**10th grade math solution exercise 5.4 part7 :**

Here we are going to see solutions of some problems that we find in the exercise number 5.4

**Question 15 :**

Find the equation of the straight line passing through the point (5, -3) and whose intercepts on the axes are equal in magnitude but opposite in sign.

**Solution : **

Intercepts on the axes are equal in magnitude but opposite in sign.

**x-intercept(a) = k and y-intercept(b) = -k**

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

**(x/k) + (y/(-k)) = 1**

**The straight line is passing through the point (5, -3)**

**(5/k) + (-3)/(-k) = 1**

**(5/k) + (3/k) = 1**

**(5 + 3)/k = 1 ==> **** 8/k = 1 ==> ****k = 8**

**a = 8 , b = -8**

**Equation of the line :**

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

**(x/8) + (y/(-8)) = 1**

**(x - y)/8 = 1**

**x - y = 8**

**x - y - 8 = 0**

**Question 16 :**

Find the equation of the line passing through the point (9, -1) and having its x-intercept thrice as its y-intercept.

**Solution : **

**x - intercept (a) = 3b **

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

**(x/3b) + (y/b) = 1**

**The straight line is passing through the point (9, -1)**

**(9/3b) + (-1/b) = 1**

**(3/b) + (-1/b) = 1**

**(3 - 1)/b = 1**

**2/b = 1**

**b = 2 **

**So a = 3(2) = 6**

**Equation of the line :**

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

**(x/6) + (y/2) = 1**

**(x + 3y)/6 = 1**

** x + 3y = 6**

**x + 3y - 6 = 0**

**Question 17 :**

**A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then find the equation of AB.**

**Solution : **

**Let us a draw rough diagram for the given information.**

Let "C" be the midpoint of the line joining the points A and B.

The required line intersects the x-axis at the point B and y-axis at A. So let us consider the point B as (a, 0) and A as (0, b)

Midpoint of AB = (x₁+x₂)/2, (y₁+y₂)/2

(a + 0)/2, (0 + b)/2 = (3, 2)

a/2, b/2 = (3, 2)

equating the x and y co-ordinates

a/2 = 3 b/2 = 2

a = 6 and b = 4

A (0,4) and B (6, 0)

Equation of the line AB :

(y-y₁)/(y₂-y₁) = (x-x₁)/(x₂-x₁)

(y-4)/(0-4) = (x-0)/(6-0)

(y-4)/(-4) = x/6

6(y - 4) = -4x

6y - 24 = -4x

4x + 6y - 24 = 0

Divide the whole equation by 2, we get

2x + 3y - 12 = 0

Hence the required equation is 2x + 3y - 12 = 0

After having gone through the stuff given above, we hope that the students would have understood "10th grade math solution exercise 5.4 part7".

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