**Equation of the Line Difficult Problems with Answers :**

Here we are going to see some how we find the equation of the line with difficult problems.

**Question 18 :**

Find the equation of the line passing through (22, -6) and having intercept on x-axis exceeds the intercept on y-axis by 5.

**Solution : **

**x-intercept(a) = b + 5, y -intercept = b **

**(x/a) + (y/b) = 1**

**x/(b+5) + y/b = 1**

**The straight line is passing through the point **(22, -6)

**22/(b+5) - 6/b = 1**

**22b - 6(b + 5) = b(b + 5)**

**22b - 6b - 30 = b**²** + 5b**

** 16b - 30 = b**²

** b**²

** b**² - 11

**(b - 5) (b - 6) = 0**

b - 5 = 0 ==> b = 5 a = 5+5 = 10 ==> a = 10 x/a + y/b = 1 x/10 + y/5 = 1 (x + 2y)/10 = 1 x - 2y = 10 x + 2y - 10 = 0 |
b - 6 = 0 ==> b = 6 a = 6 + 5 = 11 ==> a = 11 x/a + y/b = 1 x/11 + y/6 = 1 (6x + 11y)/66 = 1 6x + 11y = 66 6x + 11y - 66 = 0 |

**Question 19 :**

If A (3, 6) and C (-1, 2) are two vertices of a rhombus ABCD, then find the equation of straight line that lies along the diagonal BD.

**Solution : **

In any rhombus diagonals bisect each other at right angle.

In any rhombus midpoint of the diagonals will be equal.

Midpoint of AC = Midpoint of BD

Midpoint of AC = (x₁+x₂)/2, (y₁+y₂)/2

= (3 + (-1))/2, (6 + 2)/2

= 2/2, 8/2

= (1, 4)

(1, 4) is a point lies of the diagonal BD.

Slope of AC x Slope of BD = -1

Slope of AC :

m = (y₂-y₁)/(x₂-x₁) ==> (2-6)/(-1-3) ==> -4/(-4) ==> 1

Slope of BD :

Slope of BD = -1/1 ==> -1

Equation of BD :

(y - y₁) = m (x - x₁)

y - 4 = -1(x - 1)

y - 4 = -x + 1

x + y - 4 - 1 = 0

x + y - 5 = 0

Hence the required equation is x + y - 5 = 0.

**Question 20 :**

Find the equation of the line whose gradient is 3/2 and which passes through P, where P divides the line segment joining A(-2, 6) and B (3, -4) in the ratio 2 : 3 internally.

**Solution :**

First we need to find the point P.

P = (lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m)

p = (2(3) + 3(-2))/(2+3), (2(-4) + 3(6))/(2+3)

= (6 -6)/5, (-8 + 18)/5

= 0/5, 10/5

= (0, 2)

(x₁, y₁) ==> (0, 2) m = 3/2

(y - y₁) = m(x - x₁)

(y - 2) = (3/2)(x - 0)

2(y - 2) = 3x

2y - 4 = 3x

3x - 2y + 4 = 0

Hence the required equation is 3x - 2y + 4 = 0.

After having gone through the stuff given above, we hope that the students would have understood "Equation of the Line Difficult Problems with Answers".

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