**Equation of the Line When Parallel Line and Sum of Intercepts is Given :**

Here we are going to see how to find the equation of the line when parallel line and sum of intercepts is given.

**Question 1 :**

Find the equation of a straight line parallel to 2x + 3y = 10 and which is such that the sum of its intercepts on the axes is 15

**Solution :**

Since the required line is parallel to the line 2x + 3y = 10

Equation of the required line will be in the form

2x + 3y + k = 0

Let us change the above equation to intercept form.

2x + 3y = -k

2x/(-k) + 3y/(-k) = 1

x/(-k/2) + y/(-k/3) = 1

x -intercept = a and y -intercept = b

Sum of intercepts = 15

a + b = 15

(-k/2) + (-k/3) = 15

(-3k - 2k)/6 = 15

-5k/6 = 15

k = -15(6)/5 = -18

2x + 3y - 18 = 0

Hence the equation of the required line is 2x + 3y - 18 = 0.

**Question 2 :**

Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular from (−10,−2) to the line x + y − 2 = 0

**Solution :**

The lines PQ and AB are perpendicular to each other.

Equation of AB :

x + y - 2 = 0

Equation of PQ :

x - y + k = 0

The point P (-10, -2) lies on the line PQ.

-10 + 2 + k = 0

k - 8 = 0

k = 8

Equation of PQ is x - y + 8 = 0

By solving the equations of AB and PQ, we get the value of P.

x + y - 2 = 0 ----(1)

x - y + 8 = 0 ----(2)

(1) + (2) ==> 2x + 6 = 0 ==> x = -3

By applying the value of x in (1), we get the value of y.

-3 + y - 2 = 0

-5 + y = 0

y = 5

Hence the required point P (-3, 5)

Length of perpendicular :

= √(x_{2} - x_{1})^{2 }+ (y_{2} - y_{1})^{2}

= √(-3+10)^{2 }+ (5+2)^{2}

= √7^{2 }+ 7^{2}

= √98 = 7√2 units

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