和4x – y + 4 = 0。

解决方案：

Let us considered ABC is a triangle, in which A, B, and C are the vertices of the triangle

So according to the question the equations of the sides of the triangle are:

AB = x + y = 1  …..(1)

BC = 2x + 3y = 6  …..(2)

AC = 4x – y + 4   …..(3)

On solving eq (1) and (2), we get point B(-3, 4)

On solving eq (1) and (3), we get point A(-3/5, 8/5)

Now the altitude from A to BC is

$y-\frac{8}{5}=\frac{3}{2}(x+\frac{3}{5})$

10y – 16 = 151x + 9

15x – 10y + 25 = 0

3x – 2y + 5 = 0   …..(4)

Similarly, the altitude form B to AC  given by

y – 4 = -1/4(x + 3)

4y – 16 = -x -3

4y – 16 = -x – 3

x + 4 – 13 = 0    …..(5)

On solving eq(4) and (5) we get 

orthocenter of the triangle i.e., O(3/7, 22/7) 

问题12.三角形ABC的边AB，BC和CA分别为5x – 3y + 2 = 0，x – 3y – 2 = 0和x + y – 6 = 0。找出通过顶点A的高度方程。

解决方案：

According to the question

ABC is a triangle in which the equations of the sides are:

AB = 5x – 3y + 2 = 0

BC = x – 3y – 2 = 0

CA =  x + y – 6 = 0

On solving the above equations of AB, BC and CA we get

the coordinates of the vertices of the triangle ABC

B = (-1, -1)

A = (2, 4)

C = (5, 1)

Now the slope of BC = 1/3 then slope of AE = -3

Slope of AC = -1 then the slope of BD = 1

Slope of AB = 5/3 then the slope of CF = -3/5

Let us assume AD, BE, CF are altitudes of ∆ABC

So, the equation are

BD = y + 1 = 1(x + 1) ⇒ x – y = 0

AE = y – 4 = -3(x – 2) ⇒ 3x + y = 10

CF = y – 1 = $\frac{-3}{5}(x-5)$⇒ 3x + 5y = 20

Hence, the equation of the altitude through the vertex A is 3x + y = 10

问题13.找到顶点为(-1，3)，(2，-1)和(0，0)的三角形的正交中心的坐标。

解决方案：

Let us considered ABC is a triangle whose vertices are A(-1, 3), B(2, -1), and D(0, 0).

Here, AD ⊥ BC, CF ⊥ AB, and BE ⊥ AC

Let us assume O be the orthocenter of triangle 

So, O(h, k)

Now, AO⊥BC

(slope of AO) * (slope of BC) = -1  

$(\frac{k-3}{h+1})(\frac{0+1}{0-2})=-1$

k – 3 = 2(h + 1)

k – 2h = 5  …..(1)

And BO ⊥ AC

(slope of BO) * (slope of AC) = -1

$(\frac{k+1}{h-2})(\frac{0-3}{0+1})=-1$

3(k + 1) = h – 2

3k – h = -5    …..(2)

Now from eq(1) and (2), we get the orthocenter O of triangle is (-4,-3)

问题14。找到三角形的中心和质心的坐标，这些三角形的边具有等式3x – 4y = 0、12y + 5x = 0和y – 15 = 0。

解决方案：

Let us considered ABC be the triangle whose sides BC, CA, and AB have the equations 

BC = y – 15 = 0

AC = 3x – 4y = 0

AB = 5x + 12y = 0  

On solving the above equations we get

A(0, 0), B(-36, 15), C(20, 15) 

Now the centroid$(\frac{-36+20+0}{3},\frac{15+15+0}{3})=(\frac{-16}{3},10)$

For incentre, we have

a = BC = $\sqrt{56^2+0}=56$

b = CA = $\sqrt{20^2+15^2}=25$

c = AB = $\sqrt{36^2+16^2}=39$

Now the coordinates of incentre are

$(\frac{56*0+25*-36+39*20}{36+25+39},\frac{56*0+25*15+39*15}{36+25+39})$

= (-1, 8)

问题15：证明直线√3x+ y = 0，√3y+ x = 0，√3x+ y = 1和√3y+ x = 1形成菱形。

解决方案：

Let us considered ABCD be a quadrilateral with sides AB, BC, CD, BA as 

√3x + y = 0, √3y + x = 0, √3x + y = 1 and √3y + x = 1

So, the slope of AB = -√3   …..(1)

The slope of BC = -1/√3     …..(2)

The slope of CD = -√3        …..(3)

The slope of DA = -1/√3     …..(4)

From eq (1), (2), (3) and (4) we conclude that the slope opposite sides of quadrilateral are equal.

So, the opposite sides are parallel.

Hence, ABCD is a parallelogram.

We also conclude that the distance between (AD and BC) and (DC and AB) is equal =1 unit

So, sides AD = AB = DC

Hence, the ABCD is a rhombus

Hence, proved 

问题16。找到通过线3x + y = 5和x + 3y + 8 = 0并与线3x + 4y = 7平行的线的方程。

解决方案：

Given that the equations of lines are:

2x + y = 5  

x + 3y + 8 = 0

On solve the above equations we get the

intersection point(23/5, -12/5)

It is given that the line is parallel to 3x + 4y = 7 and passing through above point(23/5, -12/5)

So the equation of line is 

$y+\frac{21}{5}=\frac{-3}{4}(x-\frac{23}{5})$

20y + 84 = -15x + 69

15x + 20y + 15 = 0

3x + 4y + 3 = 0

问题17。找到通过线5x – 6y – 1 = 0并垂直于线3x – 5y + 11 = 0的线的方程。

解决方案：

Given that the equations of lines are:

5x – 6y – 1 = 0  

2y + 5 = 0

On solve the above equations we get the

intersection point(-1, 1)

It is given that the line is perpendicular to 3x – 5y + 11 = 0 and passing through above point(-1, 1)

So, the equation of line is 

(y + 1) = -5/3(x + 1)

3y + 5x + 8 = 0